This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
solvable. Find the amount x of salt that will dissolve in time /, if this amount is 5 lb. when 1 = 1. 23. In a certain volume of liquid, 200 lb. of salt, but no more, will eventually dissolve. When a quantity of salt is placed in this liquid, the rate at which it dissolves is T^JTC times the product of the undissolved amount by the amount that is still dissolv* able. If 100 lb- of salt are placed in this liquid at / = 0, what is the amount x that is dissolved at time /? 24. From problem 23, what is the amount of salt that is dissolved in the liquid in time t if 200 lb. arc placed in it at ( = 0? 25. From example 5, if 10 lb. of chemical U and 15 lb. of chemical M^arc placed together at / = 0, how much of compound X is produced at time I? 26- From example 5, if chemicals U and were to combine in equal amounts to produce compound x, what would be the amount of compound produced in time interval / after u lb, of U and w lb. of IV were placed together? 27, T w o chemicals U and IV combine in equal parts to form a compound -V. The rate at which the compound is produced is proportional to the square root of the product of the amounts of U and IV that are still present at the instant. If at ( = 0 there are 8 units of U and 2 units of W^, find the formula for the amount of compound formed at time /, 28- If 2 parts of a chemical U combine with 3 parts of a chemical W and the rate at which the compound is produced is always proportional to the amount of U that is still present at the instant, what is the formula for the amount of compound produced in time ty if this amount is M at / = 0 ? 29- T h e population ^ of a certain insect colony is at ( = 0, In the absence of outside influences the rate of its growth would be kp. T h e colony is, however, exposed to an influence that causes deaths at the constant rate ö. Find the formula for the population at any time t. 30. In the absence of outside influences a certain population p grows at the rate kp. However, it is exposed to an outside influence which induces growth at the additional rate {a + 4 ^ ] , where a and i are constants. If the population is at time /o, what is the formula for the population at time t? 31. T h e amount x of a certain chemical disintegrates at a rate which is always proportional to the amount that is still left. If it diminishes by 1 per cent in 10 years, and its amount is XQ at t = 0, what is the formula for x at time t? 32. T h e amount x of a certain chemical naturally diminishes, because of disintegration, at a rale kx if left to itself. In use there is a further loss by wear at a constant rate Ö . If the amount is XQ at time ( = 0, what is it at time t?
3A. Problems of flow A problem of flow is always based upon a differential equation. This statement applies to a flow of water, or of air, or of some other fluid, and to an electric current, or the flow of heat from a cooling body.
Problems of Flow
41
W h e n a liquid whose volume is V is allowed to escape from its container through an opening at a distance y beneath its surface, the rate at which it flows through that opening is k V ^ , with some constant k. T h e value of this constant depends upon the physical circumstances of the individual Ccise, in particular upon the shape and size of the opening, the character of the fluid, the system of units used, etc. This rate of flow is also the rate at which the volume V diminishes. W e draw from that fact the relation (3.5)
-dV/dt
=
k\fy.
This is not a differential equation such as we have already studied, because it involves too many variables. A differential equation i n just two variables is, however, obtainable from it, by substituting for dV its value i n terms of y and dy. T h i s value is determinable in any case from the shape of the volume F, and the location of the opening through which the liquid escapes. I n a case i n which the figure is simple, the formula for V in terms oiy may be easily discernible. A differentiation then gives dV. It is, however, often simpler to obtain the formula for dV directly from the figure without first finding V. This is done by applying the reasoning of the integral calculus, which says that dV is equal to the area of the surface of the l i q u i d multiplied by dy. Example 6.
A cylindrical tank of length 9 ft. and radius 5 ft. is mounted horizontally and is filled with oil. A t / = 0, a plug at the lowest point of the tank is removed, and a flow results to which relation 3.5 applies with k = T o find the depth ^ of oil i n the tank at any time t while the tank is draining. T h e circumference of the end of the tank is the circle + [y — 5 j ^ = 25.
For this circle, x = +
T h e surface of the oil is thus
a rectangle whose width is 2 V ' l O ) ' — and whose length is 9. Since dV equals the area of this surface times dy, relation 3.5 yields the differential equation
T h e general integral of this equation is 12(10-;'i^ =
^+r.
T o fit the condition that> = 10, when / = 0, the value o f t must be zero. T h e result can be given the form
42
Applications t
V = 10 ^
— 180
^
I n this case it was simpler to find the value of dV from the figure of the liquid, than it would have been to find the formula for V and to differentiate it, for by elementary geometry V is found to have the relatively complicated formula V = 9
25T -
25 cos"^
-
1^ +
0
-
5)
VlO^ -
If the tank had been standing on end the situation would have been different, for then the formula for V would have been recognizable at a glance. W h e n a material body whose temperature is T is immersed in a quantity of water whose temperature is T', there is a flow of heat from the body to the water, or vice versa. T h e law that applies to this is Newton's law of cooling, which states that the rate at which the temperature of the body drops, namely, —dT/dt, is proportional to the difference between the temperatures of the body and water. Thus (3.6)
-dT/dt
= k{T
~
T'].
T h e constant k depends upon the physical values involved, among them the area over which the body is i n contact with the water, the material of which the body is composed, and the units used. T o obtain a differential equation i n two variables from equation 3.6, the value of T' i n terms of T must be substituted. This value can generally be found from the fact that the amount of heat which the body loses is the amount that the water gains. T h e change i n the body's heat content is always obtainable as the product of its weight by its specific heat by the change i n its temperature. Example 7.
A body weighs 45 lb. and is made of a metal whose specific heat is 5 . W h i l e at a temperature of 300 degrees, it is plunged, at ^ = 0, into 100 lb. of water at a temperature of 50 degrees. T o find the formula for the temperature T of the body during its cooling. T h e amount of heat that the body has lost at time / is 45(g) {300 — T]. T h e amount of heat the water has then gained is 100(1) j — 501 (the specific heat of the water being 1). Since these two amounts are the same, we have V-1300 -
T] = 1001 r
-
50},
whence 7"' = 65 — {T/20). O n substituting this value into equation 3.6 we obtain the differential equation
Problems of Flow -dTldi
43
= k
T h e integral which applies is that for w h i c h T = 300, when / = 0. integral is r = -y*f|13 + 50^--t2i/20)fc£j_
This
F r o m this integral we can observe the temperature to which the body w i l l eventually cool. As t increases indefinitely the exponential term approaches zero. T h e limiting value of T is therefore ^ ^ r ^ , that is, 61.9 degrees (approximately). PROBLEMS 33, A cylindrical tank of length 10 ft. and radius 3 ft, stands on end, and is filled with oil. When a plug halfway up its side is removed the oil Rows in accordance with formula 3,5, with k = ^ir- Find the formula for the depth of the oil in the tank while the tank is more than half full. 34, A tank is 6 ft. long, and its end is a square with a 4-ft. diagonal. At / = 0 it b half full of oil, and at that instant a plug in a lower edge is removed. T h e flow obeys law 3.5 with k = At what time is the tank drained, if it lies horizontally on one side? 35, At what time is the tank of problem 34 drained if it is mounted horizontally with a diagonal of its end in the vertical position? 36, A tank is in the form of a right circular cone with altitude 4 ft. and radius 5 ft. At ^ = 0, it is filled with water, and the water is then allowed to escape through a hole to which law 3.5 applies, with k = - j ^ . Find the formula for the depth of water in the tank at time /, if the tank is mounted with its axis vertical and its apex downward, and if the hole is at the apex. 37- If the tank of problem 36 is mounted with its apex upward, and the whole is in the base, what is the formula for the time when the depthj^f water in the lank is^? 38. A tank is in the shape of a sphere with radius 4 ft. At / 0, it is filled with water, and an opening on the level of the center of the sphere then permits the water to escape. Find the formula for the time in terms of the depth k of the water in the tank, while the tank is more than half full. 39. A cylindrical tank with radius 4 ft, stands on end, and has an opening, for which k ^ "ffVj in its lower base. As water excapes through this opening, fresh water is run into the tank at the rate of 1 cubic ft,/sec. Find the formula for the time at which the depth of the water in the tank is A, if A = 9, when / = 0. 40. A 60-lb- piece of metal whose specific heat is -5- is at a temperature of 450 degrees. At f = 0, it is plunged into 300 lb. of water whose temperature is 60 degrees. Find the formula for the temperature of the body at time t. 41. A n 84b. body is made of metal whose specific heat is ^ T J . While at a temperature of 308 degrees it is plunged into 11 lb, of water at a temperature of 53 degrees. Find the temperature to which the body eventually cools.
44
Applications
42. A 10-lb. body whose specific heat is -g- is at 0 degrees when t ^ 0. It is then plunged into 100 lb- of water at 80 degrees. If law 3.6 applies with k — yf* what is the temperature of the body at time t? 43. A 5-lb. body whose specific heat is TnJ" is plunged at / 0 into 50 lb. of water. If the temperature of the body at this time is 200 degrees^ and if it eventually cools to 50 degrees, what is the formula for its temperature at time /? 44. A 50-lb. body whose specific heat is i^f, and whose temperature is 208 dc^ecs, is plunged into 20 lb. of oil whose specific heat is 3^, and whose temperature is 50 degrees. Find the formula for the temperature of the body at time t? 45. T h e differential equation for the current i which flows under an e.m.f. (electromotive force) £ in a simple electric circuit containing a constant inductance L and a constant resistance R is (3-7)
L—i-Ri at
E.
Find the formula for r\ if £ is a constant, and / = 10 when t ^ /Q. 46. F r o m problem 45, find the formula for current 1 that flows in the circuity if I >- 0, when t ^ Oj and E ^ a sin positive too. If {x, y) is any point of the curve, the slopes of the tangent and normal lines are, respectively, y' and — 1 / y ' , and these lines meet the x-axis at the points (x — y/y\ 0) and {x + yy\ 0). T h e triangle therefore has a base of length [yy' + y/y' \. Since its altitude is we have 1
yy'
+
T h i s differenticil equation can be written i n the form y' V 2 — y^ = y^ Its variables are separable, and the integral for which ^ = 3, when * = 1, is - /
-h 2 -
Vaiog
[V2
+
^]y
=
x.
T h i s is the required equation. Example 9.
T o find the equation of the curve that goes through the point (2,1), such that the area under any of its arcs and above the x-axis is equal to the length of the respective arc. F r o m the calculus we know that, for a curve above the x-axis, which extends to the right from a point (XQ, >O)» the area under an arc is given by the integral /
y dx.
T h e length of the corresponding arc is
T h e curve sought is, therefore, one for which
B y differentiation this becomes y y' = + y/y^ — 1.
+ y'^j or, i n another form,
T h e general integral of this differential equation is log \y +
V /
_
ij
=
±x-^c,
and withe determined so that)" = l , w h e n x = 2, the result can be written y +
=
^-
It is possible to improve upon this form. canceling etc., we find that
O n transposing y^ squaring,
We can now drop the ambiguous signs, since the two alternatives give the
I
I
46
Applications
same formula.
T h e equation sought is thus
This is the equation of a catenary. It is the curve i n which a rope or chain hangs when it is suspended from two of its points. I n terms of the hyperbolic functions, which are defined i n the calculus by the formulas (3-9)
svnhx
-
-
r^!,
coshjc =
i l ^
+
^""^h
the result can be expressed i n the form^' == cosh (jt — 2 ) . PROBLEMS 49. Find the equation of the curve that goes through the point (1, 3), and for which the tangent at any point, and the Une joining that point with the origin, have slopes which arc the negatives of each other. 50, Find the equation of the curve that goes through the point (4, 3), and whose tangent and normal lines always form with the x-axis a triangle whose area is equal to the product of { — - T T ) by the slope of the tangent line, 51, Find the equation of the curve that goes through the point (0, 1), and whose tangent and normal lines alw^ays form with the :r-axis a triangle whose area is equal to the negative of the slope of the normal line. 52. Find the equation of the curve that goes through the point (4, 2), such that the segment of its normal line between the curve itself and the >-axi3, is always bisected by the ^r-axis.
53- Find the equation of the curve through the point (1, 5), whose tangent and normal lines always make with the ^-axis a triangle whose area is equal to the slope of the tangent line54. Find the equation of the curve through the point (2, 5), whose tangent and normal lines always make with the >-axis a triangle whose area is equal to 4 times the slope of the normal line. 55. Find the equation of the curve through the point (-J, 0), whose tangent and normal lines cut from the >-axis a segment whose length equals the negative of the slope of the tangent line. 56. Find the equations of the curves through the point (3, 2), for which the tangent and normal lines always form with the and ^-axes a quadrilateral whose area is equal to 2xy. 57. Find the equation of the curve through the point (4, 2), for which the tangent line and the horizontal through the point of contact form with the x- and ^-axcs a trapezoid having the area 6. 58- Find the equations of the curves through the point (4, 3), for which the tangent line and the vertical through the point of contact always make with the x-axis a triangle having the area 9,
Polar Coordinates
47
59. Find the equation of the curve through the point (1, —1), for which each tangent line is at a distance from the origin that is equal to the abscissa of its point of contact. 60. Find the equations of the curves through the point (1, 1), for which any tangent line and the line from the origin to the point of contact make with the>-axis a triangle having the area ^. 61. Find the equations of the curves through the point (0, 5), for which the area under any arc and above the jr-axis is equal to 3 times the length of the respective arc. 62. Find the equation of the curve through the point (-ff, 0), along which the arc length and the function 2x^ increase at the same rate.
3.6. Polar coordinates Polar coordinates are sometimes better adapted to the statement and solution of a given geometrical problem than are rectangular coordinates.
FIG.
3.
T h e two systems are related as shown i n F i g . 3. T h e connecting relations are X = r cos df y — r sin Q. F r o m the calculus, we know that the angle ^ and the differential of arc ds are given by the formulas (3.10)
dB
tan^ = r—I dr
and (3.11)
ds"" =
r''dQ'^-\-dr''.
These are the essential formulas for dealing with the geometrical problems to be considered. Example 10.
T o find the equations of the curves through the point with the polar coordinates (1, ir/2) for which every normal line is at the distance 1 from the pole.
48
Applications
T h e perpendicular from the pole upon the normal line has the direction of the tangent line. It therefore makes with the radius vector the angle ^ or — i^j according as ^ is acute or obtuse. F o r the curves sought we have thus 1 / r = cos ^, or 1 / r = cos ( r — W e can write these relations tan 1^ = ± V ' r ^ — 1. F r o m equation 3.10 we thus obtain the differential equations r d$/dr =
±
V ^ ^ .
T h e i r general integrals are 0 = ±{Vr^
-
1 -
c o s - i (1/r)) + c,
and with c determined so that $ = T / 2 , when r = 1, the required equations are found to be 6 Example 11.
=
T/2
± {Vr^ -
1 -
sec"^ r ) .
T o find the equations of the curves through the point
with the polar coordinates {'s/2y T / 4 ) along which the arc length increases at twice the rate of the vectorial angle. T h e specific property of the curves sought is ds — 2 dd. B y (3.11), this
is
dd^ •\- dr^ — 2 dd.
We
may
write
this i n the
form
r d6 — ±dr. This equation has r = 2 as an integral- However, that integral does not go through the given point. T h e generzil integral is B = i: sin~^ + Cj and when c is appropriately determined this gives 6 = sin~^ and 6 = — s i n " ^ (r/2) + ir/2. W e can write these equations i n the forms r = 2 sin 0 and r — 2 cos 0. T h e curves are circles of radius 2 through the pole. PROBLEMS 63, Find the equation of the curve through the point with the polar codrdinates (2, x/6), for which the angles 0 and ^ are always equal. 64. Find the equation of the curve through the point with the polar coordinates (3, 3ir/2), for which 4" and 0/2 arc always equal. 65, Find the equation of the curve through the point with the polar coordinates (1, r ) , for which 1^ is always the supplement of 0/2. 66. Find the equation of the curve through the point with the polar coordinates (1, ir/4), for which ^ is always the supplement of 2(9. 67. Find the equations of the curves through the point with the polar coordinates (3, T / 2 ) , for which all the tangent lines are at the distance 3 from the pole, 68, Find the equations of the curves through the point with the polar coordinates (5, 0), for which the tangent and normal lines are always equidistant from the pole.
Some Curves of Pursuit
49
69. Find the equations of the curves through the point with the polar coordinates (l/'\/2> ~""/4), for which the distance from the pole to the tangent line is always equal to the square of the radius vector to the point of contact. 70. Find the equations of the curves through the point with the polar coordinates (1, 0), for which the distance from the pole to the normal line is always equal to the square of the radius vector to the point of contact. 71. Find the equation of the curve through the point with the polar coordinates (1, 0), for which ^ is always equal to log r. 72. Find the equations of the curves through the point with the polar coordinates (3, 0), along which the arc length increases at 6 times the rate of 6. 73. Find the equations of the curves through the point with the polar coordinates (1, — T / 3 ) , along which the arc length and the square of the radius vector increase at the same rate. 74. Find the equations of the curves through the point with the polar coordinates (4, JT), along which the arc length increases at a rate that is always 2 tan times the rate at which the radius vector increases.
3.7. Some curves of pursuit W h e n a point moves so as always to be directed toward a certain goal, its locus is a curve of pursuit. T h e goal may be fixed or moving, and outside influences may bear upon the motion of the point itself. A simple example of a curve of pursuit is the path of an airplane that pursues another while always keeping its quarry i n the line of sight. W e shall consider this case here under the following conditions: (i) T h e pursued airplane 0 flies i n a fixed direction with a constant ground speed whose eastward and northward components are a and b, respectively. (ii) T h e pursuing plane A has a constant air speed V. (iii) There is a constant wind whose components toward the east and north are, respectively, Vx and Vy. It need not concern us much that condition (i) is actually quite artificial unless 0 is fixed. W e are presently engaged in developing techniques i n differential equations, not i n obtaining results usable i n aviation. W e shall consider other curves of pursuit in Section 7.9. T o the pilot of the pursued plane 0 of F i g . 4, the path of the pursuing plane A is a curve expressible by an equation either i n the polar coordinates (r, 6) or i n the rectangular coordinates (x — x*, y — y*). With reference to the ground, A has a velocity whose eastward component dx/dt is made up of the air-speed component V cos (x — 0) and the wind component Vx. T h u s dx/dt = — Kcos 0 + Vx.
Applications
50
Similarly, A^s northward velocity component is dy/dt = - F sin
+ Vy.
For the plane 0 we have dx*/dt = a, and dy*/dt = b. these equations into
W e can combine
d{x -
= { - F c o s f l H- Vx - a\ dt,
d{y-y*}
= \ -Vsme
+
Vy-b\dt,
and by eliminating dt find that (3.12)
[VsmB-Vy-\-b]d{x-x*\
-
\Vcos6
~
^ a}d\y -y*\
= 0.
T h r o u g h the use of the relation tan B = {y — y*)/{x — x*), equation 3.12 is expressible either as a differential equation i n r and B, or as one i n [x — x*\ and \y — y*\.
Fio. 4.
Example 12.
A n airplane 0 has a ground speed with constant eastward and northward components of 144 m.p.h. and 108 m.p.h., respectively. T h e pilot observes a pursuing plane .4 at a distance of 10 miles due north. T h e air speed of A is 270 m.p.h., and there is no w i n d . T o find an equation for the path of A as it appears from 0. W e have i n this case a = \ AA, b = \08, = 0, Vy = 0, and V = 270. Equation 3.12 is therefore ¡270 sin ^ + m\d{x - X*] -
{270costf + 144lrf(;' ->'*1 = 0.
Since j j ; A : * ) — r cos B, and {y — y*] = r sin By so that d[x — AT*j = — r sin 6 d9 + cos B dr^ and d{y — y*\ = r cos B dB -\- sin B dr, the equation becomes —
+ 4 cos
H- 3 sin B\TdB +
(4 sin 5 -
3 cos B] dr = 0.
T h e variables i n this are separable, and the general integral is 5 log
5 — 4 cos B — 3 sin 0 4 sin 5 — 3 cos d
-\- log ¡4 sin
— 3 cos B] + log r =
c.
Some Curves of Pursuit
51
T o fit the condition r = 10, when 6 = i r / 2 , the value ofc must be 10 T h e resulting equation can be put into the form r =
10 V 2 ¡4 s i n g (5 -
y/l.
3cosg|
4 cos g - Ssintfj^^
Example 13.
A n airplane A whose air speed is 200 m.p.h. takes off for a fixed destination 0 which is due west and at a distance of 300 miles. A constant w i n d blows with components of 15 m.p.h., and 20 m.p.h., toward the west and north, respectively. T o find the path of A if it is always directed toward 0. Since the point 0 is fixed, we may take it to be the origin. We have, then, X * = 0, = 0, a = 0, i = 0, = - 1 5 , Vy = 20, and V = 200. Equation 3.12 is thus {200 sin e -
20) dx -
{200 cos 6 +
15) dy = 0.
Since tan 0 = y/xj we have cos 6 = V ^ ^ ^ * M - > ^ , and sin d = T h e equation is thus 200;Vx^
+
-
20
dx
200x
-
Vx'-\-y
y^
+ 15
dy =
y/'Vx^
0
T h i s is a differential equation i n which both coefficients are homogeneous functions of the degree 0. T h e method of Section 2.7 therefore serves to integrate it. T h e integral for which x = 300, w h e n ^ = 0, is found to be 10 V x ^
log {4x-\-3y\ + 8
log
+jy2 _
4x
- h
6^ _|_ 8^
= log (1200)
3y
We can write this i n the form 10 Vx^ -\-y^ -
6x + Sy]^ = 1200{4x +
3y]\
A curve of pursuit of a different kind is the locus of one end of a piece of string when the other end is drawn along a given curve. I n F i g . 5 the piece of string is shown as QP. Its length a is constant, and the curve /(x*, y*) = 0, along which point P is being drawn, is given. Since a string can pull only along its own direction, the path of Q has QP as its tangent line. Hence y' = tan 0, and therewith the obvious relations X* — X = a cos 6, y* — y = a sin tf, can be given the forms (3.13)
=
x-\-
ay
V l
+
y*
+
Applications
52
W h e n these values are substituted i n the given equation/(x*,^*) = 0, the result is the differential equation for the path of Q. T w o loci are always to be expected, since point P may be drawn i n either direction along the given curve. W h e n the given curve is a straight line, the locus of Q is called a tractrix. 14. One end Q of a piece of string of the length 3 is placed at the point (2, — 2), and the other end is then drawn along the line y = x. T o find the equation of the locus of Q. Example
FIG.
5.
T h e given curve has the equation y* = x*. values 3.13 transforms^* = x* into
T h e substitution of the
3/ V l
V l
+
+
/"2
which is the differential equation for the path of Q. by making the change of variables x-{-y = SyX—y is thereby transformed into u V l
+
(du/ds)^ =
-3
It can be integrated = u. T h e equation
Vidu/ds,
that is, into Vl8
One integral of this is u = 0. -
V l 8 -
-
u^du
=
±uds.
T h e general integral is
+ V l 8 log
Vis
+
Vis
u
-
=
±s-\-c.
I n terms of the original variables, the integral for which y = — 2 , when X = 2, is
Some Curves of Pursuit ±{x'\'y]
= A/lSlog
V2(x -
53
y) V2
-
V i s -
(x
-y)\
T h r o u g h the two choices of sign, this gives the equations of twotractrices. PROBLEMS 75. A n airplane O whose ground speed is constant at 200 m.p.h. is flying northeast. T h e pilot observes a pursuing plane A southeast of him at a distance of 6 miles. T h e air speed of ^ is 300 m.p.h. There is no wind. F i n d the equation of A^s path as the path appears from O. 76. A n airplane O is flying due north with a ground speed of 200 m,p.h., when the pilot observes a pursuing plane ^ at a distance of 14 miles directly east. T h e pursuer's air speed is 240 m,p.h. There is no wind. Find the equation of A's path as it appears from O. 77- A n airplane 0 is flying with a ground speed whose components are 100 m.p-h. westward and 110 m.p.h. northward. T h e pilot observes a pursuing plane A directly east at a distance of 10 miles. T h e pursuer's air speed is 250 m.p.h., and there is a steady wind of 40 m.p.h, from the north. Find the equation of A*s path as it appears from O, 78, A n airplane O has a ground speed whose components are 100 m,p,h. to the east and 240 m.p.h. to the north. T h e pilot observes a pursuing plane ^4 at a distance of 15 miles due west. The pursuer's air speed is 300 m.p,h. There is no wind. Find the equation of A^s path as it appears from O, 79- A n airplane A whose air speed is 250 m.p-h., takes off^ for a point 0 which is due west and 150 miles distant. A wind of 25 m.p.h. blows from the south. T h e pilot keeps the plane directed toward O. Find an equation for the path of the plane, 80. A n airplane A whose air speed is 250 m,p,h. takes off" for a point O which is 200 miles west and 150 miles south. A wind of 25 m.p.h. blows from the south. F i n d the equation of the plane's path, 81, A n airplane A whose air speed is 200 m,p,h. takes off for a point 0 which is 100 miles due south, A wind of 30 m.p.h. blows from the west. Find the equation of the plane's path, 82. A n airplane whose air speed is 200 m.p.h. takes off* for a point 100 miles due south, A steady wind blows, with components of 20 m,p.h, from the west and 10 m.p.h. from the north- Find an equation for the plane's path, 83, One end Q of a piece of string of length 5 is placed at the point (4, 3), and the other end is then drawn along the x-axis. Find the equations of the loci of Q, 84, One end Q of a piece of string of length 2 is placed at the point (0, 2), and the other end is then drawn along the line ^ = 4. Find the equations of the loci of Q. 85. One end Q of a piece of string of length 13 is placed at the point (5, 2), and the other is then drawn along the ^-axis. Find the equations of the loci of Q.
54
Applications
86. One end Q of a piece of string of length 2 is placed at the point (1, 1), and the other end is then drawn along the line x y = 0, Find the equations of the loci of Q , 87. One end Q of a piece of string of length 6 is placed at the point (3, —1), and the other end is then drawn along the line ^ + 1 ••0. Find the equation of the locus of Q , 88. One end P of a piece of string PQ whose length is «2 is drawn along the circle x^ " b^t for which b > a. Find the differential equation for the locus of and show that + = 6^ — is an integral of this differential equation. 89. A ferry boat whose speed is a 8 m.p.h. plies between opposite points on the banks of a straight river 1 mile wide. T h e river flows at 4 m.p.h. T h e boat is always kept pointed directly at its objective. Find an equation for its path. 90- If the ferry boat of problem 89 takes off for a point that is z niilc upstream on the opposite bank, what is an equation for its path? 91. Show that if the ferry boat of problem 89 had a speed of only 4 m-p-h, it could not reach a point directly opposite the starting point, and find how far downstream its landing would have to be. 92, A ferryboat whose speed is V m.p-h. plies on a river K miles wide. T h e river flows at V m.p.h. If the boat takes off for a point on the opposite bank a miles upstream^ and is kept pointed at its objective, what is the equation of its path?
3.8. Suspension cables A piece of rope or cable, like any other body, is subject to Newton's law of motion 3.2. W h e n it is at rest, therefore, the resultant of the forces upon it, or on any piece of it, is zero. T h e key to the solution of many problems concerning ropes, cables, chains, etc., lies i n this fact. y
FIG.
6.
W h e n a cable (or rope or chain) is supported at two of its points, and hangs between these points under the burden of some distributed load, we call it a suspension cable. T h e cable of a suspension bridge is an example, as is also any heavy chain burdened by its own weight. T h e curve i n which the cable hangs is determined by the distribution of its load. W e shall see how this curve can be found. I n F i g . 6, OQ is any segment of a suspension cable, of which one end 0 is
Suspension Cables
55
the cable's lowest point. T h a t point is called the vertex. T h e tangent line there is horizontal. As i n the figure, we shall take the origin at this point. T h e forces acting upon the segment OQ are its burden or load L (which is not indicated i n the figure, but which acts vertically downward) and the pulls from the adjoining parts of the cable. These pulls are along the tangents of the curve. Thus the pull P at 0 is horizontal, whereas at Q we have tan tp — y'. Since the whole segment of cable is at rest, the sum of the components of all forces acting upon it i n the horizontal, or the vertical, direction is zero. Thus, p cos ^ — P = 0,
^ sin ^ — Z, = 0.
T h e elimination of/* from these equations yields (3.14)
y' = L/P.
If the density of the load as referred to the horizontal is designated by p(x), the value of L is given by the formula (3.15)
fjp{x)dx.
Equation 3.14 is thus a differential equation for the curve i n which the cable hangs. Example 15.
A suspension cable hangs from the tops of two towers that are 200 ft. apart, and its vertex is 50 ft. below the points of support. T h e weight of the cable itself is negligible, but the cable bears a load whose density is ¡100 + x^j. T o find the equation of the curve i n which the cable hangs. Formula 3.15 gives ¡100 + x^] dx = lOOx +
L =
W,
and therewith differential equation 3.14 is /
= ^ | l 0 0 x + ^x3
T h e integral of the last equation for which y = 0, when x = 0, is y =
1
50x2 + - 1 X"'
T h e curve passes through the tower tops, whose coordinates are ( ± 1 0 0 , 50). T o fit this condition, the value of P must be 530,000/3. The curve
56
Applications
thus has the equation y =
10,600
600
If the burden on the cable is not distributed horizontally, but is distributed along the cable itself, say with the density p(j), where s is the arc length of the curve measured from the vertex, the formula for L is =
/; Pis) ds.
W i t h this formula, equation 3.14 is (3.16)
/
= ^
j\{s)ds.
T o obtain a differential equation from (3.16) we proceed as follows: Since ds = V l
dx, the derivative of (3.16) is
(3.17)
dy' =
WT7^
dx.
If it is possible to eliminate s between (3.17) and (3.16), the result is a differential equation for^', the integral of which, with^"' = 0, at x = 0, is the required differential equation for the curve of the cable. Example 16.
A chain is supported at two points on the same horizontal level and 12 ft. apart. Its slope at a point of support is and its density is given by the formula ¡1 + as\~^'^. T o find the differential equation for the curve of the chain, and then, i n the case a = 0, to find the curve itself. Equations 3.16 and 3.17 are in this case y
= -
Wx+as
~ 1),
d/
=
..ZL^
dx.
P V 1 + as T h e elimination of equation
+ asy and thus of J , yields for y' the differential
T h e required integral of the last equation is aP[y/\
+2
log ( Z + ' s / l
=2x/P.
This is the differential equation for the curve of the chain.
Suspension Cables
57
W e are prepared to integrate the last equation only if a = 0. T h e chain is then of uniform density. I n that case the differential equation is log ( y + V i + / 2 }
=v^,
and since it was given that y' = T , when x = 6, we see that P has the value 6/log 2. T h e differential equation can therefore be put into the form
Its integral for which ^ = 0, when x = 0, is thus ^ -
^
f .(i/6)Iog2 I -Cj:^6)log2 _ 9)
Iog2 ^
A n alternative form for the last equation is '
log2'''
^
PROBLEMS 93. A suspension cable of negligible weight is burdened with a load distributed horizontally with a constant density. The supports of the cable are 400 ft, apart, and 80 ft, higher than the vertex. Find the equation of the curve in which the cable hangs. 94, A suspension cable of negligible weight is burdened with a load dbtributed horizontally with the density jlOO + x^/100)}. T h e supports of the cable are 160 ft- apart, and 64 ft, higher than the vertex. Find the equation of the curve in which the cable hangs. 95, A suspension cable of negligible weight is burdened with a load distributed horizontally with the density (24 + Z t ^ ) / \ / 2 4 + x^. A t ;r =^ 1, the slope of the cable is F i n d the equation of the curve in which the cable hangs. 96. A suspension cable of negligible weight is burdened with a load distributed horizontally with the density j2 -j^e'^^^}. T h e point (10, 4) is on the curve in which the cable hangs. Find the equation of this curve. 97, A suspension cable of negligible weight is burdened with a load distributed horizontally with the density 35 sec^ (TAT/SOO). T h e slope of the cable at a point of support is -y, and these points are 150 ft, apart. Find the equation of the curve in which the cable hangs. 98. A cable of constant density hangs under its own weight without any other load. At a point of support its slope is and these points arc 10 ft. apart. Find the equation of the curve in which the cable hangs, 99. A chain of constant density hangs freely from two supports 2 ft. apart, and the slope of the chain at one of them is -j-. Find the equation of the curve in which the chain hangs. 100, T h e density of a certain string varies with the distance from one of its points O, and is p{s) ™ T I ^ ' + T h e string is hung from two pegs so that the point O
58
Applications
is the vertex, and its length is adjusted until the pull upon one of the p ^ has the hori* zontal component 1 lb. Find the equation of the curve in which the string hangs. 101. T h e density of a certain rope at distance s from one of its points 0 is p{s) = sec^ J , T h e rope is hung from two pegs so that the point 0 is the vertex, and its length is adjusted until the pull upon one of the pegs has the horizontal component ^ lb. Find the equation of the curve in which the rope hangs102. In the case of the rope of problem 101, fmd the equation of the curve in which it h a n ^ if its length is adjusted until the pull upon one of the pegs has the horizontal component 2 / V 3
lb.
3.9. The rope around a shaft W h e n a rope is passed over a shaft or around a post, a large pull P on one end of it can be held i n check by a much smaller pull p on the other
T Fio. 7.
end. W i t h a given value of P , the value of p depends upon the extent to which the rope is wound around the post or shaft, namely, upon the angle $ which the arc of contact between the rope and the shaft subtends. It depends also upon /x, the coefficient of friction between the rope and the shaft. Suppose the rope is of constant density p, and that the shaft is horizontal and of radius a. T h e n , when pull p is just enough to keep the rope from yielding to P , the configuration is that of Fig. 7. A t any point of contact Q the gravitational pull due to the weight of the rope has the intensity p, and this has the component p cos ^ in the direction of the rope. Its other component is merged in the pressure that the shaft exerts upon the rope.
The Rope Around a Shaft
59
I f we designate the normal component of the intensity of that pressure by / ( v ) , the corresponding frictional component amounts to nf{(p). These force intensities are indicated i n the figure by dashed vectors. T o obtain the components of the forces which they generate, we must integrate their respective components with respect to s. W e note that ds = a dtp. B y taking components horizontally to the right, and then upward, we thus obtain the equations 9 '0
\p cos ipsvfMp -\- f{ = i l 9 ^ ^ ' +
73.
-
V 2 - x\
57. 2xy -y
71. log r -
+ 1^0 - £Cl*-'^«*^.
51. ;r » 1 -
55. ^ = - J +
67. r sin
r=50-|-150«-'i*'»^^*'«'*'.
- 4} - •v/25
ii)--
- 5 log
5 2| = V 1 6 9 - .c^ _ 13
5 + V 2 5 -
,13 + \ / l 6 9 ' Sx
-
4
- 12,
1,000,000;- 17
62 87. J -
Applications 89. y -
-1.
93. y = xVSOO.
91. \ mile downstream. 95. ^ « V f f ! [ 2 4 + x 2 ] H _
99
101. ^ -
-
(S cos (3^5) + V Z S cos2 (3JC
log J
103. 100*""'=' lb. 105. M -
[24]'^).
log 9
-4.
^
(20.8 lb. approximately).
( l A ) log 2
107. (3/20) {«"" -1- 1 |ft.
(\/i/2)il - * | .
(0.22 approximately). (3.62 ft. approximately).
CHAPTER
4 Integral Curves Trajectories 4.1. Integrals of a differential equation T h e methods of Chapter 2 for solving differential equations (4.1)
/
=M>)
were found i n Chapter 3 to be applicable to many kinds of physical problems. It remains a fact, however, that all these methods are essentiedly special. It would be easy to write down differential equations 4.1 i n great variety, to which no one of the methods given would apply. A n d that would remain true even if more methods were to be given, as they could be. F o r the fact is, that, among a l l differential equations 4.1, those for which a solving formula, other than one i n infinite series, can (or could) be given are relatively rare. This is by no means due to a mere lack of mathematical proficiency or skill. T h e reason lies deeper than that. A n elementary calculus formula is a combination of certain functions that are more or less familiar. T h e category of these functions is relatively quite restricted. It includes the exponential, the trigonometric, and the power functions, together with their inverses, and the successive quadratures of combinations of these. N o w a differential equation may define its function through a relationship which, i n its very nature, is inexpressible by such means. I n short, it may define a function proper to itself, which is not representable by any combination of elementary functions, just as log X, for instance, is not representable by any combination of polynomials. N o w it is, of course, quite apropos to inquire what one shall take to be an integral of a differential equation, when no formula for one can be given. T h e answer is to be found i n the definition of the term "function." I n the calculus y is defined to be a function of x upon a specified x-interval, if to each X of that interval there corresponds a value of y. This does not require that y be expressible by a formula. It requires only that y be i n some way determined. Such determination may, for instance, be by a 63
64
Integral Cur\'es; Trajectories
graph, or by some process of calculation through which y is computable, either exactly or to as high a degree of accuracy as may be desired. This definition is also adopted i n the theory of differential equations. I n Chapter 1 we have already considered the direction field of a differential equation 4.1. That led to the conception of the streamlines of the field as integral curves, and thus suggested that the graph of the general integral is a family of curves one of which passes through any chosen point of a region. Analytically this statement implies that a differential equation is fulfilled by a family of functions, and that, subject to proper restrictions, there is a member of that family which for an arbitrarily assigned value xo takes on a prescribed value ^o-
4.2. The direction field of a differential system T h e considerations which led to the notion of a direction field for a single differential equation can easily be extended to the case of a pair of simultaneous equations. (4.2)
y
=fi{x,y,z),
z'
=fo{x,y,2).
Consider any region of three-dimensional (jt, y, z) space i n which the functions fi{x, yj z) and f'i{x, >, z) are real and single veilued. A t each point of such a region these functions have numerical values, and the triple of numbers 1, / i ( x , ^, z), f^ix^ y^ z), is thus determined there. These numbers, however, determine a direction in space, namely, the direction whose cosines are proportional to them. By assigning this direction to the point, and doing so for each point of the region, we impose upon the region a three-dimensional direction field. A particle, starting at any point of such a field, and moving thence always i n the direction of the field, describes a streamline. For these streamlines, as for any curve, t/x, dy^ dzy are direction numbers. Because these numbers also give the direction of the field, we see that along a streamline dx
dy
X
f\{x,yy z)
dz fiixyy,
z)
This means that the functions y and z defined by the streamlines fulfilled equations 4.2. T h e streamlines are thus integral curves. A curve i n space is given analytically by two equations i n (v, y, z). E a c h of these equations separately defines a surface. Together they define the curve which is the intersection of the surfaces. E a c h i n d i v i d u a l integral curve is thus the intersection of a surface of one family by one of another. Since each of these families is given by an equation involving a n
The Isoclines of a Direction Field
65
arbitrary constant, the pair of equations representing the family of integral curves involves two such constants. These observations have already been borne out by the actual solutions of differential systems of the form 4.2 that are found i n Sections 2.9 and 2.10. We shall return to this subject again. I n the sections immediately following, however, we shall consider further the integrals of a single differential equation 4.1 only.
4.3. The isoclines of a direction field T h e direction field of a differential equation 4.1 over a chosen region is graphically determinable by calculating the values o f / ( x , y) at a chosen and suitably dense set of points, and assigning to each of these points the direction with the slope respectively calculated. W h e n considerable accuracy is required, the number of points for which the calculation must be made is apt to be relatively large. T h e labor involved may then be substantial. Various methods, such as the following, are known for reducing this labor. For each value k which the function f{x, y) takes on in the region of the field, the equation (4.3)
fix, y) = k
defines a locus that traverses the region. This is the locus at each point of which the slope of the field is k. It is thus a locus along which the field has a single inclination. It is therefore called an isocline of the field. B y drawing i n a number of isoclines, the plotting of a field and the sketching of the streamlines may often be considerably facilitated. W e note especially that the isoclines for which k = 0 are the loci along which the integral curves have their maxima and m i n i m a . Example 1.
T o plot the integral curves of the differential equation ,
2y-2
T h e isoclines are i n this case the loci {2y — 2)/{x — 2) = k. the straight lines
These are
y - \ = ^ U - 2 ] ,
which pass through the point (2, 1). T h e isocline is crossed by all the integral curves at the same slope, and that is here twice the slope of the isocline itself. Figure 8 applies to this example. It shows in full lines some integral curves, and in dashed lines a pair of isoclines.
Integral Curves; Trajectories
Fio. 8,
Fio. 9.
The Isoclines of a Direction Field Example 2.
67
T o plot the integral curves of the differential equation ,
y =
1 -3{x+>P 1 +3lx+^l
T h e isoclines are i n this case the loci 1 -3ix+^p
^
that is, the straight lines
A l l these have the slope — 1 . It is clear that k cannot have any value greater than 1, and hence that no integral curve ever has a slope greater than 1. Some integral curves and a few isoclines are shown i n F i g . 9. ASSIGNMENTS With the use of isoclines, plot the direction field and sketch in some integral curves for each of the following differential equations. 1. y =x/y.
3. /
2. /
= ~2xy.
= ^ ^ -
A. y'
^• X
5. / = - l ^ .
6. / = \ / ? T ? .
4
7, Show that if :to is a value for which /»C^o) = 0, the line ^ = xo is an isocline of the linear equation (4.4)
y'+p(x)y
=q{x).
8. Show that if i3 any value for which p(xi) ^ 0, all the directions of the field of differential equation 4.4 along the line x ^ x\ are concurrent in the point \x\ + l/p(xi)j g{xi)/p(x\)]. This ¡5 shown for the difTercntial equation
in Fig, 10. In the manner of Fig. 10, plot the direction field and some integral curves for each of the following differential cquations4 9.y'^x-2xy.
U.y'=h-y-
, 0 . / =
12./ =
^ ^ + ! .
? ^ ^ .
68
Integral Curves; Trajectories y
X
Fro. 10.
4.4. Trajectories If two families of curves JFi and SF2 occupy a region of the {x^y) plane, one curve of each family passing through any point of the region, the curves of either family are said to be trajectories of those of the other. T h e relation between curves and their trajectories is determined by the law of their intersection. W h e n a family ffi is given, the trajectories which intersect it i n the point {x, y) at the angle i/'(x, y) has a differential equation which can be determined. Let the differential equation of the family SJi be (4,1). T h i s can be found by the method of Section 1.5. T h e inclination of its direction field at (x, y) is tan~^ /(x, y). T h e inclination of the field of the trajectories is therefore {4/{x,y) + xarT^ f{x,y)\. Hence the trajectories have the differential equation /
This equation is
= tan {yp{xyy) + larT^ f{x,
y)].
Trajectories
69
if 1^ is a right angle everywhere, and otherwise, by the addition formula for the tangent, (4.6)
y
f{x, y) + tan yp(x, y)
I
It - f{x,y)
tan^(jf,>)
T h e case i n which is a right angle everywhere is especially important. T h e two families of curves are then said to be orthogonal trajectories of each other. Orthogonal families often have interesting a n d important geometrical properties. Such families also occur i n many connections i n physics. T h e lines of electric flow in a conducting plate, for instance, are the orthogonal trajectories of the lines along which the potential is constant. A n d i n the plane flow of a n incompressible liquid, the lines of constant pressure are the orthogonal trajectories of the streamlines. Example 3.
T o find the equation of the trajectories that cut the hyperbolas of the family — 2x^ — c everywhere at the angle tan~^ (^). T h e diff"erential equation of the family of hyperbolas is ^' = 2x/y. Differential equation 4.6 for the trajectories is thus y
,
=
{2x/y) -f- 1/3 1 -
2x/'hy
which has the general integral [x-^y}'[2x-y\'
This equation gives the family of trajectories. Example 4.
T o find the equation of the orthogonal trajectories of the
curves 4x-\-ly
= cx^'\
T h e differential equation of this given family of curves is , ^
\2x + 25v Ax
T h a t of the orthogonal trajectories is, by (4.5),
^
\2x - h 25y
T h e general integral of this equation, namely log {Ax' + nxy - h 25y^} - - tan 2 gives the family of trajectories.
\
Ay
70
Integral Curves; Trajectories PROBLEMS
1, F i n d the orthogonal trajectories of the family of parabolas
= ex.
2. F i n d the equation of the orthogonal trajectories of the family of hyperbolas
3, Find the orthogonal trajectories of the family of circles x^ + y^ —
= 2cy.
4, Find the trajectories that intersect the lines of the family x + y = c at the angle 45 degrees. 5. F i n d the equation of the family of trajectories that intersect the lines y ^ ax -\- c at the angle tan~^ x. 6. Find the equation of the family of orthogonal trajectories of the ellipses x^/a^ +
7* Find the orthogonal trajectories of the family of parabolas 8. F i n d the equation of the family of trajectories that intersect the lines * + a> = c at the angle t a n " ^ 9. Find the equation of the family of trajectories that intersect the circles + = c at the angle tan~^ (x* + y^\. 10. Find the equation of the family of trajectories that intersect the ellipses y^/2S at the angle whose tangent is 11. Find the trajectories that make with the curves of the family tan~* (y/x) = c the angle { - t a n ~ ^ \ / x * +
+
~
.
12. Find the equation of the family of orthogonal trajectories of the curves 2{1 +3sin"^;'
4.5. Singular points of a direction field F r o m the fact that each point of a direction field lies upon an integral curve, it might seem natural to conclude that each point lies upon only one such curve. That, however, need not be so. Example 5.
T h e differential equation /
= 2{x^ -\-y -
1|'^ - 2x
has the general integral
TTirough the point (2, —3) therefore passes the particular integral curve
But the curve y = \ ~ x^ is also an integral curve, and it also passes through the point i n question.
Singular Points of a Direction Field
71
W h e n two integral curves have a point i n common they are, of course, tangent there, for they must both conform to the single direction of the field. A point of a field through which just one integral curve passes is called an ordinary point. A point which is not ordinary is called a singular point.
If we broaden our definition of a field to permit the inclusion of points at which the function f{xjy) is not continuous, such points may be singular points. However they are not necessarily so. F o r instance, i n the case of an equation 1,4, namely, (4.7)
/
=
Pi",
y)
(3)
6.
-
x^].
1 1 « + 4* +
;-i(:c) -
2** +
4,
4:..
-s/yTs -
yiix) » 0.
0.
• \^-2
I
-
yi(x)
-3.
-
-3.
1+ ^(4)
-
y('2)
20.
-
1.
=
>(i)
= i:r.
-iix-V^'-4>i.
18. y =
19. y
yi(x)
1.
>i(x)
=ix2 3x2,
+jrj -
-1.
>iW
=
y = , / ? l ± ^ - 3 . :)'(4) -
-12.
yi(x)
-
-3x.
4.6. Some elementary cases of the differential equation , ^ _ ox + by ^ hx + fcy Section 2.8 shows how any differential equation 4.8 may be integrated. T h e solving formula is, however, often such that the character of the family of integral curves is not easily discerned from it. T w o cases are exceptional i n this respect. If a = A = 0, and ^ = — A, the differential equation is simply y' — y/x. For this equation the family of integral curves is the family of straight lines through the origin. I n this instance, therefore, a l l the integral curves pass through the singular point. If b = h, the differential equation admits {hx -\- ky\ as an integrating
The Integral Curves as Trajectories
73
factor, and the general integral found by means of this factor is ax^ +
+
h\xy +
ky^ = c.
T h e integral curves are thus a family of similar conies centered at the origin and having their axes i n common. T h e y are ellipses (or circles) if [ak — bh\ > 0, and hyperbolas together with their common asymptotes if {ak — bh\ < 0. Figure 11 illustrates these cases. I n (a), no integral passes through the singular point; i n (b), two integral curves do so. A n y differential equation 4.8 remains unchanged if x is replaced by mx, andy by my, with any constant m. Such a replacement can be interpreted
Via. 11.
either as a change of scale, or alternatively, as a magnification (or diminution) of the geometrical figure. T h e family of integral curves therefore always appears unchanged when the figure is magnified. Figures 11 clearly have that property. T h e origin is the only singular point of the differential equation; hence every other point has just one integral curve through it.
47,
The integral curves as trajectories
I n cases of differential equation 4.8 other than those dealt with i n Section 4.6, the form of the integral curves can quite readily be inferred without any recourse to the actual solving formulas. This inference could be made, for instance, by using the fact that the straight lines through the singular point are isoclines. W e shall make the inference by considering the integral curves as trajectories.
74
Integral Curves; Trajectories
T h e trajectories that cut the straight Hues through the origin at the angle ^(x, y) have the differential equation y
,
O'A) + tan ^
=
1
—
•
tan 4f
{y/x)
T h i s is equation 4.8 if ax -\- by hx
ky
(y/x) + tan ^ 1 —
0»/*) tan ^
that is, i f k(y/x)^{b-\-k\(y/x)a b(y/^)^-\-{a-k\{y/x)-h
T h e integral curves thus cut any line y = \x at the angle iff given by the formula •^"''^
bX'^ +
[a -
k]\ - h
B y virtue of equation 3.10, we may write (4.9) alternatively i n the form U 10Ì ^ ^
b\^+la~k\\-h
^I'og'-I _
^
de
jtX2+
{b-\-k]\-\-a
where (r, 0) are the polar coordinates of the intersection of the integral curve with the line y = }ix. Thus X = tan 0. T h e discriminants of the quadratic forms i n the numerator and denominator of formula 4.9 are Di=
{b + h]^-4ak,
D2 =
{a -
k]^-{-4bh.
I n this case the numerator of formula 4.9 is not zero for any value of X; therefore the angle ^ is never zero. A l l the integral curves thus cut across every straight line through the origin and so encircle this point completely. If any one of them closes, all do so, for all are similar. T h e y are then a family of ovals such as were shown in Fig. 11 (a). If they do not close they are spirals. A l o n g these the distance from the origin always increases or always decreases if D2 ^ 0, for the numerator of formula 4.10 then maintains its sign. W h e n D2 > 0, the distance from the origin alternately increases and decreases along the spirals. Figure 12 shows families of integral curves of these respective types. C A S E 1.
Di < 0.
I n this case the numerator of formula 4.9 is a perfect square. It therefore vanishes at just one value of X, say X i . T h e straight line y = Xix is an integral curve. This line is therefore not crossed by any other integral curve except at the singular point. A s X —> X i , t h e numerical value given by (4.10) increases indefinitely. Thus r C A S E 2.
Di — 0.
as Trajectories
,
4x-f 9y • x-Ay
^ "
75
8jc+y 7x+2y
(6)
(o)
Fio. 13.
either approaches zero or becomes infinite. If D2 ^ 0, formula 4.10 maintains its sign, and log r always increases or always decreases. If D% > 0, there are 2 lines through the origin upon which ^) is a right angle. O n these lines r has a relative maximum and m i n i m u m . Figure 13 shows families of integral curves of these respective types.
Integral Curves; Trajectories
76
3x+> w Bx-2y y -
—X
6x+5y
(a)
f6) Fio. 14.
2x-Zy X
F I O . 15.
Answers to Problems
77
3. Di > 0, A N D {ak — bk] > 0. I n this case the numerator of formula 4.9 vanishes for 2 values of X, say X i and X2. T h e lines y = \\x and y = X2* are integral curves. They are therefore not crossed by any other integral cur\'es except at the singular point. A l l the integral curves are tangent to one of these lines at the origin, and approach parallelism with the other as r increases indefinitely. Figure 14 shows families of integral curves of this type, (a) applying in an instance in which D-z < 0, and (b) i n an instance i n which D2 > 0. CASE
4. {ak — bk\ < 0. I n this case Di is necessarily positive, a n d there are therefore again two straight integralsjy = Xix andy = Xox. T h e right-hand member of formula 4.9 has a derivative whose denominator is a squared quantity, and whose numerator is a negative definite quadratic form i n X. Thus tan ip always decreases as X increases, that is, a l l the integral curves, except for the straight ones, are convex toward the origin. T h e y have the straight integrals as asymptotes. Figure 15 shows a family of integral curves of this type. CASE
ANSWERS TO ODD-NUMBERED PROBLEMS 1. T h e family of ellipses 2x^ + 3. T h e family of circles 1
+ a
X
-\-+
7. The family of parabolas V y 2 tan ' X- = r. 1
9. X- +—y" 2 -
11. T h e straight lines y = ex. 13. y = 15. 17.
{>
-
+ X
5 , 2 + 4;,
+|.
I p = 4x.
-
\y^\\^^x.
19. y =^\\
-
2x^^ -
= 2cx.
log 11 + ax\ = c.
a2
1
= ^.
8x'}.
-\-
= c ~ x.
CHAPTER
5 Approximate Solutions Infinite Series Existence and Uniqueness of a Solution 5.1. Polygonal graph approximations A graph to approximate the integral curve of a differential equation (5.1)
y
=My),
for which (5.2)
= ^-o,
y{xo)
can be constructed i n the following way: Let XQ, xi, X 2 , " ' * , be any set of abscissas such that XQ < xi < X2 < • ' ' . F r o m (XQ, ^O) draw the straight-line segment with the slope /{XQ, yo) until it meets the line x = xi. Let the coordinates of this point be (JTI, ¥{), and from the point draw the straight-line segment with the slope f(xi, Yi) until it meets the line x = X2. Give this point the coordinates (x2, K 2 ) , and from this second point draw the straight-line segment with the slope / ( j f 2 , J^z). Continue i n this way. T h e graph so constructed is a chain of segments, that is, an open polygon, w i t h vertices at the points (AT,-, Yi), a n d is called a polygonal graph. W i t h YQ = yof we have for this graph (5.3)
Yi+i
=
Yi^fixi,
Yi){xi+,
-
Xi],
I = 0, 1, 2,
. . •.
T h e direction of advance along this graph coincides at each vertex w i t h that of the field of the differential equation. T h e graph may therefore be expected to diverge but little from a streamline if the intervals {xi^i — Xi) are short. It then gives an approximation to an integral curve; moreover, the shorter the intervals taken, the better the approximation. T h e approximation is always best near the point {XQ, yo). A s the construction 78
Polygonal Graph Approximations
79
advances fron^ this point the approximation becomes poorer because the divergences accumulate. A polygonal graph extending to the left from the point (;ifo, ^o) can be constructed in a similar way. Example 1.
T o construct for the differential equation 1
7J
the polygonal graph to the right from the point (0, 0) with [xij^i ~ Xi\ = ^. Also to compare this graph with the integral curve over the interval In this instance = = 0, and formula 5.3 gives Yi = — T ^ . This, together with x\ = -y, yields f{xi, Yi) = —jsF o r m u l a 5.3 thus gives ^2 = — i ^ ' Proceeding in this way, the values of T a b l e 1 may be calculated: Table 1 1
0 Yi fi^i. Yd
1
0 1
3
1 1 7
1 17
3
e 07
—
1
— T7
1 7
— 7^
1 17
1S43
T5T
•ffO 7 3
2 15 4 3
5
5 14 1
7144
20S 5
"807T 6 14 1
^144
The graph obtained is that designated by Y i n F i g . 16. Approximate solutions are primarily useful when the differential equation cannot be integrated. I n this example the differential equation is linear and therefore can be integrated. We may, therefore compare the approximation with the actual integral curve. T h e relevant integral of the differential equation is y =
that is.
-1
.-«V2
ds.
ea:V4
y =
where I{x) stands for the probability integral I{x)
=
ds.
T h e values of !{x) and of e" are obtainaljle from standard mathematical tables. Thus the numerical values of y can be computed. T o three significant figures, they compare with those of K,-, as Table 2 shows.
80
Approximations, Series, and Existence Table 2
y
1
3
T
2
5 T
-0.083
-0.177
-0.305
-0.502
-0.837
-0.087
-0.197
-0.368
-0.676
-1.30
0
1
0 0
T h e graph of > is included i n F i g . 16. T h e approximation is, i n this instance, at most moderately good, because the intervals taken were relatively large. W i t h smaller intervals the number of computations needed
F I G . 16.
to cover the given range would have been larger, but the approximation would also been better. ASSIGNMENTS 1. Construct a polygonal graph for the differential equation
y'-i-y-
h
through the point (0, —1), with [x,-+i — xi] = T , and plot also the integral curve. 2. Construct the polygonal graph for the diflTcrential equation
through the point (0, 1), with {xi+i -
x, and plot
Xi\ =
also the integral curve.
3. Construct the polygonal graph for the differential equation
/
= y +
-
- Xy
through the point (0, 1), with \xi^\ — Xi] ^ \^ and plot also the integral curve. {Note: Fig. 10 applies to this differential equation.)
Approximations for Simultaneous Equations
81
4. Gomtruct the polygonal graph for the differential equation >' =- sin {x +y] through the point (0, T / 2 ) , with
-
— x,} = -j-.
1, (The integral in this case is
y = 2 t a n - l [e^] - x.)
5.2. Polygonal approximations for simultaneous equations T h e method of polygonal graphs can be applied also to approximate the solution of a differential system (5.4)
y=fi{x,y,z),
z'
^f2{xyyyz\
for which (5.5)
z(x(,) = ZQ.
y{xQ) = yoy
T h e integral curves of such a system are found i n Section 4.2 to be three dimensional. T h e y may, however, be studied through their projections upon the (x, y) and the (x, z) coordinate planes. These projections are plane curves, one defining as a function of x and originating at the point (jffl, ^o)) the other defining z{x) and originating at the point (xo, ^o). W e may construct polygonal graphs to approximate these curves. W i t h abscissas xo, x i , X2, • * • , chosen as i n Section 5.1, draw the straight-line segment with the slope fi{xo,yoj ZQ) from the point (xo, >o) to the point where it intersects the line x = x i . Give this point the coordinates (xi, K i ) . Similarly, draw the straight-line segment with the slope fzixot yoi ZQ) from the point (XQ, ZQ) to the point with the abscissa x i , say (xi, Z i ) . F r o m the two points thus determined, draw the straight-line Table 3 1
1
Xi
0
Yi
0.5
0.37
0.19
Zi
1.5
1.58
1.75
/i(x.-, Vi, Z . )
-0.25
-0.36
/2(xi, Yi, Zi) .
0.17
0.34
Yi+i
0.37
0.19
Zi+i
1.58
1.75
-0.52 0.46 -0.07 1.98
3
-0.07 1.98 -0.63 0.70 -0.39 2.33
-0.39 2.33
82
Approximations, Series, and Existence
segments with the respective slopes Yu ^i) corresponding points {x2y Y^) and {x2, -^2)» etc. Example 2.
and / 2 ( ^ i , ^ i , ^ i ) to
T o construct for the differential system y' ==2y + z' =
z-{-
- 3 > -2z
ie^'^ -
h'^^
3,
+ 5,
the polygonal graphs for which ^(0) = ^j, and z(0) = -f, with (xi^i — *,•) T h e method described calls for the computation of the values which T a b l e 3 sets forth.
X
Fro. 17.
T h e graphs Y and Z as they may be plotted from these, and as they appear upon a single plane, are indicated in F i g . 17. T h e actual integration of the differential system is in this instance possible. B y the method of Section 2.9, the relevant solution is found to be y ~
I — se
Z =
I -t e^
-\- se —
—
, •
T h e graphs of these relations based upon the computed values set forth i n T a b l e 4, are also shown i n F i g . 17.
Power Series, Polynomial Approximations
83
Table 4 1
0
3
1
2
0.5
0.34
0.14
-0.16
-0.54
1.5
1.62
1.81
2.12
2.52
5.3. Solutions in power series.
Polynomial approximations
W h e n the function f(x,y) in differential equation 5.1 is representable by its Taylor's series about the point {XQ, yo), the solution fulfilling condition 5.2 can be obtained as a power series. T h e sum of the first {n + 1} terms of that series is an approximating polynomial of the degree n. By the change of variable S
=
X
—
XQ.
u = y -
yo;
condition 5.2 is replaced by w(0) = 0, and the differential equation is transformed into (5.6)
du/ds = / i (s, u).
T h e function f\{s, u) is then expressible i n a series in powers of s and u, namely. (5.7) k.3 - 0
and it is the supposition that this series converges for some range of values of s and u. F r o m the calculus, the coefficients Aj^k are known to be given by the formula L
ds' du
Ju = 0
I n practice it is, however, often easier to find series 5.7 by combining known power series for terms or factors of / i {s, u). A m o n g the most useful of such series are the following: 1 4-r
-
+
(5.8)
¿' = 1 + ^ + 1 2!
^2
+ i r3 _j_ 3!
84
Approximations, Series, and Existence sin z —
r
6_
3!
. . .
^5!
(5.8)
log \a ^ z\ = log a + Let u be taken i n the form (5.9) with undetermined coefficients ÛI, (22, as, * • • . W h e n this is substituted into series 5.7 and the requisite multiplications are made, all terms i n like powers of s may be collected, and their sum equated to the term of that power in the series for du/ds. W e obtain thus a system of equations which successively determine the values of a i , a^, as, * * " . The solution is then obtained by expressing equation 5.9 in terms of the original variables. A n assurance that this procedure actually does give a solution can be obtained only by proving that the power series which it yields for the solution is convergent. T h a t can be done. W e shall not, however, do it here. T h e whole matter of convergence is irrelevant if only the first n terms of the series, instead of the whole series, are considered, that is, if the method is applied to obtain a polynomial approximation. T h e method is applicable also to systems of simultaneous differential equations, to obtain either power series solutions or polynomial approximations. Example 3.
T o find for the differential equation
1 \ — X — y
the polynomial of the fourth degree that approximates the solution for which _y(l) = - 2 . T h e change of variables s = x — \, u = y -\- 2 transforms the given differential equation into form 5.6, with his,
u) =
{2-
is+u)\ —I
By the first equation of series 5.8, therefore,
• | . »
•
•
4
•
Power Series, Polynomial Approximations
85
F r o m (5.9), -1- «
=
(ai +
1)J +
{s +
u)^ =
(aj. +
2.2 + 1) V
(^ +
«)' =
(ai +
l ) 3_3 V +
J
025^ +
OiS^
2^2(^1 +
+
1).^
+
and thus. a2
{ay
4 az
+ 1)" 8
, a^iax +
1)
4
, (¿11 +
+
1)
16
O n equating this to the series for du/ds, and comparing the coefficients of like powers of j , we see that 1
+ 1
2ao = a2
3(Ja = -
4^4
=
,
1)'
(ai +
^3
^2(^1 + 1)
4
4
(^1 +
1)'
16
T h e first of these equations gives a\\ the second then gives «2, the third thereupon 03, etc. Thus, « =
+ A ^ ' + irs^
+
T ^ ^ '
+
• • • -
I n terms of the original variables, the approximating polynomial of the fourth degree is thus
Example 4.
T o find for the differential equation y' = e''y j^x
-
2,
the polynomial of the sixth degree that approximates the integral for which T h e change of variable u = y — \ gives form 5.6 with s = x, and / i = ^*'C«+i) -\-x-2. By the second equation of 5.8, with z = x\u + 1),
86
Approximations, Series, and Existence
therefore,
T o terms of less than the sixth degree this last equation would be /i
^
=
j( ^
^
-\- ^x* -\- x*u -\- • • •
O n substituting form 5.9, and equating the result to the series that
we find
fli = — 1 , 2^2 = 1, 3^3
= 1,
6^6 =
+ ai,
fl3
«
Thus
and the polynomial sought is
Example 5.
T o find for the difi'erential system /
= ^'^^^'\
z' -
(x+jy)» +
z,
the polynomials of the fourth degree that approximate the solution for which ^(0) = 0, and ¿(0) = 0. T h e second and third equations of 5.8 give ^io(v+*) = 1 + s i n ( y + ^) + : i s i n 2 ( y + z) 4 - : ^ s i n ' 0 + z) + sin (y + 2) = (y + ^) - i ( y + z)3 + Sin^
+
^ (y +
- ^(y +
sin^(y + z) = (y + ^)' + whence
••• ,
•••, +
. . . ,
•• •,
Power Series, Polynomial Approximations
87
T h e substitutions y = aix -\- a2X^ + a^x^ + a^x^ + b,x
+
¿2*2 ^ ¿^3 ^ ¿^^4 +
• ••, . . .
therefore yield 2 02 + ¿2 +
and {x+yy
+ z = bix
-f
(¿2 +
(01 +
1)^1*'' + ¡¿3 + 2^2(01 +
1)|X«+
• ••.
O n equating these series respectively to the series for u' and z', we obtain the equations ai =
1,
bi =
0,
2a2 = a i +
¿1,
2b2 = bi, 3^3 = ^2 + ¿2
+
i(oi
+
+ 1)',
3i-3 = ¿2 +
(«1
4a4 = «3 +
63 +
4^4 = ^'3 +
2d2(ai +
[ai +
6 i ] [ o 2 + ¿2],
l),
T h e required approximations are thus found to be yA = x-{'^x^-^
W ix'
+
+ %x\
PROBLEMS Find for each of the following differential equations or systems the polynomial approximation of the fourth degree for the solution specified. 1. v' = > - 6e-',
3. /
=
1 - 2x
y(0) = 3.
y(0) = 0.
2./
= ^ f ' ' y(0)
4. / - 2 ^ + > ,
1
y(\) = 0 ,
88
Approximations, Series, and Existence
5. y'
——? 3x — > — 3
7. y'
>2+8U+2}
8. y'
«2"",
9. y'
10. y' z' 11. y'
-
—
-
12. y'
\ -
>(1) -
3.
6. /
+log il
-
3 + 2 sin X,
:K ~
y(0)
= 0
:v(-2)=0.
j(0) =- 0.
z + Axy,
2 + 2> + 2>2 + xz^ + 2x^, 3 + 4x -x+y-2z,
2> + 4z + Axz, y{0) - 0 .
y{0) = 0,
z(0) - 0 .
:y(0)=0, 2 + log 11 + ¿1.
z(0) - 0.
r ' = I + icy,
^(0) - 0. ;f(0) =- 0,
z(0) -
0.
5.4. The method of successive approximation T h e solution of differential equation 5.1 which fulfills condition 5.2 is a function for which / W
~f{x,y{x))
^0.
B y a quadrature, therefore, y{x) -yQ-
f'f(x,y(x))dx
^ 0.
T h e last equation says that>'(j:) is a solution of the equation (5.10)
y=yo
+
f^'jix,y)dx.
Conversely, any continuous solution of equation 5.10 is a solution of 5.1 and 5.2, as is easily checked by differentiating (5.10) and setting x = XQ. T h e single equation 5.10 therefore effectively stands for the two relations 5.1 and 5.2. W h e n ^ i n the function /(x, y) is replaced by ^IQ, the resulting function /(*> yo) depends only on x and can be integrated. T h e formula yiix)
= >o + j^,/(^.>o) dx
thus defines the function yi{x). W i t h this formula for _>'I(A:), now, fix, yi{x)) is a function of x that can be integrated. Hence the formula yiix)
= >o +
fixyyiix))
dx
defines the function ^2(x). This process can (theoretically) be repeated again and again, and thus the functions of the sequence yMyy2ix)yyzix),
•••,
Successive Approximation
89
can be successively defined by the formula (5.11)
=>o + fjj{x,yn-i(x))
yM
n = 1, 2, 3, • • - .
dx,
W e shall prove i n Section 5.6 that, when suitable hypotheses on /{x, y) are fulfilled, the functions yn{x) converge to a limit when «>, and that this limit function is the integral of the differential equation. T h a t proof will show that the functions^i(x),^2(j^)j>3(^)) * * " , are approximations to the integral. T h e y are called successive approximations. Example 6.
T o find for the differential equation y' =
2 + 2 sin with the condition >(0) = 0. Find adso the integral of the difTcrcntial equation and condition, and compare its Taylor's series with y^{x), 16. Find the formula for > A W for the differential equation y' = cos x — y^ with the condition ^(0) = 1. Find also the integral, and compare its Taylor's series with
17. Find the formula for y\{x) for the differential equation > + — ^ , with the condition ^(0) = 3. F i n d also the integral, and compare its Taylor's series with
18. Find the formula iov yz{x) for the differential equation >' = 7^ + x^y + 1, with the condition ^(0) = 0. 19- Find the formula for yz{x) for the difTcrcntial equation y' = \y — the condition ^(1) — J-^.
A i with
20, Find the formula for^'sW for the diflrcrentialcquationy =^ \y^ — 2y}/{2 with the condition y(0) = 1.
—
2^1,
5*5* Successive approximations for difFerentiai systems T h e method of successive approximations is also applicable to a system of simultaneous differential equations. I n the case of a system 5.4 with the conditions 5.5 it shapes up as follows: T h e solution of the system consists of a pair of functions y{x) and z{x)y which fulfill the equations of the system. By quadratures they are therefore seen to fulfill the equations
z = ¿0+
I'f2{x,y,
z) dx.
JXa
T h e functions yo, ZQ) and fzix, yo, integrated. T h e equations :yiW = >o + zi{x)
=
ZQ
-\- f
ZQ)
involve only x a n d can be
vo, ^0) dx, f2{x, yo, zo) dx
92
Approximations, Series, and Existence
therefore define the functions )'i(x) a n d zi{x). yzix) =>o
+
zi{x)) dx,
j'ji{x,yi{x),
2 2 W = ^ 0 + l'ji{x,yi{x),
define the functions Example 8.
W i t h these the equations
zi{x)) dx
and Z2W, etc.
T o find for the differential system y' = x-\- z,
z' = \ -
y,
the successive approximations to the integral for which ^(0) = 2, a n d z(0) - 1. T h e formulcis that are successively found arc 2H-X,
zi = 1 +
\)dx
jy2 = 2 +
2 + x,
^2 = 1 + j '
\ -\
= 2 +
1
>3
^3=1
jy4 = 2
Z4
= 1
- x\
-
x\ dx = \
-
dx^2
+
-x-ix'^\
x-^^x\
x] dx = \ — X — ix^\
+ 1 +
X
-
3! 1
— X — —
2!
' 1 X* +
— X*.
4!
T h e actual integral is i n this instance ^ = 2 + sin X ,
z — cos X
— X.
Existence of a Solution
93
I n power series form these formulas are y
=2-[-X-~X^-\-~-^X^
-
2!
^4!""
'
• '
6!
,
^
It is easy to see that approximations to these series were obtained. PROBLEMS 21. Find the formulas for yi{x) and Zi(x) for the differential system
y - ^ - 1 ,
z'-2>-z-2,
with the conditions ;'(0) — 2, z(0) =• 0. 22. Find the formulas for;'8(x) and zi{x) for the differential system
with the conditions >'(0) = 1, z(0) = - 1 . 23. Find the formulas for ys{x) and z^^x) for the differential system
z'^y-2,
-z-1, with the conditions ;r(0) =» 3, z(0) = — 1 .
24. Find the formulas for yh(x) and zeW for the differential system y' = —z/x,
z' ^ y -~ z,
with the conditions >(0) = 1, z(0) — 0.
5.6. A proof of the existence of a solution Thus far we have relied upon the geometrical evidence of the integral curves for an assurance that a differential equation 5.1 has an integral fulfilling a condition 5.2, even when no formula for such an mtegral c a n be given. Subject to suitable hypotheses upon the function f{x, y) we shall now prove that such an integral does i n fact exist. Such a proof coidd be given by several different methods. W e shall give it by the method of successive approximations. T h e hypotheses upon /(x, y) shall be the following ones : that i n some region of the (jt, y) plane of which (XQ, yo) is an interior point, (5.12)
(i) (ii)
\fix,y)\<M, {fix,y) -
fix, y*)\
^N\y-y*l
O f these, relation (i), i n which M may be any positive constant, says .that /(x,y) is bounded. Relation Çii), i n which N may be any positive constant, shall hold whenever both (x, y) and (x, y*) are points i n the region.
94
Approximations, Series, and Existence
Equation 5.12(ii) is known as a Lipschitz condition, i n honor of the G e r m a n mathematician R . Lipschitz (1832-1903). It is worth observing that any function having a partial derivative w i t h respect to y fulfills a Lipschitz condition i n any region i n which fy{x, y) is bounded, but does not fulfill such a condition i n a region i n which/^{Ar, y) becomes infinite. This can be seen at once from the mean value relation of the calculus /(^.>)
-=fy{x,yx)\y-y*\,
i n which^1 is some value between^ a n d ^ * . T h e relations (5.13)
X
—
X(i\ ^
A,
y - ya\ ^
Mh,
confine the points {x, y) to a rectangle centered at (XQ, ^'O) - T h e size of the rectangle is determined by h. It is small enough to lie within the region i n which relations 5.12 apply, if A is small enough. We shall suppose such a value of A to have been chosen, and shall henceforth refer to this rectangle as R. I n it the range of variable x is (5.14) W e begin the proof by showing that all the graphs of the functions _y„(x) given by formula 5.11 remain within rectangle R for all x of interval 5.14. This is easily done by mathematical induction. T h e fact is obviously so for^o- W e need therefore only show further that if it is so f o r ^ „ _ i ( x ) it is also so {oTyn{x). Suppose it is so for^n_i(x). T h e n \f{x,yn—\{x))\ < A f , by (5.12), and, by (5.11), (5.15)
yn{x) - yo
f^^ f{x,yn-i{x))
- / ! l o g y - x ; - ! ! -y\
=0.
6.3. Singular solutions again A diff"erential equation 6.1 may have a singular solution, for its general integral, or some part of it, may have an envelope which is not itself a curve of that net. A t the points of a n envelope the differential equation is fulfilled by fewer directions than at some neighboring points. This can be seen from F i g . 19, in which the circles are the curves of a general solution. T h e i r two common tangents are envelopes, and hence are singular solutions. A t a point Pi the differential equation is fulfilled by two directions which make the angle 6i with each other. A t a point P2 nearer the envelope the angle 62 is smaller, and as the point approaches the envelope at P the angle between the two directions reduces to zero; that is to say, the two directions reduce to one. T h e envelope is thus included among such loci as there may be upon w h i c h the differential equation defines a reduced number of slopes, namely, upon which it has a reduced number of roots y'. T h e condition for that is (6.4)
dF(x,y,y')
ay
0.
102
Further Equations of the First Order
T h e éliminant oiy' between this equation and (6.1) includes any existing envelope. It cannot be assumed that every locus given by the éliminant is an envelope, for there are other loci also upon which the differential equation defines a reduced number of directions. T h e line of centers of the circles in F i g . 19 illustrates this. O n this line only the perpendicular direction fulfills the differential equation. Each locus obtained from the
X
FiO. 19. éliminant must therefore be tested to determine whether or not it is a solution. Example
6.
T h e differential equation for the family of circles i n F i g . 19
is \x -
Viy\ny'^
+
\\ -
{x+yy'\^
= 0.
For this equation, equation 6.4 is =0,
{x-Vly]^y'-y\x+y/]
and the y' éliminant is y\y -
Vlx][Vly
-
x]'' = 0.
T h e éliminant thus gives the three l o c i ^ = 0, )" = y = (l/\/3)x. B y test in the differential equation the third of these loci is found not to be a solution. It is the line of centers. T h e other two loci are solutions.
Integrations by Differentiation
103
T h e y are singular solutions, because the general solution
does not include them. Example 7.
F o r the differential equation { / - ^ 2 i y + / + i =0,
y 2 _
the y' éliminant is = 0. This does not fulfill the differential equation. Thus there is no singular solution. PROBLEMS Find a singular solution for each one of the following differential equations. 21. / y ' - W - ' + T b ' - ' l
22. {3y -7\y^
=0.
23.
{xy' - ; y p - y 2 -
25.
{x^ - y\y''^ - xy' -\-x^ = ^.
-4{y
-2]
=0.
24. xy'* - 2yy'^ + 2x^ = 0.
1 =0.
26. jc^"' - /
+ ^ -
2 = 0
X
27. y 2 -
4xy' + 4> = 0.
28. >' -
29. xy' - y' cos~' (y') + V l 30. f i + / i
-
log \xy' - y\ - 1.
> =• 0.
log {1 + y i - y i i -t-xj 4 - y -
6.4. Integrations by differentiation.
1 = 0,
The Clairaut equation
B y definition a solution of a differential equation 6.1 is a function ^(x) tor which the expression F{x, y{x), y'ix)) is identically zero. T h e derivative of this expression is therefore also zero, that is, F x ( ^ , > W , > ' W ) + F^{x,y{x-),y'{x-))y'{x) + F^{x,y{x),y'{x))
"
^
= 0.
This signifies \S\aXy{x) is a solution of the equation (6.5)
FÂ^yy.y')
+ Fy{x,y,y')y'
- f F^{x,y,y')
= 0,
and that it is therefore a solution of simultaneous equations 6.1 and 6.5. T h i s fact can sometimes be used to accomplish the integration of equation 6.1. If >' is determinable i n any manner from equation 6.5, the substitution of that value for^i' i n equation 6.1 converts that equation into an integral. W h e n equation 6.5 involves too many variables to permit a determination of^', we may eliminate^ between equations 6.1 and 6.5. A value of y' determined from the;) éliminant also converts (6.1) into an integral.
104 Example 8.
Further Equations of the First Order F o r the differential equation 2/ + > -
2 / logy = 0,
equation 6.5 is y' -
0.
2 log/idy'/dx)
This is a differential equation f o r ^ ' whose integral is * —
= jlog^y')^.
F r o m this r e s u l t / = e^^^^^'y and / = e~^^^''. These values convert the given differential equation into the respective integrals
and Example 9.
2e^'l\
-
2e~^^'{\
+
Vx
-
c]y = 0
-
c\y = 0.
F o r the differential equation 2yy'^ + 2x^ = 0,
xy'^ -
equation (6.5) is =0.
[4xy'' -6yy''\^+{6x^-y"] dx
There are too many variables in the last equation to permit a determina tion oîy'. T h e éliminant of y between the two equations is dy' 'Tx
T h e factors of the last equation give, respectively, y' = ex, and y'^ = 6x^. These values convert the given differential equation into, the general integral c^x^ -
2c^y + 2 - 0 ,
and the singular integral 9y^ = 4 V 6 ; f 3 _
There are two types of differential equation 6.1 to which this method is peculiarly well adapted. (i) T h e equation that can be made explicit for y, namely, (6.6)
> =/(^,/);
(ii) T h e equation i n which x and y occur only i n the combination [xy' - y\, namely, (6.7)
f(x/
-y,y')
=0.
A n equation 6.7 is called a Clairaut equation i n honor of the French mathematician A . C . Clairaut (1713-1765).
Differentiation with Respect to y
105
For a Clairaut equation, equation 6.5 is simply {dy'/dx) • Fj/(xyy, y') = 0. T h e first factor of the last equation gives y' = c. A Clairaut equation is thus converted into its general integral by merely replacing y' by c. Example 10.
T h e differential equation \ =0,
W - y \ ^ - y ^ -
is a Clairaut equation. Its general integral is therefore \cx — y\^ — 1=0. Equation 6.5 is i n this instance
dx
Uy
-y'
-"y]
—
=0.
T h e second factor of the last equation gives ^' = xy/jjc^ ~ this value of y' the differential equation is converted into the singular integral x'^^y''
= 1.
PROBLEMS Solve the following differential equations. 31. xy' -y
+ / 2 = 0.
32. 1 6 / -
—
{xy' -y\^
33. 2y - A ~ log {y^+ 1} = 0.
34.
35. xy'^ - yy'"^ + 2 = 0 .
36. / 2 Jr\x-
37. 2y - 1
1
38. y
V^y"" + 1 39. y< - x.2..»2 y + "^.xyy' - y
0.
y — log y' =» 0. \\y' -y-\.
1 = 0.
-Ixy' .'2 -
1
40. \zx - 1 l y ^ - 2{> -
41. 2x>./2 = 4 x V ' + 1.
43. 16{:c + 1
X
- 0.
1 l y + 4 =• 0,
42. i l -\-r^']y = (y - 1]*.
- 4{4> - ! } / - !
= 0.
44. ^ + 2 + 2 1 o g (y/2} - xy' = 0.
45. Axy - 2xy'^ + ^
= 0,
46. y * + l / + > } { 4 y 2 + l | - 0 .
6.5. The method of differentiation with respect to / If we think of ^ as the independent variable, and recall that dx/dy = 1 / y ' , the rccisoning of Section 6.4 shows that any solution of a differential equation 6.1 is also a solution of the equation obtained by the differentiation of (6.1) with respect t o ^ , namely, of the equation (6.8)
FAx,y,y')
~ + Fy{x,y,/) y
+ FAx^y,/)
7^ = 0. ^y
A n y value o f y that can be determined from (6.8) converts (6.1) into an integral.
106
Further Equations of the First Order
If y' cannot be obtained from (6.8) because too many variables are involved, we may eliminate x between equations 6.1 and 6.8. A value of ^' obtained from the x éliminant also converts (6.1) into a n integral. T h i s method is peculiarly well adapted to a differential equation that can be made explicit for x, namely, (6.9)
^ =/(>,>').
Example 11.
T o integrate the differential equation ilog{l
- l o g / - x
+ 2 = 0.
T h e derivative of the given equation with respect tojy is 1
dy' r 2 - ~
1
dy
+ i
and integrates to give y' = tan {c — y\. differential equation into its integral log sin
^1
(c —
+
=0,
This value converts the given
X —
2 = 0.
Example 12. T o integrate the differential equation
| x 2 - 2 x l { y 2 + l) + 1 = 0 . T h e y derivative of the given equation is . . lv'^ + 1 {x^-2xiy- = 0.
Ó.Ó. The Riccati equation A differential equation (6.10)
/
= io(^) + ii(^)> + ? 2 ( x ) / ,
i n which q2{x) ^ 0, is called a Riccati equation, i n honor of the Italian mathematician Count Riccati (1676-1754). T h e reciprocal of any solution of this equation, w = l / y , is a solution of the associated equation (6.11)
w' =
-q^ix)
— qi{x)w — qo{x)w^.
T h u s the integrals of either one of these differential equations c a n be obtained by accomplishing the integration of the other equation. A n equation 6.10, or 6.11, may be of a type that is integrable by a method of Chapter 2. W h e n put into form 2.1, it may be linear, or it may have separable variables, or it may have coefficients that are homogeneous and of the same degree. Except for these, and a few other rare elementar)' cases, the Riccati equation is not integrable by any systematic method other than that of power series. It is sometimes possible to find a solution of a Riccati equation by trial. One may try such functions as ^ = ax^, or y = a^^, with undetermined constants a and h, or some other function that the appearance of the differential equation itself suggests. Success i n finding a solution is rewarding, for, if any integral can be found, the general integral is obtainable from it. Suppose ^I(x) is any particular integral of diñ"erential equation 6.10. T h e change of variable (6.12)
y =yi{x)
-
u
with an arbitrary constant cu transforms the equation into A _L ' ^ i " ' _L r \ yi {x) = qo-\- qiyiix) u
'^^ii ^ 2/ ^ + 9 ^ 1 [x) Ü
2ciq2yi{x) a
i
^• u
B y such cancellations as can be made because yi{x) fulfills (6.10), this transformed equation reduces to +
{qi + 2q2yi\u = ciq2.
This differential equation for u is linear. of a form
Its general solution is therefore
U = Ci^
57.
58. xy' => [ 6 - 4 log x} -
[5 -
4 log x\y + (1 - log x j / .
59. _y' = 3x' + - + 3x).2. X
61. /
2 - 2x'
= 2x* +
y -Vy
^ - 1
60. ^'
=
62.
=> sin^ X + _)> cot X
X
+
+
Find the integrals of the following differential equations that fulfill the given conditions. •
63. y'
1
2 +X
64. y = 9x + {3 65.
y
66. >' -
X
I S x b + (9x -
•X
tan
-7
X +
sin
y{-\)
= ^
..2
y
1 +x^
31/,
x l l +x^j
X
cos X
'
>(1) = 0.
- y"^ cot X ,
;y ( -
0
=
1
110 67.
Further Equations of the First Order - 5** + 6*^7 + 2**/,
68. y' "x'
69. /
-\-
= tan
-f.
X
X + S i n jr
70.
;r(0) -
cos Jc
+ /
^-(0)
cot x,
( - ^) = \ 4/
-
-1.
6.7. Some properties of the integrals of Riccati equations If xo is any point of an interval (6.16)
a-^x-^b,
upon which ?o(*), ?i(*)» and 5 2 ( ^ ) are continuous functions, the point ( ' 0 ) ^ 0 ) with any ordinate ^yo lies within some region i n which the function i{x,y)
= qo(x) +
qi(x)y + q2{x)y^
and its partial derivative y) are bounded. T h e function therefore fulfills a Lipschitz condition 5.12, and accordingly every point (xo, ^o) has just one integral curve passing through it. A Riccati equation therefore has no singular solutions. A R i c c a t i equation may have some integrals that are continuous over the whole of an interval 6.16. However, it always has some that are discontinuous, a fact that is easily seen. For, with any chosen XQ, differential equation 6.11 has an integral «;o(^) for which wo{xo) = 0. T h e reciprocal of wo{x) is an integral of equation 6.10, and it obviously has the vertical line X = xo as an asymptote. Example 16.
T o find the solution of the Riccati equation
which is discontinuous at x = 1. T h e differential equation has constant coefficients. constant solution, and by trial this is found to bejJi solution, as found by the use of^"! = 2, is y = 2
—
It therefore has a 2. T h e general
-
T h i s is discontinuous at x = 1 if ci + 12^ = 0. thus -
1
T h e required solution is
Integrals of Riccati Equations
111
T h e R i c c a t i equation has important relations with the equation (6.17)
pMu"
+ pi{x)u' + p2{x)u - 0,
which is the linear differential equation of the second order. Thus if i;(x) is any solution of the R i c c a t i equation
Po{x}^ PQ{X)
on a n interval upon which po{x) ^ 0, the formula (6.19)
u = el"^'^'^'
gives a solution of equation 6.17. Therefore equation 6.17 can be solved by solving equation 6.18. Conversely, if v{x) is any solution of the differential equation (6.20)
v" -
v' + qo{x)qi{x)v = 0,
qiix) H — 7 qiix)
which is of type 6.17, the formula (6.21)
=
-
q2{x)v{x)
gives a solution of Riccati equation 6.10. Example 17.
T o find a solution of the differential equation u" -
{x +
l j u '+
{x -
lju =
0.
T h e related Riccati equation 6.18 is r,' = ~{x-\]
+
{x-\-\U
- ri\
It has 71 = X as a solution. By formula 6.19 the given differential equation therefore has a solution u = e'^^^. Example 18.
T o find a solution of the R i c c a t i equation
,
2H-X x[l -\- x\
2 x\\
X
—
x^
12
X
T h e related equation 6.20 is
^
x{\ -\-x]
x\\ + x l
T h i s has f = as a solution. By formula 6.21 the given equation therefore has a solution)' = — 1 / | 1 +
112
Further Equations of the First Order PROBLEMS
71. Find the solution of the equation 1
2 x
which 15 discontinuous at * = 4. 72. F i n d the solution of the equation 3 4 - -
1 2 - -
x
which is discontinuous at x
x
1
73. Find the solution of the equation
for which y{0) — -a".
Determine its discontinuities, if any.
74. Find the solution of the equation x'y + y\ for which ^(1) = 0.
Determine its discontinuities, if any,
75. Find a solution of the differential equation
76. Find a solution of the differential equation u " + u' log X + -
=0.
x
11. Find a solution of the differential equation u " — u' cos X + u sin
0. x
=
78. Find a solution of the differential equation u"
—
2u' tan X
— u —
0.
ANSWERS TO ODD-NUMBERED PROBLEMS 1, y' = X-
T h e general solution is 5
3- x^ — 4^ = 0. 5. ^ = — 3x, 7. ^ + 3.
—
x}^^ = x^ + f.
T h e general solution is \ / x ^ — 4)^ = x + c-
The general solution is 2 ' s / 3 x + ^ = log x +
T h e general solution is 3 y/y
— 3 =
9. ^ = (n + ^jw — X, with any whole number n. The general solution is
cos (A: +
= x + c.
+ {x + 4 p
Answers to Problems {y - 2x^ - c][y - log X - c\
11. 13.
xy --x^
log
V-^-c
15.
— h 4 log X y
- c
2x X
+
y -
113
0. = 0,
— c
= 0
— c
X
11 J *
19. y\21y ~ 5(x 23.
= 0.
2S.y
= 1.
x2
- > = 0.
31.
— ^ + tf* " 0,
and ^ =
^ = 2 — log cos \x + ¿1,
4;- + Ix -
= 0,
39. c* - c^x^ 4- 2£xy 41. ^
-2^^
+ ^
= 2cx + c\
51.
59. jvi
IX,
61. yi
{1 + i l x ,
and 2y^ = x. 49. 16^ = c(2x - c\\ 2\/y
y =
3x -
X
57. :y fi +
tan {x^ + c]. jv = x 2
X
+x2tan
{-i*' 65. > =
11
- 3
69. y
X
~ X
* + l tan
X
cot
X +
-8x X -
4' 1
•> = ' ' + ? T - i =* cxp
and 27^ = 2x*.
+ c + 2] = 0 .
2
67. > - 7 tan f*' -
u
and 14;^ + 1 p + 4|x + 1} = 0.
and 27>'* + 125x' = 0.
3 x - x 2
75.
0,
Scix^
-2x -
71. >
1 -
+ 2 H x * - v ^ -
=0,
55. y -
63.
1} -
= 0,
f ^ f V ^ + 2 \/>
3 +
and > = 1
and 1 6 / = xV
and > - 0.
53. 5 « - 2 / ^ +
73
- 0,
4c{4y -
43. 16c^{x + 1} -
47.
and y = 0,
K> and 2;»' = 27x.
2c - s / x
45. 4c^ Vx
and y = 2.
and 2y^ = 27x^.
35. c^x - c > + 2 = 0, 37. 4 / -
= x ' ~ l
29. y = sin x.
27.
33.
=x-V\.
21.
It has no discontinuitiefl. 77. u
lin X
2
1:2^'''*'
CHAPTER
7 Some Solvable Differential Equations of the Second Order Approximations.
Applications
7.1. DifFerential equations of the second order A relation (7.1)
which actually involves^", is an ordinary differential equation of the second order. T h e simplest type of such equation is that of the form y"
=
fW-
B y successive quadratures the last equation yields
and i n which ci and C2 are arbitrary constants. A solution of a differential equation of the second order is called the general solution, or the general integral, if it involves two arbitrary constants i n such a way that these constants are not replaceable by just a single constant. A n y differential equation of the first order that is obtainable from the given equation of the second order by an integration, and that involves a n arbitrary constant, is called a first integral. W h e n a first integral c a n be found, the problem of integrating the original equation is reduced to that of integrating the first integral. F o r that, the methods and observations of the preceding chapters can be drawn upon. W e shall consider a few types of difi"erential equations 7.1 for which first integrals can be found. 114
Equations in Which Either ^ or ;c are not Present
115
7.2. Equations in which either / or x are not present A differential equation 7.1 in which y is not present, namely, F{x, y\y") = 0, is i n effect F(x, y\ dy'/dx) = 0, that is, it is a differential equation of the first order for^'. A n integration of the equation is therefore possible and yields a first integral f{x, y\ c) = 0. Example 1.
T o integrate the differential equation \x + 2]y''-\-y'^
-
1 - 0 .
As a differential equation for^' the given equation has separable variables. A n integration gives
y + 1
that is,
>» = 1 +
T h i s is a first integral. integral
{x-\-2]'-ci
B y a quadrature of the last equation the general
[x -\- 2 + ci
+ C2
is obtained. A differential equation 7.1 in which x is not present can be treated quite similarly. Since dy dy' dx
dy
,dy' dy
the equation Fiy, y\y") = 0 is, i n effect, y\y' dy'/dy) = 0. This is a differential equation of the first order for^' as a function of;'. A n integration of it as such yields a first integral f{y, y', c) = 0. Example 2.
T o integrate the differential equation y"2
y
yy" y
By setting^" = y' dy'/dy, the form of this equation is changed into dl dy
dy
This is a Clairaut equation 6.7 f o r ^ ' as a function oiy. It therefore has the first integral c\ + c i j — >' = 0, and the singular integral^' —i:)'^.
116
Solvable Equations of the Second Order
Integrations of these relations yield the general integral y and the further integral)» = 4/¡jt + f j .
= Í2Í^^' —
íi,
PROBLEMS Find the general integrals of the following differential equations 1. xy" + y - * -= 0.
2. xy" -
3. y" - /
4. yy" + {1 + 2 / | / 2 -
V l +
5. xy'' + xy''' + / 7.
= 0.
'
+
2yy' = 0.
y \y' = 0.
8. 1 + xy" - y' ~ y"^ = 0 . /
A\ - 0.
6. {;y + 1 b " + {> + 2 b ' 2 - 0.
cos ^ + sin ^ \y" + {2 cos y — y
10. y 3 _ ^ y y ' +
{1 + x'^Wy'
/
9. ¡1 "
- / ^ J - 4 / / = = 0.
= 0.
7.3. The differential equation of a 2-parameter family of curves. Checking a general solution A family of curves whose equation (7.2)
^{x,y,cuC2)
-
0
involves two arbitrary constants is called a two-parameter family. A n y curve of this family characterizes)' as a function)'(x), for which the expression ^{.x, y{x), c\y C2) is identically zero. T h e first and second derivatives of this expression are therefore also zero. Hence y{x) also fulfills the equations ^x{xyyy ci, C2) + ^y{x,y, c i , C2)y' = 0,
(7.3) *z.x(^,:y, ci, C2) + 24>x.v(^,>, cu C2)y' H- %,v{x,y,
€2)/' = 0,
that is, y \s a solution of simultaneous equations 7.2 and 7.3. T h e éliminant of ci and C2 from these three equations is the differential equation of family 7.2. T h e final step of elimination i n this finding of the differential equation can be avoided by arranging the several equations so that the differentiations themselves bring about the eliminations. Thus if equation 7.2 c a n be made explicit for one of the constants, that constant is eliminated by the differentiation, a n d does not appear i n the first equation 7.3. If that equation can then be made explicit for the remaining constant, the second differentiation completes the elimination. T h e final equation 7.3 is then the differential equation of the family. Example 3.
T o find the differential equation for the family of a l l circles having the radius 3.
The Differential Equation of a Family of Curves
117
T h e coordinate equation of this family is (7.4)
= 9,
\y-C2]'
and the derived equations 7.3 are thus
1
= 0.
{y-c2]y"
T h e éliminant of ci and cz from these three relations is the differential equation (7.5)
9/'2_ {l+y2j3^o
If equation 7.4 is first made explicit for C2, thus,
the first differentiation yields
V 9 -
{x-
ci\
If the last equation is now made explicit for ci, thus, x ± 3y'/y/l y'^ ci, the second differentiation yields, without any further step, the result
(1
+y'T
=1.
T h i s is, of course, the same as result 7.5. T h e graph of the general integral of a differential equation 7.1 is a twoparameter family of curves. T h e correctness of an integration of such a n equation m a y therefore by checked by using the integral obtained to re-derive the differential equation. Example 4.
T o check that y^ of the differential equation
—
cix^y
—
cz = 0 is the general integral
T h e derivative of the relation given is 2y/
-
ci{xy2xy}
= 0.
If this is first made explicit for c i , the second differentiation yields
118
Solvable Equations of the Second Order
T h e denominator may be dropped. Since the result is then the given differential equation, the questioned relation hcis been checked to be a n integral. PROBLEMS F i n d the differential equations of the following families of curves. 11. ylogy
=cix +C2.
13. y -
12. y =- {cix
+ C2\^.
15. ciy + Ci+
log \xy\ + 0.
14. y -cie'
+ C2X^ -
16.
2
17. log 19. 7 -
+/} X -
21. [xy]*
= ci t a n - i (y/x) + C2.
18.
+ ci\x - y\ + o,
y'(xo)
yoy
can be obtained as a power series, Polynomial approximations of the integral are thus also obtainable. T h e method is that of Section 5.3. W e seek an integral i n the form 7
= flo +
ailx
—
XQ] 4-
azlx
—
XQ]^ +
asix
—
XQ]^
+
•
#
*
Conditions 7.11 require that Jo = >o, and ai — yo'. T h e remaining coefficients are determinable by substituting the series (ory into the differential equation, a n d equating the coefficients of like powers of {x — xo]. Example
9.
T o find as a power series the integral of the differential
equation y"
=
, y
_
y
/2
Approximate Integral Curves
123
which fulfills the conditions y{0) = O.>'(0) = 1. Since XQ =
ya = 0, and yo' — 1, the series sought is i n this instance
T h e substitution of this series into the differential equation converts the equation into 2(i2 H" (ia-^ +
\7.a^x^ +
* ' '
=
— 1 — 4a%x
+
|1 -
6^3 + 4 ^ 2 ^ ! : ^ ^ +
•••.
T h e equating of coefficients of like powers of x yields the relations 202 = - 1 . 6fl3 =
—4(22)
12^4 = 1 - 6fl3 + 4a2^
T h u s ai = —^, ^3 =
• • • , and accordingly,
= —
PROBLEMS Find (up to terms in ;c^) the integrals of the following differential equations that fulfill the given conditions.
45. y" = xy' sin y,
46.
y{^) = 0, >'(0) -
2.
= jy + cos {xy), ).(0) - 0,y(0) = - 1 .
47. y =
+ /
+ 4, >(o) = o,y{o) = 0.
48. / ' = { / ^ x y p . 49. y = >' log:y + X .
:v(o) - o , y ( o ) = 2 . :yCO) = 1,>'(0) = 3.
50. y =. 1 + {jy - 2i^!'^2|^
^(oj ^ 2.y(0) = - 2 .
7.7. Approximate integral curves W e shall give two methods by which a graphical approximation to the integral of a differential equation 7.9 fulfilling conditions 7.11 may be obtained. 1. T h e differential equation may be replaced by a differential system 7.10. T h e relation _)'o' = fiixo, yoi ZQ) then determines ZQ. T h e polygonal graphs that approximate the^- and z of the resulting system METHOD
124
Solvable Equations of the Second Order
are then constructable upon the basis of any chosen set of abscissas x „ b y the method of Section 5.2. O f the two graphs so obtained, the one through the point (xo, yo) approximates the designated integral curve. Example 10.
T o construct a polygonal graph to approximate the integral of the equation / ' + { 2 x - h 2 ! y + l 2 x - l ) > = 0, for which
>(0) - 0,
jv'(O) = 2.
T h e given differential equation and conditions are replaceable by the system y = z, z' = - ) 2 x - l b - { 2 x - | - 2 l z , w i t h the conditions ^(0) = 0, ^(0) = 2. O n the basis of abscissas w i t h the differences {x.+i — x,j = i , the method of Section 5.2 yields for Yi and Z,- the calculated values of T a b l e 6. Table 6 1 T
0
1
8 T
5 T
1
3
7 T
9
2
1
6
Yi
0
0.5
0.75
0.88
0.91
0.89
0.83
0.76
0.68
0.59
0.51
Zi
2
1
0.54
0.14
-0.09
-0.23
-0.30
-0.34
-0.35
-0.34
-0.30
Yi'
2
1
0.54
0.14 - 0 . 0 9
-0.23
-0.30
-0.34
-0.35
-0.34
-0.30
Zi' - 4 y
0
-2.25
-1.62
-0.93
-0.55
-0.30
-0.16
-0.03
0.06
0.15
0.39
0.61
0.70
0.74
0.71
0.67
0.60
0.54
0.47
0.41
T h e polygonal graph Y thus obtained is shown i n F i g . 20. T h e actual solution is i n this instance^ = 2xè~'. Its values at the points x,- have been included i n the table for purposes of comparison, and the integral cur\^e^ is also shown i n F i g . 20. 2. A differential equation 7.9 associates a numerical value y" w i t h each triple of values x, y, and y'. It thus associates with these values a curvature K, given by the calculus formula METHOD
(7.12)
K
=
¡1 + y 2 | « '
and a radius of curvature,
(7.13)
p
=
fl
+y''\^ ft
Approximate Integral Curves
125
A graph approximating the integral curve through the point (xo, yo) with the slope ^^o' may therefore be constructed as follows: A t (xo, ^o) the normal to the integral curve has the slope — 1/yo'A l o n g this normal, with the direction of increasing y taken as positive, lay off po, the value obtained from (7.13) by assigning to x, y, and y' the values xo, yo, and yo'. This locates the center of curvature Co of the integral curve at (xo, ^o)- W i t h Co as center, draw a circular arc through (XQ, yo). Let Vi and Yi be the ordinate and slope of this arc at any suitable chosen
X
FIG.
20.
point xij where x\ > XQ. T h e construction may now be repeated with the point (xi, Yi) and the slope Yi i n place of the original ones, etc. T h e graph obtained is a chain of circular arcs which at each abscissa Xi advances i n a manner that fulfills the differential equation. I n a certain way it therefore approximates the integral curve. T h e approximation is best near the point (xo, yo), and the smaller the differences {xi+i — are taken, the better is the over-all approximation. Table 7
y
1
1
3
5 T
S
0.54
0.32
0.07 - 0 . 1 8
0.71
0.50
0.23 - 0 . 0 4
-0.55
-0.78
-0.96
-1.72
-3.36
-5.33
0
T
1
0.97
0.87
0.73
0.96
0.86
0 -0.26 -1.13
Yi Yi' Pi - 1
1
T
1
-1.07 -13.6
3
7 T
9
6
2
1
-0.42
-0.63
-0.80
-0.27
-0.62
-0.86
-1.09
-1.07
-1.07
-1.07
-0.99
-0.81
78.5
11.6
5.07
3.24
1.95
126
Solvable Equations of the Second Order
Example 11.
F o r the differential equation and conditions y+JV-O,
;-(0) = 1. / ( O ) = 0 .
the solution is ^ = cos x. This solution and its approximating g r a p h based upon the calculated values of T a b l e 7 are shown i n F i g . 21. F o r the
X
Fro. 21.
construction of this table the values of ¥{ and Y'i were determined by measurements of the figure, and the values of Pi were computed b y formula 7.13. ASSIGNMENTS 2. ConstrucE by Method 1 a graph to approximate the integral curve of the difTerential equation
/ ' + {2* + 2 } /
+ (2x -
l b - 0.
for which >(0) = 1,>'(0) = 1. 3. Construct by Method I a graph to approximate the integral curve of the differential equation I2x - x ^ i y for which
= hy'
¡2 ^ x^\y + J2 - 2xly = 0,
(0 = - A -
4. Construct by Method 2 a graph to approximate the integral curve of the differential equation for which ^(0) = l , y C O ) = 1.
Some Geometrical Applications
127
5, Construct by Method 2 a graph to approximate the integral curve of the differ ential equation /' for which >(0)
=
-T,/CO)
-
3>' + 2;- = 0,
"0-
7.8. Some geometrical applications W h e n a curve is specified i n terms of its curvature, it is in fact characterized by a differential equation of the second order. Its coordinate equation can then be found by integrating that differentiéd equation. Example 12.
T o find the equation of the curve whose curvature is always equal to the sine of its inclination, and that goes through the point ( — 1 , 0) with the slope 1. T h e inclination has the tangent^'. Its sine is therefore ) ' ' / \ / l +>'". Since the curvature is given by formula 7.12, the curve is one for which /'
y'
|i + y ' i ^
(1 4-/^1^^
This is a differential equation i n which ^ is not present, and which is therefore of the first order for^'. Its variables are separable. A first integral is thus log
1
= 2x + ci.
/2 +y
T h e condition y* = 1, when x = — 1 , is fulfilled by W i t h ci thus evaluated, the first integral yields
= 2 — log 2.
e'+' dy = —.
—
dx.
and by a quadrature of the last equation y = sin~^ {tf*'*"VV^} + a. W i t h C2 determined so that y — 0, when x = — 1, the equation of the curve sought is found to be ^ = sin ^
TT
V5
PROBLEMS 51. Find the differential equation of the curves whose centers of curvature are always on the x-axis. Integrate this differential equation, 52. Find the differential equation of the curves whose centers of curvature are always on the^-axb. Integrate this differential equation.
128
Solvable Equations of the Second Order
53. Find the equation of the curve that goes through the point (0, 3) with the slope zero, and whose curvature is always equal to twice the cosine of its inclination, 54- Find the equation of the curve that is tangent to the ;r-axis at the point (1, 0), and for which the product of the slope by the curvature has the constant value — 1 , 55, Find the equation of the curve along whose normal the center of curvature and the jr-axis are equidistant from the curve, and are on opposite sides of the curve. 56, Find the equation of the curve that goes through the point , 0) with the slope •ff, and along whose normal the center of curvature and the ^--axis are equidistant from the curve, and are on opposite sides of the curve, 57, Find the equation of the curve that goes through the point (1, 0) with the slope zero, and for which the product of the curvature by the sine of the inclination has the constant value -j, 58- Find the equation of the curve that goes through the point (0, 1) with the slope 1, and for which the product of the curvature by the cosine of the inclination has the constant value —
7.9. Curves of pursuit W h e n an airplane P flying along a certain path is pursued by another airplane Q that is always directed toward it and that flies at k times its speed, the path of Q is a curve of pursuit. T h e equation of this curve is determinable as follows: Let the coordinates of Q be (x, y) and those of P be (x*, y*), as i n F i g . 5. T h e n because Q is always directed toward P we have y' = {y — y*)/{x — X * ) . T h i s relation and its differential give us the two equations {x~x*\y'
(7.14)
jx - x*| d/ -y'dx*
= =
[y-y*], -dy*.
T h e equation of the path of P and its differential, and the fact that the speed ds/dt is k times ds*/dt give us, further, the three equations f{x*,y*)
(7.15)
fAx^y*)
= 0,
dx* + / „ * ( x * , > * ) dy* = 0, Vdx^
+ dy^ = k Vdx*^
+ dy*\
F r o m equations 7.14 and 7.15 the four quantities x*, >•*, dx^, and dy* can (theoretically) be eliminated. T h e éliminant is the differential equation of the curve of pursuit. Example 13. A n airplane P flies northeast.
twice its speed.
A pursuing plane Q has T o find the equation of the pursuer's path.
Curves of Pursuit
129
If the origin is chosen on the path of P , equations 7.15 are y*
=
X*,
dy* =
dx*,
dx =
2V2
Vl
dx*.
T h e éliminant of x*, >*, dx*, and dy*, between these and equations 7.14 is -
1P Vl
+
-
= 2V2{y
x]y".
This is the differential equation of the path of Q. T o integrate this equation we set ^ — x = u, which transforms it into a'2 1)2 = 2 "N/Z UU", and this equation does not contain x. Hence, when u" is replaced by u' du'/du, the differential equation appears as an equation of the first order i n u' and u. T h e variables i n this are separable. A first integral is thus V2
+
2u'
+
1
+ V2
u
V~2
Vu
which can be put into the form ci Vu
rr - V i -
Vu
2(r 1
«' = 2 V 2
T h e general integral is accordingly 2ead sliding on it is in the position for which 9 — of the bead at any later time t
cos 9 .
at f » 0.
Find the coordinates
ANSWERS TO ODD^NUMBERED PROBLEMS \. y = \x^ + a log X + C2.
3. >
2 tan ^ \ci€^] + C2
5- > — log {log X -\- a] -\- C2y and y " c. 9. y=
7, y %in y = c\x + rj-
-
C\X + f2 - 1 ClX + C2
./2
11.
{1 + l o g ; - î / ' + — y
15. : r y V + / y
+
27. 31.
{1
+ 0.
= 0.
19. | ^ _ ; , i y ' _ ( y _ i j 2 23.
n. yy" - yy' -
= 0.
= o.
= O - ^ i l o g 11 -
X sin y + y sin x =
cix
21. 2 U + : y l / ' + {1 - / M i l + / i 25.
x].
Inexact.
1 29. ^ = - + C2 -
+ cj.
X
1^ -
[ x + c i l y e ' + c i .
33. > = log [x^ + cix + C2], 35. y = 37. y -
z=
{ci+
2xU-''.
CiX + C2 —
X'
.3
Ix + c i l s i n - ^ {x + C 1 Î + - \ / l -
z = ^ 4- sin
+^2,
+ i'll-
39. y' = x^ ~ yz, 43. :y' = log ^,
U +f,j2
z' = xj-.
2' = > + 2z.
41. /
z' - x ^
45. ;v = 2x + i x * +
4 «
+ /
-0,
Answers to Problems 47. y ~ 2x' -{-^x* -\- • • • . SI. yy"y'^
I -0;
49. ^ =- 1 + 3;t +
{x +
a]^-\-y
-
Ci
53. ^ = 3 — -J log cos 2x. 57. y
-
59. V l
{V('
-
+
65. x «> 4r, 67.
X
=
69. ^ = 71.
X
1 w
;
"
-
4)^^ -
-
2x)\\
cos-» (3 -
2M
^
63. y = 6 [ V 2
'
- i
= 2»"
+
61. v * i
-x)
1)(2
— log X -
2} -
_u
^—1-
4*1
^^'^ + 40
-l6/^ ?5
16
{12;
= 16( -
+3
V3|** -
8 sin (2/),
3.
y =
-8
«2.
+ 8 cos (2/)-
-
4)« -
32}
fx*
+
CHAPTER
8 The Linear DifFerential Equation with Constant Coefficients 8.1. The complete equation T h e differential equation (8.1) i n which a,
y"
+
ay' + i> =
or its derivatives, is called the linear differential equation of the second order because each of its terms is of, at most, the first degree in y and its derivatives. If wi{x) and W2{x) are, respectively, solutions of equations of this type w i t h the same coefficients a and by thus (8.2)
h, f{x)
do not involve
/(x),
y
wi'
+
awi
+
bwi
= fi{x)y
W2"
+
aw2
+
bw2
=
/2(jf),
we see, upon multiplying these equations respectively by any constants ai and 02 and adding them, that the combination (8.3)
y =
aiwiix)
-\- a2W2(x)
is a solution of equation 8.1 with (8.4)
/(;.)
=
aifi(x)
+
a2f2{x).
Clearly a corresponding statement can be made if there are more than two equations 8.2. Hence a differential equation 8.1 in which the function fix) consists of a number of terms, as in (8.4), can be integrated by integrating each of the simpler equations 8.2, and combining their solutions. Several important conclusions may be drawn from this observation. T o begin with, let us consider the differential equation (8.5)
u" + au' -\-bu
= Oy
which is obtainable from (8.1) by replacing/(x) by 0. If ui(Af) and are any two particular integreils of this equation, the combination (8.6)
U
ClUlix)
+
134
C2U2{x)y
U2{x)
The Complementary Function
135
w i t h arbitrar)' constants ci and C2 is also an integral. It is the general integral if the constants ci and are not replaceable by just a single arbitrary constant, that is, if the ratio of ui{x) to « 2 ( ^ ) is not a constant. W h e n differential equations 8.1 and 8.5 are considered together, (8.1) is generally referred to as the complete equation, and 8.5 as the reduced equation. T h e genera] integral 8.6 of the reduced equation is called the complementary function.
If i n equations 8.2 we take fi(x) = f{x), and fiix) = 0, we may choose for wi{x) any particular integral yi{x) of equation 8.1, and for W2{x) the general integral of equation 8.5. T h e combination (8.7)
y = yi{x)
+
ciUi{x) + czuzix)
is thus also an integral of equation 8.1. T h e general integral of a complete equation can therefore be found by finding a particular integral, and adding the complementary function to it. Thus the task of integrating a complete equation is resolvable into two parts, namely, the determination of the complementary function, and the determination of a particular integral. T h e simplest type of equation 8.1 is that in which the coefficients a and b are constants. Such an equation is always integrable by quadratures. T h e case i n which ¿ = 0 is especially simple. A quadrature can then be applied directly to obtain a first integral, and this first integral is a linear differential equation of the first order.
8.2. The complementary function One method of integrating the reduced equation 8.5 is already known to us. W e found, in Section 6.7, that by setting
this equation is transformed into the Riccati equation v' =
—b
— av — v^.
This Riccati equation, since it has constant coefficients, has a constant solution. B y the trial of y = m, it is found that this is a solution if m is a root of the equation (8.8)
m^ +
am -\- b =
0.
This is called the auxiliary equation. If its roots mi and mz arc distinct, we obtain i n this way two integrals «i = e^^' and «2 = e^^'. W i t h these integrals, formula 8.6 gives the general integral. If mo = m i , we obtain only one integral, «i = Í " * ' ^ in this way. However, the Riccati equation then also has the integral = m\ -\- \/x, and &2 leads to u^ = xe^^'. T h e
136
Linear Equations with Constant
Coefficients
complementary function is thus (8.9)
u = CKT"*!' H- C2f"'^'y
if
mu
u = citf"'* + C 2 « " ' * ,
if
m2 = m i .
W i t h this result once established, a procedure for recovering it quickly i n any instance is easily formulated. By substituting e"*' for u i n the given differential equation, we are led at once to auxiliary equation 8.8, the roots of which must be found. W i t h these roots, formulas 8.9 yield the general integral. T h e particular integral for which (8.10)
u{xo) = « 0 ,
U'(XQ)
uo'
=
is obtainable, as usual, by determining the constants ci and C2 i n (8.9) to fulfill these conditions. Example 1.
T o find the integral of the differential equation u" -
2u' -
2« = 0,
for which w(0) = 1, and u'(0) := - 1 . T h e substitution u = e"' converts the differential equation to f"'|m2 -
2m -
2)
=0.
B y striking out the exponential factor we obtain the auxiliary equation, the roots of which are m i and m2 = 1 — T h e general integral is thus B y differentiating the general integral we find that «' = ¡1 + \/3)tfitf!i+V3I*+ |1 - \ / 3 | c 2 ^ U - v ^ l « .
T h e conditions to be fulfilled are therefore ci-\-C2
=
These yield ci = ^ integral is therefore
1, —
{1
+
I/'N/S,
V3]ci
and
¿"2
+
{1
-
V3]C2
= i + i/'s/Sy
-
-1.
a n d the required
u = Example 2.
T o find the integral of the differential equation « " - 1 « ' + iftr« = 0,
for which u ( l ) = 0, and u'{\) = 2.
Complex Roots.
Euler's Formulas
T h e auxihary equation is i n this instance ~ - f ^ = 0. mi = f as a double root. T h e general integral is therefore u —
cye
137 It has
-\- Czxe
By differentiating the general integral we find that
T h e conditions are thus = 0,
{ci-\-C2]e^
F r o m these conditions, ci = —2e~^^ and integral is therefore u =
= 2.
Ihi-^^c^U^ C2
= 2e~^^. T h e required
2{x-\]e^^^'-^K
PROBLEMS Find for each of the following differential equations the integral that fulfills the respective conditions. 1. u" -
3u' + 2u = 0,
2. u" + u' - TU = 0,
u(0) -
9u = 0,
4. u " -
7u' +
5. u" -
4«' -
u = 0,
6.
2
a' + 2u = 0,
7. a " -
2u' -
80u = 0,
8. u" -
2 V ^ u ' + 3« = 0,
9. u " -
7u' + lOu = 0.
10. 9 « " -
6u' + u = 0 .
11. u" + 3u' -
u(l) = 1, u'(0 = 1= 0,
u =0,
12. 2u" + u' = 0,
1.
uCO) = 1, u'(0) = 0.
3. u " -
u" -
0, o'(0) -
u ( - l ) = 0, u ' ( - l ) = 2.
u(l) = 2, u'(l) =
-1.
u(0) = 1, u'(0) = 2.
uC-1) = 0, u ' ( - l ) = 0.
uCO) => 0, u'CO) = 0.
u(0) = 7, u'(0) = 14. u(3) = - 1 , u'C3)
u(0) = 2, u'(Q) =
4. -3.
u ( - 4 ) = 1, u ' ( - 4 ) = 1.
8.3. The case of complex roots.
Euler's formulas
Formulas 8.9 are valid for complex as well as for real values of mi a n d mi. However, they do not give the integral i n a real form when these roots are complex. Alternative real formulas are therefore desirable i n that case. If mi = a + 1^3 with /3 0, the R i c c a t i equation has a + as a complex solution. Its general integral, as given by formula 6.14, is then V
= a — 0
tan
{0x +
c].
138
Linear Equations with Constant Coefficients
F r o m this value of v we obtain the general integral of equation 8.5: w — - „ - « + I o 8 c o « ItfaH-cJ
T h i s result can be put into any one of the following forms: u — C2^"* cos {j8x + c i l ,
(8.11)
u = CiC^'sm
{/3x + c i | ,
sin
u —
+
C2 cos ^x\.
T h e first of these forms is obvious. T h e third is obtainable from the first by using the cosine addition formula, a n d renaming the arbitrary constants. T h e second is obtainable from the first by replacing c\ by ci — i r / 2 . W h i c h one of forms 8.11 is the most convenient to use w i l l ordinarily depend upon the circumstances of particular problems. Example 3.
T o find the integral of the differential equation u" -
6 « ' + 13« = 0,
for which u(0) = 2 and u'(0) = - 1 . T h e roots of the auxiliary equation are i n this instance 3 ± 2 i . T h u s a = 3 a n d j3 = 2. B y the third formula 8.11 the general integral is therefore u = e^'{ci sin 2x + ^2 cos 2x\.
T h i s general integral has the derivative u' = e^'{{3ci — 2c2) sin 2x + (3^2 + 2ci) cos 2x\.
T h e given conditions are accordingly fulfilled if ^2 = 2 and 3c2 -\- 2ci — 1. T h e required integral is thus u = e^'{ - i sin 2x + 2 cos 2x Example 4.
T o find the integral of the differential equation M" +
2u' +
4« = 0,
for which U(T/2) = 0 a n d U'(T/2) =- 1 . I n this instance the roots of the auxiliary equation are — 1 ± i/2. T h e first formula 8.11 therefore gives u = C 2 ? ~ ' c o s l(x/2) -\- ci]. This has the derivative w' = —cze *cos
+ ci
—
1
- f 2*
•
X
Complex Roots.
Euler's Formulas
139
and the conditions therefore take the form cos 5 + ,
C2e
cos
c^e
-
1
^ C2e
= 0,
72 sm
=
-1
T h e first of the last pair of equations is fulfilled if ci = 7r/4, and the second is then fulfilled if t:2 = 2e'^^. T h e required integral is therefore
2^4 Some important formulas of mathematical analysis follow from the results we have obtained. T h e integral of the equation u — 0, for w h i c h u(0) = 0 and u'(0) = 1, is found from (8.11) to be sin and from (8.9) to be
-
— e"*'].
Similarly, the integral for which u(0) = 1
and a'(0) — 0 is found to be both cos x and i\e*' + that (8.12)
1 sinx = - {^"-^-"1,
W e see thus
1 2
2i
I n their converse form these evaluations are (8.13)
e^' =
cos
X -\- i
sin
tx
x,
=_ cos
A: —
4 z•sm
x.
Relations 8.12 and 8.13 are called Euler^s formulas, i n honor of the Swiss mathematician Leonard Euler (1707-1783). PROBLEMS F i n d for each of the following differential equations the integral that fulfills the respccdve conditions.
13. u" - 4u' + 5u = 0,
u{0) = 4, u'(0) = 2.
14. u" + 2u' + lOu = 0,
u(l) = 0, o'(l) = 4.
15. u" - 3u' + V-u = 0,
u(0) = 1, u'(0) = 0.
16. u" - lu' + 2u - 0, 17. u" + 25u = 0, 18. u" + 2 y/l
u(2) = 1, u'(2) = 0.
u(3) = - 1 , u'(3) = 0.
u' + 4u = 0,
aI
VV2/ 19. « " + 2 V 3 «' + 4u = 0,
J = 0, u' (
)
V V i /
u(0) - 0, a'(0) = 0.
1
140
Linear Equations with Constant
20. u" -
12 log 2]u' +
21. u" -
4 « ' + 53« -
22. u " - 2 { 1 23. « " -
u
o(l) - 1, u'(l)
2.
0,
-
=
2ru' + 2^^*« = 0,
u(l) = 0. «'(1) =
«(1) = 1, «'(1) =
- 1 , u'
uC-3)
+ \ / 2 l u ' + | Y H - 2 \ / 2 | U =0,
24. u" + 17« = 0,
8.4.
|4 + log^ 2\u = 0,
Coefficients ^ 0.
= 0, u ' ( - 3 )
=0.
2T.
1.
Differential operators
W h e n a differentiation is denoted by D instead of by the less simple d/dxj and is used to denote the second derivative, such expressions as fp'{x) 4- ctp(x)j and = Example 6.
+ 1 + cie' + C2e^.
T o find the integral of the differential equation + 2y' + 10;r = i « + 2,
/'
or which y(0) = 1 and y'{Q) = 0. T h e roots of the auxiliary equation are — 1 ± 3i, therefore the given equation is \D-\-i
3i] ¡Z) +
-
1 + ?>i]y =
+
2.
T h e successive equations by which this equation can be integrated arc (Z) + 1 -
3i]Y
=
+ 2,
+
3i]y -
Y(x).
T h e first of these has an integral 2 2 — 3i
1
—
3i
and therewith the second has an integral y = - J ^ Í * -|- i. T h e complementary function is e~'lci cos 3x + f2 sin 3x|, T h e general integral is therefore y = + i + ^"'Ui cos 3x + cz sin 3x1.
Finding a Particular Integral
143
F r o m the general integral and its derivative we see that the conditions to be fulfilled are T^
+ i + '^i = 1,
3^2 = 0.
T V - ^ J +
These determine ci and C2, and the integral required is thus found to be > = TV«* + i + e-'{U Example 7.
cos 3x-\-U
sin 3x\.
T o find a particular integral of the differential equation iy = xe' + sin 2x + 5x.
y" -
I n place of this equation, we may consider the simpler equations 8.2, which i n this instance are wi"
~ iwi =
W2"
—
-1^2
~
sin 2x,
^z" — i«^3 = Sx.
These, respectively, have the integrals Wi — ^xe' — -V"^*>
1^2 = ~ A" sin 2x,
wz = — 20x.
Their sum, 4^ — 3
321 4 . —\ — — sm X 9I 17
— 2\Jx.
is a particular integral of the given diflferential equation. PROBLEMS Find the general integral of each of the following differential equations. 25. y" - y' -2y
= 3«^
26. y" ~ Ay = x^.
27. >" + > = Ae'.
28. y" + 2y' + 5y = 3. e~'
29. v " + 2y' + y =
31. y" -
2y' -
x+2
•
30. y " -
8> = 9**^ + l O * " ' .
2v' + 2y = sin x.
32. y" - y =^ - ; f + 2 —
+
x""
Find for each of the following differential equations the integral which fullills the respective conditions. 33. y" +2y 34. y" -
= 4,
y{Q) = 3 , / ( O ) = 1.
2y' •\-2y =
:y(0) = 0, ^-'(0) - 0.
144
Linear Equations with Constant Coefficients
35. y
+9> = 8sin:c,
36. y" + 2y' -f-
= -1,
> » 2 cos^- -
1 +
37. >" - h ' + h = 38.
+ 2;.' -
1.
2T
sin^'
= 1, / ( I )
+ 2 U ^ yiO) = 3. / ( O ) - 4,
+
>(0) - 0, /(O)
3;^ -= - 3 ; c - + x + 7.
= 2.
8.6. A second operational method A n alternative method by differential operators can be used whenever the roots of the auxiliary equation f re unequal. T h i s method is suggested by the following analogue from algebra. If k is any number not equal to either mi or m2, the value of ^ that fulfills the equation [k y =
is
mi\{k
-
m2\y
=/
{k — mi) (A — m 2 )
A resolution into partial fractions gives the last equation the form {//('"I - ' " 2 ) 1
y = -
Thus
3» = ^1 + > 2 »
{k -
k — m
+ ,
|//(m2-mi)j
k — m2
where
mi]yi
=
m2l>2 =
{k -
mi — m2
/ m2 — mi
Consider now the analogue of this procedure for differential operators T h e relations (8.23)
{/) - m i i j v i =
fix)
{D -
mi — m2
m2]y2
=
fix) m2
— mi
are linear differential equations of the first order. T h e results of operating upon equations 8.23, respectively, by \D — m2l a n d [D — m\] are the equations yi"
+ ayi' + hyi =
{fix)
-
m2f{x)\,
7fl\ — y2
+
ayi + ¿^'2 =
-1
l/'W
- m i / W ) .
T h e sum of the last two equations shows that \y\ + ^ 2 } is an integral of differential equation 8.1. W e can therefore find a n integral of equation
A Second Operational Method
145
8.1 by finding and adding integrals of the simpler equations 8.23. T h e formula obtained i n this way is (8.24)
y(x) =
/
J
Example 8.
-^^^^
¿-'"1'
dx +
/
mi — m2
J
e-"^^ dx. m^ — mi
T o find the general integral of the differential equation jy" + >' - 2V
T h e roots of the auxiliary Equations 8.23 are therefore
=
e'
1
equation are mi = 1, and m2 — — 2
3(1+^)
'
3(1-1-^}
T h e y have, respectively, the integrals y\{x) =
1
(
log ^ll
3
e'
y2{x) = - 7 + J ^ - ' - ^ ^ ^ l o g ¡1 + . ' 1 . 6
3
3
T h e sum of these is a particular integral of the given differential equation. A d d i t i o n of the complementary function yields the general integral > = i (T* log
\ +en
-\e'' 3
log{l + °'
+ ' 3
6
+ cie'
Example 9.
+
C2e~^''
T o find a particular integral of the differential equation y"
-ly'
^-ly
^ e sin^ x.
T h e roots of the auxiliary equation are 1 + i . Equations 8.23 are therefore \D-
—t
\ ~ i\yi = ^
^ sin^ X , 2 '
|/) -
'
I
1 + i\y2 ^ ^ • •"• 2
2 sin^ x.
T h e y have as solutions yi = >2
—
{— sin^ * + 2i sin x cos x + 2}, — sin^ X
— 2i
sin * cos j ; + 2 ) ,
from the sum of w h i c h , we obtain the desired particular integral y = h'{2
- sin^xj.
146
Linear Equations with Constant Coefficients PROBLEMS
Find the general integral of each of the following difTerential equations. 39. y" -
40. y" + 5 / + 6> = 12x».
8>' + 15> =
42. y" — y — sin x.
41. / ' +:y = 43. y" + ly' - 2y = e' + 45.
+4/
47. y" -
+ 13;- =
5>' -
=
44. :y" 46.
X.
-
2> = 3*^'
-
+ 4;- = cos 2x.
48. > " + > ' - 6> = X + «2x
7;> = 1.
49. y" - ly' -\-2y
51.y'-:y
r'.
x\
50. / ' + /
5
1
52. :v
1
-2> =
3^
4^
i l + x ^
8.7. A comparison of formulas Formulas 8.24 and 8.22 each give a particular integral of differential equation 8.1. But 8.24 does so by two independent quadratures, whereas 8.22 does so by two successive quadratures. Although i n any given instance the evaluation by one of these formulas may be simpler than that by the other, the content of the two formulas is the same. T h a t can be shown as follows: A n integration by parts applied to formula 8.22 changes its form into
m\
—
J
mi
—
m^
f{x)e-^^'dx
It is Ccisy to see that this is formula 8.24. It is sometimes desirable to choose for the indefinite quadratures i n formula 8.24 those that reduce to zero at a given point JCQ. T h e y are the definite quadratures from XQ to x. W i t h the variable of integration denoted by s they make the formula appear as lis)
y{x) =
Jo mi
ds -
—
Jxo mi — m2
ds.
Since X is now not the variable of integration, the factors and may be put under the integral signs. T h e formula can then be written compactly with a single integral sign as (8.25)
yM
=
fis) xo
^1
—
ffl2
T h i s gives the particular integral for which >'(xo) = 0 and>'(xo) = 0,
The Method of Undetermined Coefficients
147
ASSIGNMENTS 1. Consider formula 8.22 when nti the form (8.26)
yix) "
f'f{s){x
mu and show that it can then be reduced to
- .r}*'"'^^') ds,
if m2 = mi.
2, Consider formula 8.25 when the auxiliary equation has the complex roots a ± tß^ and show that it can then be reduced, by the use of Euler's formulas 8,13, to the form ,W -
(8.27)
äs.
8.8. The method of undetermined coefficients M a n y problems i n engineering and science depend upon differential equations of type 8.1 in w h i c h / ( x ) is a polynomial, an exponential, a sine or a cosine, or some product of such factors. F o r such an equation a n integral can be found without making any quadratures. Suppose f(x) is a polynomial of the degree n. W e shall then take y as a polynomial of the same degree (8.28)
yix"-^
y^g^^--\-
with undetermined coefficients. T h e substitution of (8.28) into the given differential equation converts the latter into a n equation between polynomials. B y equating the coefficients of like powers of x we obtain a set of relations from w h i c h the values of ^o, ?i> * ' ' » ?n can be successively found. Example 10.
T o find a n integral of the differential equation y" H- jy' -
6> = 2x^ +
-7x-\-2.
Substitution 8.28 with n = 3 converts the differential equation into the equation -6?6*» +
1 - 6 ^ 1 + 3qo\x^ +
1 -6q2 - f 2qi + 6qo\x +
{-6qz
+
?2 +
2qi] = 2x^ + Sx^ -
7x + 2.
By equating coefficients of like powers of x we obtain the relations -6qo
= 2,
—6^1 + 3^0 = 5, —6q2 + 2qi + 6qo = — 7 ,
— 6^3 4- 72
+ 2qi =
2.
148
Linear Equations with Constant Coefficients
F r o m these equations qo = — i ; qi — — 1 ; q2 =
and q^ =
—
Thus
is an integral. A n equation 8.1, i n which f{x) is of the form €*Yi(x), is transformed by the change of variable y = e^v into v" +
(a + 2k\v' + | i +
If is a polynomial, method above. Example 11.
+ k^\v =
can be determined as a polynomial by the
T o find an integral of the differential equation /'
e-'^^\\2x^ -
- W
9x -
V-l-
T h e change of variable y = e~'^'^v transforms the given equation into 2v' = 12JC^ -
v" -
9x
-
I n the transformed equation the coefficient h is zero. W e may therefore perform a quadrature upon the equation directly, to give it the form = 3x* -
v' -2v
\x^ -
^x.
If we now substitute for v a polynomial 8.25 with n = 4, we are led to the equations =
3,
4qo =
0,
-2qo -2qi
+
-2q2
+ 3^1 =
-2?3
+ 2?2 -f
-2?4
F r o m these
v = -h*
Hence
-
?3
=
= 0.
-
- e'^^H ~ix^
-
-f,
+ 2x + 1-
3x^ -
^x^2x-\-I],
is an integral. Example 12.
T o find an integral of the differential equation y"
1 x ^ - 1 ) ^ ^ + {3A:+
-2y'-^y=
4)^'.
A pair of equations 8.2 are i n this instance wi" W2"
-
2wi'
+
«;i -
e^^'lx^ -
2w2' -\- W2 = e'i^x
Ij,
-\- 4].
Undetermined Coefficients Continued
149
T h e respective changes of variables wi = e^vi and W2 — them into + 2oi' +
v{'
=
-
1,
transform
v^' = 3* + 4.
T h e first of these has an integral v\ = — 4x ~\r 5. T h e second, w h i c h can be solved by two quadratures, has an integral v% — ^x"^ + 2x^. By multiplying v\ and v^. by the appropriate eponentials to obtain w\ and KJ2, and then adding w\ and w^, we find the integral V =
-
4x +
51.2^ +
\kx^ +
2^2|^.
PROBLEMS Find, by the method of undetermined coefficients, an integral for each of the following differential equations. 53. / ' + / - ! > ' = 3*2 54. y" + / 55. y" -
-
6> = Jf.
ly' = - 3 5 * ^ + 76x^ -
56. y" -^-y = \2x 57. >" + y
-6y
58. y" + 2y' 59. y" - h ' 60. y" -y' 61. y" -
-
By = 9e^^. \20x^ + 24^2 + 5x -
iU^\
= 12*V^-.
3 / +|>
= 27«'^^^
6y = 2e^ + 3^' + 5JC + 1-
Ay' + 2> =
-2y=
65. 66. y" -
42x + 6.
= \x^ -\-\U'.
-\-h
64. / ' ^y-
lAx'^ -
2}*"^
- y=
62. y" + Sy' 63. / '
8x + 1.
123*2 _ 43^ _,_ \2\r'^^ +
\\^x^ + 4x + S}*** + + 24 + 3e^ +
2/ -
3> = - i4x + 4}^' -
\\\x +
19
{6x + Sjir^''.
4f-^ {8x -
2)«-' -
6^^' -
(5x -
6|f-K
8.9. The method of undetermined coefficients continued T h e method of undetermined coefficients can also be applied to an equation 8.1 i n w h i c h i s of the form /(x) = P i W sin
¡7:^
+ 5) +
^ 2 «
cos
(7X
+ 51,
i n which 7 and 5 are constants, and F\{x) and Pi{x) are polynomials. I n this instance we shall take y i n the form (8.29)
y - Qi(x) sin (7^: + 6) + Q2(x) cos ¡7^ + S j ,
Linear Equations with Constant Coefficients
150
w i t h Q i and Q2 as polynomials with undetermined coefficients
Q2W
= ?2,0X"
?2.1«""
+
+
•••+
?2.n.
T h e degree n shall be the larger of the degrees of P i and Pz, unless the instance is one i n which a = 0 and b = 7^. I n this exceptional instance n must be taken greater by 1. Substitution 8.29 converts the differential equation into an equation between sums of functions of distinctive types. B y equating the coefficients of like functions, we obtain relations from which the values of qtj can be determined. Example 13.
T o find an integral of the differential equation y"
+ 2>' + 5> = 3 cos
T
X
4
I n this instance n = 0, 7 = 1, and 8 = — T / 4 . Substitution 8.29 converts the given differential equation into the equation H^i.o
— 292.0) sin
X —
ir 4
{2qi,Q +
+
492.0I
T
cos
" - 4
3 cos ' * ~ ^ By equating the coefficients of like functions we obtain the relations 4?i.o — 2^2,0 = 0,
2^1,0 + 4^2.0 = 3.
These simultaneous equations give 71,0 — ^ 0 and 72,0 = %• , 3
T
X
+
4
Thus
T
-COS
Jf — 4
is an integral. Example 14. y"
+>'
T o find a n integral of the differential equation +
87
=
IIOJC^ +
21*
+
9!
sin
3x
+
x cos
3x.
I n this instance n = 2, 7 = 3, and 5 = 0. Substitution 8.29 with these values converts the differential equation into the equation l(-?i.o -
3^2.0)-«^ +
(-
qi.i
-
372,1 4- 271.0 -
1272,0)*
+
( - 7 1 . 2 - 372.2 H- 71.1 - 672,1 + 271,0)) sin 3x + {(371.0 - 72.o)x^ + (371.1 - 72.1 + 1271,0 + 272.o)x H- (371,2 — 72.2 + 671.1 + 72,1 + 272.0)) cos 3x =
\\0x^ + 21jr + 9| sin 3x + x cos
3x.
Undetermined Coefficients Continued
151
By equating the coefficients of like functions we obtain the relations — ?i,o —
3^2,0
= 10,
3?i.o -
?2,o
0;
— ? i . i — 3 ? 2 . i + 2yi.o — 12^2.0 = 21,
— 6^2,1 + 2yi,o = 9 ,
~ ? 1 . 2 — 3^2.2 + — ?2,2
391,2
+
6^1.1
+
?2.1
+
2^2,0
= 0-
T h e first pair of these relations is a simultaneous system which yields fji.o = — 1 and 9 2 , 0 = — 3 . Therewith the second pair yields ^ i . i = 7 and q2,i = 2. T h e third pair then yields 9 1 , 2 = — 1 3 and ^2,2 = ~ 1 T h e integral thus obtained is y =
l-x"^
Example 15.
-\.Jx
13j sin 3x +
-
{-2,x^ +
2x -
1 i cos
3x.
T o find an integral of the diff"erential equation _
^ e, 2 x
4y' -\-Sy
^) + 2 cos
sin(. +
I
A:
+
-
T h e change of variable y — e^^v transforms this diff'erential equation into v" -\-
V
= sm \x
+ 2 cos
~
'+5
Although the polynomials i n the right-hand member of the transformed equation are of the degree zero, we take n = 1, because this is an instance i n which a — 0 and b = 7^. B y the method described, we then find that 2?i,o ~ 2, and — 2 ^ 2 . 0 = N o relations are found for 91,1 and 9 2 , i T h e values of these coefficients are therefore arbitrary. W e may assign to them the value zero. Thus V
=
X
1
sin \x -\- -
- X
2
cos x-^^n
and the integral found is 2x
X
sm I X + ~ I
1 ~
-
X
cos
PROBLEMS Find, by the method of undetermined coefficients, an integral for each of the fol lowing difTcrcntial equations.
152
Linear Equations with Constant Coefficients
67. / ' + 3 / +7> 68. y" -
2)-' -
-
15 cos
3;- = sin X
X.
12 cos x.
-
IT
69. y" + 2;-' + 5/ -
3 sin
70. y" -
-*'cos«.
2/ + > -
71.
X
4
+ 2) sin X +
72.
-
4 / + 4^ = sin X -
73.
+ 6y' + 16> = sin 2x + cos
74. y
-
ix^ + 4x| cos
3/^'. 2JC
+ 4 sin 4x.
4y' + 5> =- «2"^ sin x
75. y" -\-Sy = - 1 5 ^ ' sin 3x -
29^"' cos x.
76. / ' -
2 / + 2> = e'{2 sin x + 4 cos x\ -
77. y
2 / + 6> = «*{4 sin x + sin 2x -
-
jf.
78. / ' + 3y' + 4;. = } Jx^ + x| sin 2x + 79. / ' -
3 / + 3> = x ' + ^'1 x^ -
80. / ' -
4 / -\-Sy = e'\{x'^ -
lOx cos 2x.
4 sin 3xi. \x^ + 2x2]
4xj sin X -
2x.
2x cos x|.
\2x + 1) sin 3x + ( - I B x * + 12x + 18) cos 3x1 + «2'{4sin X + 2 cos x|.
8.10. DifFerential equations of higher order than the second A l l the methods of this chapter are applicable to linear differential equations of higher order if these equations have constant coefficients. T h e general form of such a differential equation if it is of the reduced type is u^"l + aiKt"-^l + 0 2 « ' " " ' * ' +
• • • + a„u = 0,
with u'"^ standing for d^u/dx^. T o obtain the complementary function, we substitute e"^' for w, and thereby convert the differential equation into the auxiliary equation
For each root rrij of the auxiliary equation, the function e"^'' is a particular integral. If all the roots are distinct, the complementary function is
and involves the n arbitrary constants c i , (^2) * ' ' , ^n- F o r any root m that is multiple, say of the order r, the functions r " " , x^"**, • • •, x'^^^"" are particular integrals. If the coefficients of the differential equation are real, as we have been supposing, a complex root m = a + of the auxiliary equation occurs only i n conjunction with its conjugate ih =
Equations of Higher Order
153
a — i/3. B y the use of formulas 8.13, the integrals e"^' and e^' can be replaced by the real ones, e"' cos 0x and e"' sin jS.t. T h e complete linear equation of the order n is
and is expressible by the use of differential operators i n the form {D -
mi] ID -
m2\ - • • {D -
m^ly
=
f{x).
A particular integral of this equation is therefore obtainable by integrating successively the equations -milri =
f{x\
— 7722 Y2 =
Viix),
\D — tn^^Yz =
y2{x), »
{D-~m,]y
T h e method of undetermined and 8.9, can be applied without higher order if its function f{x) kind mentioned in Section 8.8. particular integral. Example 16,
=
coefficients, as described in Sections 8.8 modification to a differential equation of is made up of terms and factors of the This method can then be used to find a
T o find the general integral of the differential equation
T h e auxiliary equation is m^ -
2m^ + 6m=^ -
32m + 40 = 0.
Since this equation can be given the form [m -
2\^{m^ +
2m +
iO\ =
0,
it is seen to have m = 2 as a double root, and to have also the roots — 1 ± 3i. Thus the reduced equation has the integrals i " * and xe"^^, along with cos 3x and sin 3A:. T h e complementary function is u = cie'^ + c^xe'^ + Czé~^ cos "hx + c^e"'
sin 3A:.
T o find a particular integral of the complete equation we make the change of variable y = e^w. This transforms the given equation into ^ [ 4 ] _^ 2«^'^' + dw" -
22«;' + 13«J = 2x^ -
3x + 1.
154
Linear Equations with Constant Coefficients
T h e substitution w = qox^ + qix'^ + qzx + 73
is now found to fulfill this equation if qo = 2, qi = — 6, 72 = " " 3 , a n d 73 = 4. Thus y = tf*{2x^ - 6x^ - 3x + 41 is a particular integral. B y adding the complementary function, the general integral is obtained. PROBLEMS Find the general integral of each of the following differential equations. 81. u''> - 2u" - u' + 2u = 0.
82. «1^1 + u " -
83. u^^l + 6 u " + 5u' -
84. uf'I -
85.
-
86.
-
12u = 0.
2ul''J _ 9 „ " + 2u'
4«' -
4u = 0.
6 u " + 3ii' + lOo = 0,
+8u=0.
- 21u" + u' + 20u = 0.
87. a'^J + 2uf^] + l O u " + 18u' + 9u = 0, 88. uf^J -
2u'^' + 5u" -
89.
2u" -
8u' + 4u = 0. 90.
+ u" + 8u' -
91. u'^l - f 6 u " + 8u = 0.
92.
+ 6 u " + 5u = 0.
93.
94, y^^i + 3>" -y'
-
+ 2/' -
3u' + lOu = 0.
3>' =
X.
95. y [31
3;-" + 3^' - ;r = 5x -
97. y^^^ -
7>" + \9y' -
98.
3y" +y'
-
96. ^1^1 -
lOu == 0.
-3y
=* H
Sy" + 16>' = 4x + 2,
13)- = tf2'{2 cos x + 6 sin x}.
-\-5y = ^'(4 cos 3x + 33 sin 3x}.
99. y^*i + 4^f^l + 8;-" + Sy' + 4;» = 1. 100.
-
8>'31 + 26>" -
40;-' 4- 25> =• 5,
ANSWERS TO ODD-NUMBERED PROBLEMS 3. u = if2.='f--'> + * - ^ f ' - i M
I. » = 5. u = [(2 -
/2:
'\/2)x + l U " ^
9. u = i2 - x ^ 3x-9 13. u = f-'|4 cos
X —
7. w = 0. 11.
6 sin x}
15. u = ^''^-{cos 2x -
f s i n 2xj.
17. u = —cos [5x — 15
19. u a 0.
21. u = ^2^^^^' cos {7x - 7 } .
23. u = - Z i ' t ' ^ - ^ ' cos i^rx -
25. y =
27. y =
+ ,i,2r
2g'
ci
-
sin x + ¿2 cos x.
Answers to Problems 29. > -
155
U + 2 } « - ' log {x-\-2\ + {cix -h czU-'.
31. ^ -
+ C2f.-2*
2*-' +
-
2 + cos V 2 * + — ^ sin
33. jy
-N/2 35. ) r - sin
+ ^ cos 3x + 2 sin 3x.
X
37. > ^ x V + S i x + l l * ' ^ ^ 41. ^ = ^e' +
39. > = 43. y = e- —
X +
^2 cos x.
1.
+ f2*
45. y - T^x —
sin
+ e~^^\ei sin 3x + ci cos 3x)
47. ;y = 49-
y = ^x^ +
51, J- =
X
+ -s
e^{ci cos jf +
-
-1 + ^ log {1 +
sin
x\.
log |1 + ^} + cie' + C2fi~'.
- 4*2 _ 12.
55.
>
57. >
59.
> =
61. y
63.
53. ^ =
65. y 69. >
4x' + 24x +
Sin
X
—
T
1
4
2
+ 2x
71. >
cos
7 j sin
73- > = Tff{si" 2x -
{2x*-i-x^
=i
67, > = sin
+ 4*-'.
2x* + 3*^
=- X * -
-
+ \2x + \U bx
_ X +
2 cos
x.
4
X -
•
5x + 8} cos
X.
2 cos 4x1.
75. _y ^ {sin 3J: + 2 cos 3x)f* + [2 sin x — 5 cos x]e 77. >
5^(sin x + sin 2JC + sin 3jej.
79. y
ijjt' +
+ 4JC + 21 + ^{2 sin X +
81. u
cos x).
83- tf - cid^ + c»r-^^ + €^^^^
85. « * + C2^e * + ^3 cos 3x + Ci sin 3x,
87. a ™ 89.
.2x sin x. + C2«"'2x ^^ cos X + cz€^^
u =- ci^
91. u ^ ci cos 2x + ^2 sin 2x + f3 cos ^2x 93. :y -
- - i x ^ - |x + o + -:2^ + -:3^"^^
95. > - x^ + 97- > =
+ ^4 sin \ / 2 x .
X
sin
X +
3 + cie^ + ^2«^^ + c ^ x V . ^itf* +
C2fi^^ cos 2x + c^^^ sin 2x,
99, ^ = T + ci€~' cos X + c^fi"^ sin X + czx$~^ cos x + axe^^ sin x.
CHAPTER
9 Applications of Linear Differential Equations of the Second Order 9.1. Simple harmonic motion A particle moving on a straight line is said to be in simpU harmonic motion if its acceleration is constantly directed toward a fixed point of the line and is proportional to the distance (displacement) from that point. T h e fixed point is called the central point of the motion. T o be directed toward this point the acceleration must be negative when the displacement is positive, and vice versa. It is therefore a restoring acceleration. W h e n the origin is taken at the central point, and the displacement is denoted by the acceleration is d^y/dt^ and is proportional to y with a negative constant of proportionality, say —ß^. T h e differential equation for simple harmonic motion is thus (9.1)
d^dt^-^-ß'^y
-
0.
This is a differential equation of type 8.1, with / i n the place of x. Its auxiliary equation is + = 0, and has the complex roots +ßi. By (8.11), therefore, its general integral is given by any one of the formulas y = C2 sin {ßt H- ci\,
(9.2)
y = C2Cos {ßt + ci\, y — c\ sin ßt + C2 cos ßt.
T h e constants ci and cz can be determined to give the solution w h i c h fulfills the conditions (9.3)
:v(/o) =>o,
v{t) = vo,
w i t h any assigned values of /QJ >0J and VQ. It is clear from formulas 9.2, for instance from the first one, that, as / increases, y goes through a cycle of positive and negative values, and that 156
Simple Harmonic Motion
157
this cycle is repeated again and again. T h e motion is therefore oscillatory, that is, vibratory, and periodic. Since the largest and smallest values of a sine or a cosine are 1 and — 1 , the largest value of ^ given by either of the first two formulas 9.2 is \c2\. This m a x i m u m displacement is called the amplitude of the motion. T h e motion completes a cycle i n the time it takes the value [0t -\- ci\ to increase by lir. This interval, 2ir/0, is called the period of the motion. Thus Period = Irr/^.
(9.4)
T h e number of cycles completed in unit time is the reciprocal of the period. This is the frequency of the motion. Thus Frequency =
(9.5)
27r'
T h e character of the variation of y w i t h time / is shown by F i g . 23. indicated values cj and C2 are those of the first formula 9.2.
The
Fio. 23.
Example 1.
A particle is i n simple harmonic motion with y = "3, and y = 0, at / = 0. These same values apply to y and v again 3'^ second later, but not before then. T o find the formulas for y and v at any time. If the second formula 9.2 is used to describe the motion, the conditions given take the form C2 cos
Ci =
3,
—C2^
sin
ci =
0,
C2
cos
= 3.
T h e second of these three equations is fulfilled if ci = 0. T h e first is then fulfilled if C2 = 3, and the last if/3 = AT. T h e required formulas are thus = 3 cos 4ir/,
v =
— 127r sin Airt.
Applications
158 Example 2.
A particle in simple harmonic motion passes through the central point at ( = 1, and at intervals of second thereafter. W h e n it does so at / = 1, its velocity is 10, T o find the amplitude of the motion. T h e interval between passages through the central point is a half period. T h e period is thus % second; hence, by (9.4), /3 = 5ir, T h e first formula 9.2 therefore gives y = €2 sin I Sirt +
],
v ~ Sirc2 cos {Sirt +
1•
T h e conditions^(1) = 0, v{\) = 10 require that €2 sin {5ir + €i]
= 0,
T h e y are fulfilled by ci = — 5T, and
5ir^2 cos {Sir + ^ri) =
= 2/x.
y = (2/T) sin Sirj/ -
10.
T h e formula is thus
Ij.
and from this the amplitude is seen to be 2/T. PROBLEMS 1. A particle in simple harmonic motion with the period T has the position > « 2, and the velocity P = 4, at ( — 0, Find the amplitude of the motion, 2. A particle in simple harmonic motion with the amplitude 8 passes through the central point with the velocities ± 1 0 , F i n d the frequency of the motion, 3. A particle is in simple harmonic motion with a frequency of 10 cycles/sec. and the amplitude 7, What is its speed as it passes through the central point? 4. A particle is in simple harmonic motion. Its velocity as it passes the central point is twice its velocity as It passes the point y = Find the amplitude of the motion, 5. A particle is in simple harmonic motion with the amplitude 6 and the period x / 4 . With what velocity does it pass the point ^ = — 3? 6. A particle is in simple harmonic modon with the period 4. At f 0 its position is given b y ^ ^ and at / =» 1 it is given by = ^ i Find the formula for the position at any time. 7. A particle is in simple harmonic motion with the period p. When its position is given by J' = its velocity is ±.v\. What is the amplitude of the motion? 8. A simple harmonic motion has the frequency 10. At / = / i , the velocity is and ji second later the velocity is Find the formula for the velocity at any time, 9. The motion of a particle is along a straight line. Its acceleration is always directed away from the origin on this line, and is equal to the distance &om the origin. Find the general integral of the differential equation for this motion. 10. If in the motion described in problem 9, ^ = 2 , formula for the distance at any time?
and r = 2, at ^ =* 0, what is the
Free Vibrations
159
11. Ifj in the motion described in problem 9, the position and velocity of the particle at ( = 0 are y S and :^ = — 5, find the position which the particle approaches as a limit when the time increases Indefinitely12, If, in the motion described in problem 9, the position and velocity of the particle at f = 0 are 7 = 3 and v ^ 0, and at / = 1 the position is ^ = 5, find the formula for the position at any time.
9.2.
Free vibrations
M a n y mechanical systems include a heavy particle which vibrates with respect to some coordinate when it is displaced from its equilibrium position. W i t h the varying coordinate taken to be^, Newton's law of motion 3,2 and the relation 3,3 give - 5 = / ,
m
g dt-
where / is the force acting i n the jf-direction. T h e following discussions a, bj and c apply to such mechanical systems. a. A coil spring (or an elastic string or wire) resists stretching by a force T which is proportional to the amount by which it is already stretched. Thus, if its unstretched length is /o, its backpull when it is stretched to the length / is (9.7)
/o(,
T = k\l-
with some constant k. This value T is also the pull which must be applied to stretch the spring to the length /, for the stretching ceases when this pull just balances the resistance of the spring. If the spring hangs vertically from a point of support and a particle of the weight w is suspended from its lower end, it stretches to a length / i , where (9.8)
w -
k[h
-
/o|.
I n equilibrium, then, l\ is the length of the system. F r o m w and l\ the value of k is determinable through relation 9.8. T h a t is the value of k in (9.7). Consider the particle now when it is displaced by the amount^ from its equilibrium position, as shown in F i g . 24. T h e resultant of the upward force of the spring and the downward pull of gravity is T ~ w. B y (9.7) and (9.8), this resultant is k[l - h], that is, {w/{h - h)]\l - h\. Since li ~ I = y, the resultant is —wy/{lx — /o), and the equation of motion 9.6 is therefore g__
160
Applications
Because this equation is 9.1 with = g/{l\ ~ lo)j the vibration of the suspended particle is seen to be simple harmonic, with the equilibrium position as the central point. Example 3.
A certain spring, shown i n F i g . 24, stretches by 6 i n . under a pull of 4 lb. A 1-lb. particle is attached to it, and is released at / = 2, with the velocity 0, from a point 2 i n . above the equilibrium position. T o determine the formula for the position of the particle at any later time. I n the foot-pound-second system, equation 9.7 holds with 7" = 4, and {/ — lo] = M ' T h e spring is therefore one for which k = 8. F o r the S
1
I 1
•
y
i Fio. 24.
1-lb. particle the equilibrium position is, by (9.8), that i n which {/i — lo] = -ff. T h e equation of motion (taking g = 32) is thus d'^y/dt^ + 256;» = 0,
Its integral, for which y{2) = i , v{2) = 0 , is = i cos ¡16/ - 32}. F r o m this evaluation of y the amplitude of the motion is seen to be -B- ft., and the frequency is 8/ir cycles/sec. b. A body floating i n a liquid is acted upon by the usual downward pull of gravity, but also by an upward force called its buoyancy. This buoyancy is due to the supporting action of the liquid, and is equal to the weight of the liquid which the body momentarily displaces. T h e position i n which the body Hoats in equilibrium is that in which the buoyancy just balances the weight. If the body is displaced from the equilibrium position it bobs up and down with an oscillatory motion. For a body which is so shaped and ballasted that the amount of liquid it displaces is proportional to the depth s to w h i c h it is immersed, the buoyancy is given by a formula B = ksj with some constant k. T h e weight w and the depth so of immersion in equilibrium are thus connected by the equation w = ksQ. W h e n the body is displaced so that its depth of immersion is as in F i g . 25, the resultant upward force is B ~ w.
Free Vibrations
161
This is k\s — SQ], that is, {w/so){s — SQ]. It is therefore —{w/so)y, y = SQ — Sj and the equation of motion is thus d'y/dt^
=
-
if
(g/so)y.
This determines the motion of the body to be simple harmonic.
^
• —j
t
'
^ Fio. 25.
F r o m an analysis of Fig. 26, the differential equation for the motion of a simple pendulum of length / is found to be d'0
(9.9)
—
dt
g
+ ^ sin I
= 0,
where 6 represents the angle between the pendulum and the vertical. Equation 9.9 is not of form 9.1. T h e motion of a pendulum is therefore
FIO.
26.
not simple harmonic. However, it is nearly so if the angle of swing (measured in radians) is small, as it is, for instance, in a pendulum clock. T o show that the motion is nearly simple harmonic we draw from the third of equations 5.8 the relation » - sin ^ = — 3!
-
—
5!
+ i : 0^ - T T 5^ + • 7! 9!
É «
162
Applications
This is an evaluation of the change that is made if sin 9 is replaced by $. T h e relative change, namely, the ratio of {$ — sin 0] to $ itself is thus ^ (9.10) ^
^
e—
smB $
1 ^ = — ^2 3!
1 5!
. e*
42
I n (9.10) each quantity within a brace is positive when 0 ^ir. W e see, therefore, that the left-hand member of (9,10) is less than the first term on the right. T h e relative change made i n replacing sin 6 by $ is thus less than 6^/31 This is less than 0-5 per cent when \d\ < 0.175 (i.e., 10°), and less than 2 per cent when 6 < 0.35 (i,e-, 20^). W i t h only such relative errors, therefore, equation 9.9 is replaceable by the differential equation d^e g dr
I
T h e variation of 0 i n accordance with this is simple harmonicPROBLEMS 13. A coil spring ha5 the unstretched length 2 ft. When a 4-lb. particle hangs from it the equilibrium length is 4 ft. T h e particle is pressed down until the spring is 6 ft. long, and is released from that position, while it is at rest, at time i = 0. Find the formula for the position of the particle at any later time /, 14. A piece of steel wire 100 ft. long stretches by 6 in. when a 50-lb. particle is suspended from it. What is the frequency with which this suspended particle oscillates when displayed vertically from the equilibrium position? 15. A 1-ft. cube of wood is ballasted so that it floats with a horizontal base in water. In equilibrium it is just half immersed. If the block is raised 1 in. from the equilibrium position, and is there released while it is at rest, what are the amplitude and period of its resulting oscillation? 16. If the block described in problem 15 is pressed down until its upper face is at the water surface, and is then released while it is at rest at time f = 15, what is the formula for its velocity at any later time i? What b the highest position to which it will rise? 17. A simple pendulum swinging in a small arc without resistance completes the cycle of its motion in 1 sec. What is the length of this pendulum?
}^
18. A certain coil spring is stretched 1 in. by a pull of 10 lb. A particle of weight w lb. suspended from it oscillates with a frequency of 5 cycles/sec. What is w? 19. A buoy in the shape of a right circular cylinder is ballasted so that it floats with its bases horizontal. Its period of vertical oscillation is 1 sec. What is the depth of immersion at which it floats in equilibrium? 20. A simple pendulum swinging in a small arc without resistance passes through the vertical position at t ^ 2, It has at that instant an angular velocity {dB/di) ofta radians/sec. What is the formula for its position 6 at any later time /?
163
Damped Vibration
21. T h e length of a simple pendulum is ft. At ( = ti, the pendulum has the position 5 = "5 radian, and its angular velocity is zero. What is the value of 6 at the instant % second later? 22. A pull of 5 lb. stretches a certain coil spring by 2 ft. A particle of weight 5 lb. hangs from it in equilibrium, and while in this state it is given a hammer blow which gives it a velocity of 6 ft./scc. Find the formula for the position of the particle / sec. later.
9.3. Damped vibration W h e n a mechanical system whose free vibrations would be simple harmonic moves in the presence of a force which resists the motion, the motion is said to be damped. If the resistance is proportional to the velocity it appears i n the right-hand member of the equation 9.6 as a term —rv, namely, as —r dy/dt, w i t h some positive constant r. W e shall call r the coefficient of resistance. If the restoring force of the system is —ky^ the equation of motion 9.6 is (9.11)
-•—2=
-ky
g dt'
r
'
dt
which is d^y
(9.12)
dy
^ 2 +^^^
+ ^^ = 0^
with (9.13)
w
w
T h e general integral of equation 9.12 is given by formula 8.9 or 8.11, w i t h ^ and / in the place of u and x. I n these formulas mi and m2 are the roots of auxiliary equation 8.8, and a and /3 are the real and imaginary parts of m i . As we shall see, the character of the motion depends upon the relative sizes of a and 6, that is, of r and k. CASE
(9.14)
1.
a^ < 4b.
I n this case the roots mi and
m2
are complex, with
/3 = i -N/46 - a^.
a = -a/2,
The general integral of the equation of motion is therefore given by any one of formulas 8.11, that is, by y = ae"'
(9.15)
cos {^t
y = ca^'^'sin
+
+
y = e"^{ci sin ^t +
C2 cos
164
Applications
T h e first of these formulas shows, through the factor cos {^t + ¿^1!, that the motion is oscillatory with the frequency ^/2ir. It has i n the place of an amplitude constant the factor cj^"'. This decreases as t increases, since a is negative. T h e motion is therefore a vibration w i t h a fixed frequency, but w i t h an amplitude that diminishes toward zero with incresising time. T h e graph of such a motion is shown i n F i g . 27. T h e exponential tf"', that is, is called the damping factor. A n increase i n r produces a n increase i n a and a decrease i n ^. Increasing the resistance therefore hastens the damping, and decreases the frequency of the motion.
Fio. 27. Example 4.
A certain coil spring is stretched 1 ft. by a pull of 3 f l b . A 5-lb. particle is suspended from it, and is set to vibrating. T h e air resists the motion w i t h a coefficient of resistance ^. T o find the frequency of the motion. T h e value of k for this spring is T h e equation of motion is accordingly _5^^__29 ndt^
~
~
8
_ 1 _ ^ ~ Sdt'
T h e general integral of this equation is y = ^26-2^/5 cos {^t + Ci]. F r o m this formula for> the frequency of the motion is seen to be 12/5T. CASE
2.
a^ ^ 46.
I n this case the roots of the auxiliary equation are
Damped Vibration
165
T h e y are, therefore, real and negative. Hence (9.16) V =
micif"*!* + m^2t^^\
if a > 4Ô,
and (9.17) V ^
{C2 + mici + miC2/)^'"*',
if a'^ = 4b.
Since a l l the exponentials i n these formulas decrease toward zero as a limit when t increases, the values of y and v are damped out. N o right-hand member of a formula 9.16 or 9.17 vanishes for more than one value of T h e moving particle therefore passes through its equilibrium position once at most, a n d changes its direction once at most. T h e motion is clearly non-oscillatory. Example 5.
A certain coil spring is stretched 8 i n . by a pull of 1 l b . A 1-lb, particle is suspended from it, and the system is immersed i n oil for which the coefficient of resistance is -y. T o determine the motion for w h i c h ) ' = 1, and v = 0, when ( — 0. T h e equation of motion (with g — 32) is
Its general integral \&y = given conditions is
\ d^y
3
1 dy
32 dt^
2^
2dt
-1- C2e~^^\
y = "^e
and the integral that fulfills the
— se
I n the right-hand member of this formula the first term is obviously always larger than the second when / > 0. T h e y both approach zero as t increases. T h e moving particle thus subsides, without any change i n the sign of y, to the equilibrium position. Example 6.
T o determine the motion of the system of example 5, for which)* = 3, and v = — 4 4 , when / = 0. T h e integral of the equation of motion that fulfills these conditions is y -
-e-*'
+ 4e-'^',
v = Ae-'' -
48^-*^'.
T h e moving particle passes through the equilibrium position )» = 0, i n this instance when ~e~^^ + 4^~'^' — 0, that is, when t = ^]og2. It changes its direction when Ae~^ — 48^^'^' = 0, that is, when / = ^ log 12. Thereafter it subsides to the equilibrium position.
Applications
166
PROBLEMS Solve the following problems, assuming the units to be feet, pounds, and seconds. Take g - 3223. A certain coil spring is stretched 2 ft. by a pull of 1 lb. For a 1-lb. particle suspended from it the coefficient of resistance is x V F i ^ d the formula for ^, if > 3, and p = 0, when / = 0, 24. For the system of problem 23, find the time when when ( = 0,
= 0, if;' = 0, and v = \2,
25. A certain coil spring is stretched 3 in. by a pull of 1 lb. A 2-lb. particle is suspended from it, and for this the coefficient of resistance is 1. Find the formula for i[y(0) = ^'o, and v(0) — VQ. 26. A certain coil spring has a 3-lb. particle suspended from it. This stretches it ^ ft. T h e coefficient of resistance is x- Find thcformula for y^ if >(0) = 0 and v{0) = 7. 27. For the system of problem 26, find the formula for v if >(0)
3 and r(0) = 2,
28. For the system of problem 26, find the frequency of the motion, and find what the frequency would be if the resistance were not present. 29. Derive the differential equation for the approximate motion of a simple pendulum swinging through a small arc, if there is a resistance of which the coefficient is r. 30. A simple pendulum is 32 ft. long, and has a bob weighing 4 lb.
What Is the
smallest coefficient of resistance for which this pendulum will not oscillate? 31. A simple pendulum has a bob weighing 1 lb. It is to swing in a small arc with a coefficient of resistance xtIts frequency is to be -y. What must its length be? 32. A particle moves on a horizontal line against a resisting force which is proportional to its velocity. N o other force acts upon it. Find its position at any time if y ^ yoi and v ^ VQ, ahen ( ^ 0.
9.4. Forced vibration without clamping.
Resonance
A mechanical system i n w h i c h a l l the forces are internal, as i n those considered i n Sections 9.2 a n d 9.3, is called a free system. Its motions are said to be free motions. A system upon w h i c h an outside force acts is called a forced system. W h e n a force /(/) acts upon a system w h i c h i n the free state obeys equation 9,11, the dilTerential equation for the resulting forced motions is g dt
dt
W i t h a and b determined by relations 9.13, this equation is (9.18)
g
at
+ a
dt
J+
w
Forced Vibration without Damping.
Resonance
167
This is a complete differential equation. It can therefore be solved by finding a particular integral and adding the complementar)' function to it. T o find a particular integral, any one of the methods of Sections 8.5, 8.6, 8.8, or 8.9 may be used. T h e case in which j{t) is a pulsating force which varies harmonically in accordance with a formula (9.19)
/(/) = Ksin \yt + 5}
is of special interest. F o r a system without damping, whose free motion would accord w i t h equation 9.1, the equation of motion is then (9.20)
-J+^^y dt
=
w
~sm{yt-{-d
T h e character of this motion depends, as we shall see, upon the relative values of/3 and 7, and especially upon whether or not ^" and 7^ are equal. 1. 7^ jS^. A particular integral of the differential equation (9.20) , may be found i n this case, by the method of undetermined coefficients of Section 8.9, to be CASE
T h e general integral is thus (9.21)
y =
^ s i n ¡7^ + 51 +
sin {^t + d}.
For each possible motion, c\ and have specific values. A motion 9.21, for which C2 = 0, is simple harmonic. It has the same frequency and phase as the impressed force, but ordinarily it has a diff'erent amplitude. A motion for which C2 7^ 0, on the other hand, is compounded of two motions with different frequencies. It is therefore less simple. If the two frequencies arc commensurable, namely, if ^/y is a rational number, the motion is periodic and executes a certain pattern i n the course of a period. T h e pattern is, however, more intricate than that of simple harmonic motion. Figure 28 shows the graph of such a motion for which ^/y = 4. T h e period of this motion is 27r/7. A motion for which 5^ 0, and ^ and 7 are incommensurable, is not periodic but has an aspect of being erratic. However, even such a motion is bounded, for neither of the sine functions in formula 9.21 takes on a value numerically greater than 1. Hence l^^l never exceeds the value
168
Applications
T h e range of values w h i c h ^ may take on is, however, large when {/3^ — 7^) is small. 2. 7^ == RESONANCE. A particular integral of differential equation 9.20 is in this case found to be CASE
2ßw
cos {ßt + a).
and the general integral is accordingly (9.22)
y
=
~gKt
2ßw
cos \ßt + 51 + C2 sin {ßt + Cl]
T h e motions which accord with this formula are quite dissimilar from those of Case 1, for the multiplier of the cosine term in (9.22) contains / as a factor. T h e amplitude of the motion therefore increases indefinitely as
Fro. 28. t increases, which is to say that the motion becomes increasingly violent. A n actual mechanical system is i n this case pushed to the bounds of its physical limitations, the point where it must break down. This phenomenon is called resonance. It results when the impressed force is of the same frequency as the free vibrations of the system. A marching army is said to break step i n crossing a bridge to forestall such strains as may result from actual or approximate resonance between the frequency of the footbeat and that of the free vibrations of the bridge. T h e application of a vibrating force 9.19 to a particle suspended from a spring, as i n F i g . 24, affords a simple instance of a system in forced vibration. A way i n which this can be realized practically is to make the point of support S of the spring itself move vertically in a simple harmonic manner. T h u s if S is raised by the amount F, the length of the spring w i l l be
Forced Vibration with Damping { y + / i — y]then be
169
T h e equation of motion for the suspended particle will
w at
¿1 — ' 0
If y = A sin ¡7^ + 5 j , the previous equation becomes
0
This is clearly of form 9-20, PROBLEMS Solve the following problems, assuming the units to be feet, pounds, and sccondsTake g = 32, 33, A particle of weight 2 is subject to a restoring force which is numerically equal to of its displacement. It has also the outside force / = sin 2l impressed upon it. Find the motion for which ^ = 0, and v = -^^ when / = 0. 34- Find the motion of the particle in problem 33 for which)' = ^-^andv
= 0, when
/ = 3x/4. 35, A particle of weight 32 is subject to a restoring force which is numerically equal to 9 times its displacement. It has also the outside force / = 2 cos 3f impressed upon it. Find the motion for which ^ = 2, and = 0, when t = 036, A particle subject to a restoring force vibrates freely with a frequency v greater than 1, When the outside force / = cos / is impressed upon it, the motion for which y — yu and v = 0, when ( = 0, is simple harmonic. Find the value of ^ i , 37, Find the general integral of the difTcrential equation fy , „ . . . . . 7 - ~ + /37 = 2 sin 7' + 5 sin ^ I,
(9.23)
assuming that the motions arc not those of resonance. 38, Show that there are two values of y for which the motions defmed by differential equation 9.23 are resonant. Find a particular integral of the differential equation for each of these values of 7.
9.5. Forced vibration with damping W h e n a damping resistance is present i n a forced vibrating system the law of motion is given by (9.18) with (9.19), that is, by the differential equation (9.24)
^
at
+ ^ ^ 4. at
= ^ sin ¡7/ + Ô|, w
170
Applications
with a > 0, T h e roots of the auxiliary equation are then either real and negative, or complex with a negative real part. T h e complementary function therefore approaches zero as a limit when t increases indefinitely. Every motion of the system accordingly subsides into the motion that is represented by the particular integral. This is called the steady-state motion of the forced system. As it is found by the method of undetermined coefficients, its formula is y =
rl2
,
-
2 2>
-y') sin (T¿ + 5) -
cos (7¿ +
5)1.
W e can apply to this formula the general trigonometric relation (9.25)
^ sin i> ±
fíeos
=
+
fí^sin
± t a n - ^ (5/^4)1,
to give it the form (9.26)
= ^ s i n | 7 i + ( 5 - 61)!,
with (9.27)
K =
,
^'^
=y
« ; V ( e - T ' ) ' + «'7'
5i = t a n ' ^ ^--^
F o r m u l a 9.26 shows that the steady-state motion is simple harmonic, w i t h the same frequency as the applied force. T h e motion is, however, not i n phase with the force, for, by (9.19) / = 0, when / = (nx — 5)/7, w i t h n any interger; whereas, by (9.26) = 0, when / = \m — (6 — 5i)l/7, that is, at times w h i c h are later by the amount 61/7. T h e amplitude K of motion 9.26, though it is given by 9.27, can also be put into the alternative form (9.28)
K = w
If — ^ 0, the value of K always decreases when 7 is increased, that is, when the frequency of the impressed force is raised. O n the other hand, if — 6 < 0, the value of K first increases to the m a x i m u m (9.29)
-
^'^
w
and then decreases. T h e m a x i m u m is attained when 7 = V 6 — For this value of 7 the system is said to be in resonance. T h e resonance
Forced Vibration with Damping
171
amplitude 9.29 is evidently large if a is small. Unless sufficient damping is present, therefore, the breakdown of a mechanical system at resonance results. T h e frequency at w h i c h an applied force is i n resonance w i t h a damped system is 1
(9.30)
^TZl.
This is somewhat less than the frequency of the free vibration of the systern, w h i c h is (1/2T) '\^b — PROBLEMS 39. F i n d the formula for y in the steady-state motion given by the differential equation dh
dy + 2 ^ + 10, = cos 2/.
40. Find the formula for y in the steady-state motion given by the difTercntial equation + 2-~ + II7 = sin yt. dt^ • -dt if 7 has a value for which the system is in resonance. 41. Find the general integral of the differential equation
^ + 3 -
+
2,=sm
and put it into the form that displays the phase difference 2i between the force and the steady-state component of the motion. 42. Find the formula for y in the steady*state motion given by the differential equation
when y has such a value that the phase difference b\ between the force and the motion has the value ir/2,
43. Find the general integral of the differential equation
Observe that all the motions here represented subside completely-
172
Applications
44. Find the general integral of the differential equation ^ at with k > 0. vibration.
+
sin ßt,
=
Observe that as / increases all the motions subside into a simple harmooic
9.6. The forced vibration of a system with a repelling force If a particle that is subject to a central repelling force that is proportional to the distance is acted upon by an outside force 9.19, the differential equation of motion is again of form 9.24. However, the value of è is now negative. T h e auxiliary equation therefore has real roots, and of these one is positive and one is negative. A particular integral is again given by
Fio.
formulas 9.26 and 9.27. formula (9.31)
y
=
29.
T h e motions of the particle are thus given by the
Ksm{yt-\-
8 -
5i\
-\- cie""^* +
Cie""^'.
N o w one of the exponentials i n this formula increases w i t h t. A n y motion whose formula 9.31 includes that exponential is therefore unbounded. U n d e r it the particle flies off in the positive or the negative direction. T h e motions whose formulas 9.31 do not include the increasing exponential are different. They subside into the simple harmonic steady-state motion given by (9.26). A simple example of a mechanical system of this type follows. A smooth straight rod is mounted to rotate i n a vertical plane about one of its points 0. It carries a heavy particle which is free to slide eilong i t without friction or resistance. T h e system is shown i n F i g . 29, where y denotes the angular velocity of the rod in radians per second. T h e rotation of the r o d generates a centrifugal force upon the particle, w h i c h amounts to {ivy'/g)yj and repels it from 0. Gravity is a n outside force
Motion Under an Intermittent Force whose component along the rod \s —w sin {yt + 6]. motion of the particle along the rod is therefore ID d\
17.3
T h e equation of
wy^ y ~ w sva. [yt -\- h\J g
gdt'
that is, (9.32)
= - 5 s i n ( 7 / + 5!.
dt
Some of the motions defined by this equation are unbounded. I n them the particle flies off from one or the other end of the rod. T h e other motions subside into the simple harmonic steady state (9.33)
y -
i
- % s i n [yt ^-h]
2y
T h e polar coordinate equation r — ¿4 sin ^ represents the circle with its lowest point at the pole and with the radius {}'i)A. Equation 9.33 isof this form. I n the steady-state motion the particle therefore describes a circle in the vertical plane, the lowest point of the circle being at 0, and its radius being g/Ay \ PROBLEMS 45. Find the formula for^ in the motion defined by equation 9.32, for which ^ = 0, and 0 = 0, when ( »» 0. 46. Find the formula for D in the motion defined by equation 9.32, for which y — g/3y^, and v = g/^y^ when / = 0. 47. Find the formula for y in the general motion defined by the differential equation (9.34)
+ 6
-
16> = - 3 2 sin 4t
48. Determine what the relation between ^(0) and v{0) must be in order that the motion defined by differential equation 9.34 may subside into the steady state. 49. Find the formula for y in the general motion defined by a differential equation 9.24, in which b ^ 0 and a > 0. Observe that every motion in this case subsides into a simple harmonic steady state, but that these do not all have the same central point. 50. Find the formula for y in the general motion defined by a differential equation 9.24, in which 6 0 and a » 0. Observe that some motions are unbounded and the others are simple harmonic.
97.
Motion under on intermittent force
A force /(() that must be described by different formulas i n different parts of the total time interval is said to be intermittent. A particular
174
Applications
instance of an intermittent force is one which acts during only a part of the time. T o such a force may be assigned the value zero while it does not act. A n intermittent force may well be discontinuous. E v e n when / is discontinuous, however, formulas 8.25, 8.26, a n d 8.27 give an integral for diflferential equation 9.18, w h i c h is continuous and has a continuous derivative. T h e general integral is obtainable, as usual, by adding the complementary function. Example 7.
F o r a certain particle the differential equation of motion is
at
+
2 - + 5> = -fit), at w
w i t h fit) = e^ from / = 1 to / = 2, and otherwise /(/) = 0. T o find the integral of this equation for w h i c h ^(0) = 0 and p(0) = 6. T h e auxiliary equation has the roots — 1 ± 2i\ hence formula 8.27, w i t h X and fix) replaced by t and ig/w)f{t), gives a particular integral. I f we choose to = 0, since the conditions apply at that time, the formula is
2w Jo
N o w , because/(j) = 0 when j < 1, and when j > 2, this formula is more explicitly 0,
(9.35)
^
yp(t) =
\2w
when
0 g / ^ 1,
ds,
when
1 ^ / g 2,
e^'-^ sin 2it - s) ds,
when
( § 2.
sin 2it~~s)
j'^
^
2w
T h e general integral is y = ypiO
+
sin 2t + C2 cos
2t],
and gives p = ^
dt
+
e-^{i-ci
-
2c2) sin 2t +
(-^2
+
2ri) cos
2t].
Since yp(t) a n d its derivative are both 0 at / = 0, the conditions to be fulfilled are = 0, —C2 + 2ci = 6. T h e solution required is therefore y = >p(0 + 3(r~' sin 2t.
I n the instance of this example quadratures 9.35 for^p(/) can be made. T h e resulting formulas are
Motion Under an Intermittent Force 3tf~' sin 2/,
when 0 ^ Z g 1,
^2-'[sin 2(t -
-f- {g' -
Sw
175
1) + cos 2(/ -
1)11 + 3 2.
T o find a particular integral of the differential equation ^ - y - m .
i f / ( 0 = tjromt = OtoZ = l ; / ( 0 = l , f r o m Z = 1 toZ = 2; a n d / ( 0 = 0, when t > 2. F o r m u l a 8.25 is i n this instance VW = i
pis)
-
ds,
W-
- 2.
PROBLEMS 51. In the difTcrcntial equation
+ 2
I+
2> = /(,).
the function f(t) has the value e"^ from / = 0 to / ^ 1, and is otherwise zero. the formula for ^, if ^ «= 0 and y' = 0, when / = 0,
Find
176
Applications
52. In the difTcrential equadon dS
dy
the function /(/) has the value 1 from i 1 to / = 2, and is otherwise zero. There is a solution for which ^ =* ^, and ^' =» 1, when ( = 1, Find the formula for this solution when ( > 2. 53, Find a particular integral of the differential equation
dt2 if / ( O has the value sec / from ( = 0 to / = ir/4, and is otherwise zero, 54. In the differential equation d'^y
dy
+ i :-
dt^ ' dt
6> = / ( / ) ,
the function f{t) has the values
fit)
= { 0,
when
0 ^ r ^ 1,
when
1 < / < 2,
when
/ ^ 2,
e~^\ Find a solution of the equation.
55, T h e differential equation of problem 54 has a solution for which > =• 0, when J » 2, and which docs not increase indefinitely in numerical value as t increases. F i n d the formula for this solution when 1 ^ / ^ 2. 56. In the differential equation
2 + 3 T + tv=/W.
dt^ ^ ^ dt
the function /(/) has the value / when ( > 1, and is otherwise zero. T h e equation has a solution for which > = 0, when / = 0, and for which also y ^ 0 when t ^ 2. Find this solution.
9.8. Simple electric circuits W h e n an electric circuit has an electromotive force E impressed upon it, a current i flows and a charge q accumulates i n it. A simple circuit is diagrammed i n F i g . 30. Its elements arc a resistance an inductance L , and a capacity C. T h e values of these are constants. I n terms of an appropriate system of units (t i n seconds, E i n volts, i i n amperes, q i n coulombs, R i n ohms, L i n henrys, and C i n farads) the laws that apply are:
(9.36)
I =
do
Simple Electric Circuits
177
and di
(9.37)
g
^ 7 / + ^ ' + ¿ = ^'
These are derived i n the elementary theory of electric flow. If the value of i , as given by (9.36), is substituted into relation 9.37, the result is the equation 1
dq
(9.38)
This is the differential equation for the charge q. If, alternatively, equation 9.37 is differentiated and dq/dt is then replaced by its value 9.36, the result is the equation (9.39) This is the differential equation for the current i. Differential equations 9.38 and 9.39 are of the same form as equation 9.18 for the motion of a particle. T h i s reflects an analogy between the
WW R
L
30.
FIO.
phenomena of electric flow and those of mechanical motion, particularly between those of alternating currents and those of vibrating particles. Under an electromotive force E which pulsates in simple harmonic fashion, there may be a manifestation of resonance, of a steady-state flow, etc. Example 9.
F o r a certain electric circuit L = ^Vj ~ 4, and C = -r^ffT o find the formula for q,ifq = go, and i = to at Z = 0, and £ = 0, when t > 0. Differential equation 9.38 is i n this instance 1
do
25 dt 2 + 4 dt + 676? = 0.
178
Applications
Its general integral is q =
s i n l 2 0 / + o +1-0]/ +>o|^-''.
27. & _= ,-4(/3 |2 cos At -
29. - 2 + - + f e = 0. dt^ w dt I
31. / = 32/(T2 + 1).
33. y = - Y s i n 2/ + 11 sin /.
35. y = it sin 3/ + 2 cos 3/.
sin yt +
37. jy
39. )- =
41. > =
1
14 sin 4/j.
sin - ( + C2 sin {j3; + ci|.
2/ + t a n " ' I
sm
2 -\/l3
1
VlO
stn
/ -I- ' 4
—21
tan-'3
cos | i + Uit + c s l ^ - ^
43.
45. > =
g 4T
¡2 sin yt -
+ ^-T'l'
47. > = I sin |4f + tan-^ J} + CK^' + C2^-^'.
49. y =
y^ + -
51. y =
sm 7/ + 5 + tan when
cos ¿1,
^ '|cos (/ — 1) — cos t\y 53. y
-
—at 0 S / ^ 1,
when
i > 1-
M i n / + cos / log cos /,
when
/ ^T/4,
(ir/4) sin i — -ffOog 2) cos /,
when
i > 7r/4.
55, y = 59. i" = 5 sin 60/ +
57. q = TshslA sin 140/ + 3 cos 140/}
sin 10
-\/llr+
61. j" = 40/ cos 180/ + ci sin 180/ + 63. R ^ AO.
fa cos 10 -^/u
cos 180/.
t
CHAPTER
10 The Linear Equation with Variable Coefficients 10.1. Ordinary and singular points T h e general linear differential equation of the second order, i n its complete form, is (10.1)
po{x)y" -hPiix)/
+P2(x)y
=/«,
and i n its reduced form is (10.2)
Po{x)u" + piixW
+
p2{x)u =
0.
W e shall suppose that the functions po{x)j piix), and p2(x) have no common factor. W h e n such a factor is present, it shall be divided out of the equation. A point X at w h i c h (i) the coefficients poix), pi{x), and p2{x) are continuous and (ii) poix) ^
0
is called an ordinary point for the differential equation. A point which is not ordinary is called a singular point. Throughout this chapter we shall suppose that the differential equation is being considered upon a n x-interval that does not contain any singular point. As we have already observed i n Section 8.1, if ui{x)y and U2{x) are any integrals of equation 10.2, the linear combination {ciUi(x) -\-czU2{x)\ with arbitrary constant coefficients ci and C2 is also an integral. If this combination cannot be expressed with the use of only a single arbitrary constant, it is the general integral. F o r any equation 10.2, u{x) ^ 0 is a n integral. This, however, is rarely of any use, and is therefore called the trivial integral. Other integrals are non-trivial. If yi{x) is any particular integral of the complete equation 10.1, the general integral is obtainable from^i(j>f) by adding the complementary function. Aside from the equations with constant coefficients, there arc only a few minor classes of equations 10.1 or 10.2 that are solvable by methods other than that of power series. 180
The Euler Equation
181
10.2. The Euler equation T h e differential equation (10.3)
xol Y
\x -
+ a\x - xo\y' + by = f{x),
in w h i c h a, b, and xo are constants, is called the Euler equation. It has a singular point at XQ. Hence we suppose that ;i: > J:O, or J: < XQ. The change of variable
and
X
= xo -\- e",
X
=
—
XQ
e*,
if
X
if
x
XO, XQ,
transforms differential equation 10.3 into d^y , , + ¡a ds^ • '
,,dy 11 - f +by=f{xo ' ds
±
e')
This equation has constant coefficients and can therefore be integrated. Its integral, when expressed in terms of the original variable x, is an integral of (10.3). Example 1. \x -
T o find the general integral of the differential equation \ \Y
-
4{x -
l|y -
Uy
= x^ ~
3x2 +
3;^ _
8.
T h e change of variable x = \ + e' transforms the given equation into d^y ds^
-
^ dy
5-f " ds
14> = - 1 ± e^\
By the method of undetermined coefficients the general integral of the last equation is found to be y = s + sjie
+
cie
-\- cze
T h e integral i n terms of x is therefore > = i Example 2.
íVí:^ - 1 r +
- 1 r + C2\x ~~
1
r'.
T o find the integral of the differential equation x^u" -
3xu' + 13u = 0,
for which u(l) = — 1 , and u'(\) = 1. I n this instance xo = 0, and the imposed conditions apply at a point where x > 0. T h e appropriate change of variable is therefore x = e'. It transforms the equation into
182
Linear Equations, Variable Coefficients
F r o m this equation the general integral is found to be u — x^{ci sin (3 log x) + C2 cos (3 log x)\.
T h e assigned conditions are fulfilled by C2 = — 1 , and required integral is thus
= 1.
The
u = x^{sin log x^ — cos log x^]. PROBLEMS Find for each of the following differential equations the integral fulfilling the given conditions, or the general integral for the indicated values of x. 1. fjf + 2 | V ' -
20a -
2. \x + 2} V
-
A\x + 2 } « ' -
3. \x -
3} V
+
U -
4. |x -
5}V'+7{jf -
5. xY
-
2xy' +ly
6. x-y"
-
*y' + 5^ -
8. , Y
+^'+
0,
* > 0. 14« = 0,
^\u' + a = 0,
2i,
X < 3. x 0.
" cos X + 17. xy"
+
cos
{ J : ^ sin J : ) ; ' ' +
2 +
+
I. [x^ cos JT + a: sin J : ) ^ = cos x.
6
6JC
1 20. y"
Jf =
y =
4x
X,
0.
2)2 Ax^
{1 -
1 + log ^ X log
{x log x|-2 y = 2x.
X
21. u" + w' tan x -\- u sec" 2JC
,
^" ^ lT7^^'
AT
2|1 -
=
0.
x^\
_
11 + x 2 p ^ "
4x |1 -Tx^i-
10.4. The adjoint difFerentiol equation.
Integrating factors
After a differential equation 10.1 is multiplied by a function v{x), it is exact, by (10.4), if (10.5)
\pQv}" -
[pivY ^-Piv
=
0.
184
Linear Equations, Variable Coefficients
T h i s is a differential equation for f, which is, more explicitly, (10.6)
po{x)v" +
\2po'{x) -
Pi{x)W
+
Wix)
-pi'ix)
-\-p2{x)\v
= 0.
Equation 10.6 is called the adjoint of the given differential equation 10.1 or 10.2. A function v{x) is thus a n integrating factor for a given differential equation, if and only if it is an integral of the adjoint equation. T h e adjoint of equation 10.6 is i n turn found to be equation 10.2; thus each is the adjoint of the other. Example 4.
T o find the general integral of the differential equation \x^ -
x\/'
+
¡2*2 + 4;f -
3 1 / -h Sxy = 1.
T h e adjoint of this equation is {x"" -
x\v" -
\2x^ -\]v'
-\- |4x -
2]v = 0,
B y the trial of A:"*, this equation is found to have x^ as an integral. Thus x^ is an integrating factor of the given differential equation. T h e general integral, as found by the use of this factor, is 1
y =
1 7 — T +
- 5 + *^2«
PROBLEMS Find an integrating factor for each of the following differential equations, and b y the use of it find the general integral. 4x
23. u" +
Bx -S
7 «' + 2x - \
2x - \\
u - 0.
24. {3x-\-x-\u" + |12 + 4* - X 2^ I..' H' -
25. {x
-
26. u " 27. u" +
|4 + Sx\u - 0
1 X
u' tan x -
1
u{2 +
+x
tan x + tan^ x\ = 0.
— ~ u - 0. x{\ +x\
28. ¡2 sin J: — cos x\u" + j? sin x + 4 cos x\u' + lOu cos x = 0, 29. y
+
1 -
1
y
-
1
1
> -
30. / ' + I2x + 11/ + i2x + 2\y -
|1 - x\e-2JC
2*-^
The Existence of a Solution 10.5.
185
The existence and uniqueness of a solution. equation
The Riccati
It was observed in Section 6.7 that any solution of the R i c c a t i equation , • ^^^H
p2{x)
pl{x)
po{x)
po{x)
^
^^^^^^^^^^^^^^
2 ^?
yields, through the formula
an integral of differential equation 10.2, and vice versa. W e can use this fact to show that, when XQ is any ordinary point, differential equation 10.2 has an integral for which (10.7)
U{XQ)
=
Mo,
u'{xo) = ttu',
and that this integral is unique. If U(j ^ 0, the Riccati equation has an integral UO'/UQ. T h e formula u{x) = Wo exp /
T)(X),
for which
TJ{XO)
=
7i(x) dx
Jxo
gives a n integral of (10.2) which fulfills (10.7). If UQ = 0, let ai(x) and U2(x) be integrals for which «i(xo) = 1, «i'(xo) = 0, and «2(^0) = — 1 , "2'(^o) — Wo'have just shown that such integrals exist. T h e i r sum, u(x) = {ui{x) + "2(^)1, is a n integral that fulfills (10.7). Suppose now that two integrals fulfill any given set of conditions (10.7). Their difference is an integral C/(x), for which (10.8)
U{xo) = 0,
U'ixo) = 0.
Therefore both the integrals U(x) + u\{x) and «i(x) fulfill the conditions u{xo) = 1, u'{xo) = 0. N o w each of the functions + ui'(x) '—' — Lh{x) Uy{x)
V'(x)
+
^ and
ui'(x) U,{X)
is a n integral of the Riccati equation, which at XQ has the value zero. There is only one such integral; therefore the two functions are equal, that is, U(x) + ui(x) = kui{x). Since this equality applies at XQ we see that A — 1; hence that U{x) = 0. A n y two integrals fulfilling a specific set of conditions 10.7 are thus identical; the integral is unique. W e note especially that the integral fulfilling condition 10.8 is merely the trivial one. It was shown in Chapter 6 example 17, that an integral of a differential equation 10.2 is sometimes obtainable through the use of the Riccati
186
Linear Equations, Variable Coefficients
equation. T h e method can even be given a somewhat more general form. If i}{x) is an integral of any Riccati equation (10.9)
7} - <Tix)7)-,
po{x)
- 3x2. 2
y -
1
y -
{x + 3if-'.
{x' + 2x2 - g^j^/ _,_ j3^2 _
= 0.
2;r = 2x2 + 2.
10.10. The method of variation of parameters W h e n two linearly independent integrals ui{x) and 1/2W of the reduced equation are known, the integral of the complete equation may be found as follows: Set (10.19)
y = gl{x)ui{x)
g2{x)u2{x),
with undetermined functions ^i(;>c) and gzix).
Then
and / '
= glUl"+g2U2"
+
+
+52'w2i'
T h e substitution of these values into differential equation 10.1 reduo the differential equation to -^polglUi
PolglW
T h e last equation is fulfilled glUi
4- ^ 2 ' « 2
that is, if ^i'(^) =
=
+g2U2]
+ ^ 2 ' W 2 ) ' -hpllglUi gi{x)
0,
=/(*)•
and gzix) are such that giW
-"2(^)/(;^)
po(x)W{uu « 2 ; x)
+ g2U2
g2'(x) =
=
fix) Poix)' ui{x)f{x)
poix)W(ui,
U2; x)
194
Linear Equations, Variable Coefficients
W i t h the choice of any point XQ on the interval tinder consideration, therefore, we may take
^iW (10.20)
--/•
U2is) po{s)W{uu « 2 ; s)
J Xt
Therewith formula 10.19 yields an integral. Example
11. T o find a n integral of the differential equation
{x^ - 3x +
\]y"
-
-
X -
2}y'
+ ¡2* -
= ^(x^ - 3x + I j ^ .
3]y
T h e reduced equation is found by trial to have the integral ui = e'. Therefrom the second integral U2 = x — x'^ c a n be found by formula 10.18. T h e n by (10.19) and (10.20), ~e'
that is,
W
-
-
2 1 ..3 _ 2x'-
y =
s']e-^ ds-t
-
[x 4JC -
x'']
4+
s ds.
4' -
-
* cos - -
1,
j
-I
= — . 1 -1- x"* 1 + x^ 24xV -
^
» t a n - i x.
4x' sin x^,
j -
x^
10.12. Simultaneous linear equations T h e method of Section 7.5 can be used to integrate the simultaneous differential equations (10.23) It is not even necessary to begin by putting these equations into forms that are explicit for^' and z'. Equations 10.23, together w i t h their derivatives, constitute a system of four equations from which z", z\ and z can (theoretically) be eliminated. T h e éliminant is a differential equation for y. W h e n the integral of this differential equation has been found, the éliminant of z' between the two equations 10.23 is a n equation for z. T h e roles oîy and z can, of course, be interchanged. Thus the éliminant of y", y\ and y from the four equations is a differential equation for z, and the éliminant of^»' from equations 10.23 then gives Example 13.
T o integrate the system ey
¡1 -
e'\y'
z' +
[A -
z' -
e'\y
z A y
= 0, ^ e"'
sin x.
These equations, together with their derivatives, are four equations w h i c h have as their éliminant of z " , z', and z the equation y"
Ay = e~'{cos * — sin x
T h e general integral of the last equation is y — —-^^—^Isin X -\- 3 cos x\ + ci sin 2x -1-
cos 2x.
T h e éliminant of z' from the given equations is z — y' -\- e'y — e~' sin x.
F r o m this éliminant, and the evaluation of^ eilready made, it follows that z = A{2?"' -
3} cos X - - ^ ( 6 ^ " ' + 11 sin * + [cxe' -
2c2] sin 2x -\- { c 2 « ' ^-2ci\ cos
2x.
The Laplace Transformation Example
197
14. T o integrate the system 2xy
+
{x -
2x^)2' +
3xz' -\-2xy -
[2x
22 =
4-
+ 2jz =
{3x -
2]e',
2|tf^,
These equations a n d their derivatives have as the éliminant of y"j y'j and y the equation xz" -
z' ^
[x -
\ \e'.
T h e general integral of the éliminant is 2 = tf* +
Cix"^
+
C2.
T h e second of the given equations then yields y ^
^-cAx''2x}
+ ^2
X
PROBLEMS Integrate the following differential systems, 77.
y' xy' -
78. {5x -
z' -2Sy z ~ x = 0. 3z' + ISy - xz + Zx = 0. 10;- + 2xz = y' - 2xz =
1 }>' + 2 x V -
79.
u' - z' Su' - z'
80.
4u' - x^' - 6u - U -3u'+xr' + 5u+
81.
-
1 2
-X.
8u + 3z = 0, 2 = 0. 1
= 0. z=0.
2 x 1 / + xh' + 4^ = 1 - x'. ' 2 y — xy — xz — X . ~ z — 2
82.
X,
z' +
1
1 +-
z+
y = 2t\
X
xy' -\-~z = 2xe'. X
10.13. The Laplace transformation For a differential equation 10.2 whose coefficients are linear i n x it is sometimes possible to obtain an integral i n the form (10.24)
u{x) =
(^\'^v{t)dt,
where x\ and X2 are constants and v{t) is a function that can be determined.
198
Linear Equations, Variable Coefficients
L e t the differentisil equation be (10.25)
{aox + boW + {aix + bi]u' + [a^x + 6 2 } « = 0,
It is converted by substitution 10.24 into the form (10.26)
fj'xe'JVit)
dt + j'\"R{t)V{t)
dt = 0,
where (10.27)
V{t) = [a^-^
axt •\-at]v{t),
and (10.28)
^(0 = aot" + ait -r 02
A n integration by parts changes (10.26) into [e'^Vm:
-
lJ*e"{V\t)
-
R(t)V(t)]
dt = 0.
This is fulfilled i f F ' - R{t)V = 0, that is, i f (10.29)
logK(0 ^
SR{t)dt,
and if xi and X2 are roots of the equation V{t) = 0. Example 15.
T o find a n integral of the differential equation xu" -
{x-\\u'
[2x + 11« = 0.
-
F o r m u l a 10.28 gives i n this instance
t ^ - t - 2
and thus, by equation 10.29, V{t) ^ \t- 2)'^(/ + 1 1 ^ T h e equation V{t) = 0 has the roots ATI = — 1 and X2 = 2. equation 10.27, v{t) =
( 2 ~ / - 2
j/-2l'^[/ +l l ^
F o r m u l a 10.24 therefore gives the integral
Also, from
Answers to Problems PROBLEMS Find an integral for each of the following differential equations. 83. xu" ~- \x 85. xu" -
4|u' -
\Ax -
3}«' +
87. x u " + 3u' -
84. xu" -
3« = 0. l\u = 0.
f3x -
xa = 0.
89, x a " + I2x + 2 l a ' + a => 0.
i2x -
3)a' - 4u
86. xu" + 5u' -
(x + 3}a =
88. x a " -
l}a' - a
{2x -
90. 2 x a " + j2x + 3 } a ' + a
ANSWERS TO ODD-NUMBERED
PROBLEMS
1. a = f i ( x + 2 i » + f 2 ( x + 2l sin log {3 — x| + ci cos log ¡3 — x}
3. a =
+ f log X + i + cix + = i U o g 7. y = {x -
1 i« + \cx log (x -
9. > - i | x - 1} +
11. : y -
^ 1
20
13
13- li • ^1
10
1) + cat
{ l o g ( x - 1)
. , 2 Sin log X *
- 1
+1).
1
cos loK 20 ^
X
X
15. a ^ ¿1 CSC X +
cot x.
17. Inexact. 19. a
21. a ="
23. a
1 -
2x
1 +2x
^1
cos X log
2x -
1
1
(1 - 2 x )
1 + sin X 1 — sin X
X
+
+x \
— - e'^' + cix /
31. a «• cxp X
+ £2 cos X .
log (x — 1)
X log X — 1 1
+ f2
2x - 1 ,
27. a = ci
29.
£i log
^
—
C2
f2« 1 dx C2X33. a -
exp \ -2e']
200
Linear Equations, Variable Coefficients
35. u « ^ - ^ i " " ' - — ' .
37. u = cxp {3e^\.
39. w" -\-9w'0.
41. w" -\-\w
-2w
43.
3*"' + 1.
X
5
= e'.
45. « ; " + -5
-
47. u - ^ + C 2 ^ .
49. u -
51. y "
f 1^^*"* + c^e
53. u — cix sin x + f 1 sin x.
55. u -
o i l
57. > -
+ fa*.
- + «1 —
X 59. y '
- {x^ + 2x + 2} -\-cif' j
+
^2«"'.
X
^ rfx + c j ^ .
61. u - fix** + C2«'.
63. > -
Ix -
65. > - i x ( l o g x|'.
67. y -
- 4 x ^ + x + 1 - x M o g x.
X* 69. ^ — 2 + f 1 sin - J +
11^"*.
x» cos —•
71. J, - 4 + eV^[ci sin \ / ^ + a cos 73. u - f 1 log X + .
75. y - tan * x H
V l
logx 77. > -
- ^ x +
ci-r^' +
.
ca*-'',
79. u — ^{ci sin X + C2 COS x|, z — ^'{(Sci — j) sin X + (ci + 3^2) cos x|. 81.:)'^ . 83. u 87. u
i + cj** +
log X,
* -|-fi(2 - x ^ l +ct{l
+2Iogx
-xMogx).
85. u « / / * ^ ( ' - 11
•
CHAPTER
11 Solutions in Power Series
11.1. The reduced equation at an ordinary point T h e method of power series of Section 7.6 is applicable i n particular to a linear differential equation 10.2. If the equation has an ordinary point at xo, and has coefficients that are representable by power series i n — XQ], its integrals are also so representable. T h e change of variable x — XQ = z makes the point x ~ XQ correspond to 2 = 0, and transforms the differential equation into another linear one. W h e n XQ 7^ 0, we shall suppose that this change is made. I n proceeding we shall therefore assume that XQ — 0, and accordingly that Poix) = pQ,G -\-pO,lX
(11.1)
-\-p0.2X^
+
• • •,
Pi{x)
=pl,Q-\-pl,lX-\-pi,2X^-\-
• • • ,
p2{x)
= /'2.0 + / ' 2 , 1 * + / ' 2 , 2 * ^ +
• • • .
^0.0 ^
0,
T h e form of an integral is then (11.2)
u{x) = Co + cix + czx"^ +
I n the general integral, CQ and ci UQ' i n the integral for which u(0) T h e substitution of series 11.2 ferential equation 10.2 converts series. By equating coefficients system of relations
••• .
are arbitrary, whereas = «o 3.nd a — = UQ, U'{0) = UQ'. with undetermined coefficients into dif(10.2) into an equation between power of like powers of x, we then find the 2/^0,0^2 -\- pi,oCi + p2,QC0
2 • 3yjo.oC3 +
{2po.i + 2pi,o]c2 +
{pi,i
-\-p2,o]ci
•i'p2,ico
= 0, = 0,
These successively determine cj, C3, ¿4, • • * , i n terms of and ci. If it is deemed desirable, equation 10.2 can first be transformed into an equation (11.3) w" - q{x)w = 0, 201
202
Solutions in Power Series
by the change of variable u = ci = 0 ,
i c i —
fn =
0,
F r o m these it is found that ^2 = — {^o + i ^ ^ i l ; ^3 — = {^CQ + ^ V i h • • • . T h e particular integral required is determined by the values Co = 2 and ci — 0. W h e n z is replaced by {>: — 7r/4j, the particular integral is found to be
PROBLEMS Find (up to terms of the fifth degree) the integrals of the following differential equations that fulfill the respective conditions. 1. « " -
{x -
2. u" +
¡11 -
'*
3.
ISx^lu' -
\3x + 4x^\u' +
u" -
xu' +
4. a " + a ' sin x 5. a " 6.
Vl
il -
7. a " + 8.
{1
a(0)
-
+2^2
I* +
u(0) =
\2 -
u' +
(9 -
6x]u = 0, {1 -
=. CO, a'(0)
u(2) = 1, u'{2) = 0.
6x + x^Ju = 0,
= 1, a'(0)
a(0)
i^^^x
+
{1 + log (1 -
u(-l)
=
- 1 , u'{-\) = 1
= 2.
= 0,
Y^^la
-3.
a(0)
x)W
-
= 0, a'(0)
= 2.
u log (1 + x ) =
0,
= 2.
{4 + X + 6x2}«' _ x^ja" +
«(0) = 2, u'(0)
{13 +24JC + 12*2)a = 0,
+ % V ) a " -
- 3 , a'(0)
X -
3x^]u = 0,
{1 -
|1 +
|3 _ 6x +
4^ _i_ 4^2j„ ^ 5x^)u' +
{1 -
^(QJ ^ ^^^^ ^/^o) 5x +
x^lu =
=
o.
0,
= ci.
11.2. A proof of convergence T o assure that a series 11.2 does in fact give an integral, it is necessary to show that it converges upon some interval (11.7)
-h<x
sec x =
11.4,
The ordinary point at . By the change of variable x — 1/z, such a differential equation is transformed into one which has an ordinary point at z = 0. T h a t accounts for the special forms of series 10.12 If the series (10.13)
M = Co + " + X\
-2 + X
+
*
•
*
x^
is substituted into the reduced equation, and the coefficients of the several powers of 1/x are then equated to zero, a system of relations from w h i c h (^2, C4, • • * , are determinable is obtained. T h e values of co and c\ are left arbitrar)'. F o r the complete equation the method determines the coefficients of a particular integral (10.14)
, = p +
+
T h e series so obtained for the integrals converge upon any range x > H upon which all of series 10.12 are convergent, and/>o(x) 9^ 0.
207
Solutions in Power of Example 3.
T o find the general integral of the differential equation 2]
,
2
Ax- -\-2x-\- 10
i n powers of 1 /x. W h e n the given equation is divided by x"^ its coefficients have the forms 10.12. I n the case of the reduced equation the relations obtained by substituting series 10.13 are 2c2 H- 2c(i = 0, 6cz + 4ci = 0,
12c4 4- 10 = 1 + x + 2x^ + - f • • • . B y (11.18), the second integral is thus uzix) = Example 7.
(1 +
X
+ 2x=^ +
+
- . •) -
2ui{x) l o g x .
T o find a pair of integrals of the differential equation x V
+ x[2 + x]u' +
( i + Ix -
2x^1« = 0
about the point XQ = 0. T h e indices are i n this case equal, and have the value — H integrcil found is „ 1 = x-^{\
- x
+ x2-|x3+
The
• • .}.
B y virtue of this evaluation of u i , formula 11.20 yields fix)
=
box'H-i
+ 3 x - i x 2 +
• • -1,
and the substitution of form 11.21 into equation 11.19 leads to the relations ¿1 +
¿0 =
2bi -
2bo =
9bz -f- 3*2 -
2bi =
462 +
—bo, 3bo, -ibo,
The Complete Equation
213
O n assigning to ¿0 the value 1, the second integral is found to be «2 = * - ^ l l - 2x + Ix^ -
• • -1 - «iW l o g x .
+
PROBLEMS Find two integrals for each of the following differential equations about the point *o = 0. 31. x V - x{2 32. x^l\ -
5x\u' + \x -
x^\u" + *(3 ~2x
33. x^{2 + 3x^\u" + x[2 34. u" sin^
X
u' &in
X
6x^\u = 0.
~ 3x^W 4x -
— xu
\3x + 3x^ + 2x^]u = 0.
3x'^}u' -
12 + 2x -
5x^ -
ix^]u = 0.
=0.
35. x^u" - xe'u' + « = 0. 36. 3 x V ' + 6xu' + ( i + Sx^la = 0.
ny.
The complete equation
I n the case of the complete equation 11.19, i n w h i c h fix)
=
X^\fo + / l X + / 2 X 2 +
- - -1,
fo9^
0,
the form of a particular integral depends upon h. 1. Neither index exceeds {h is i n this case an integral of the form CASE
(11.22)
y = x^{ao +
a^x
+
—
1 j by a positive integer.
There
a^x'' +•••),
the coefficients of which c a n be determined. 2. Just one index exceeds (A — 1) by a positive integer. L e t us say ¿1 = A + where m is zero or a positive integer, a n d ¡¿2 — h] is not zero or an integer. T h e change of variable CASE
(11.23)
>-=>•*-
amUi(x)
log
X,
then transforms the differential equation into one for which has a n integral of form 11.22 whose coefficients can be determined. 3. E a c h index exceeds (A — 1 ( by a positive integer. L e t us say ¿1 = A - f m i , and A2 = A - f ^2. There are then the following alternatives: CASE
3a. T h e reduced equation has two integrals of form 11.16. T h e change of variable, SUBCASE
y = yif -
amiUiix)
log x -
a^^zix)
log x.
214
Solutions in Power Series
transforms the equation into one for 11.22.
w h i c h has a n integral of form
T h e reduced equation has an integral ui{x) of form 11.16, and another of the form M 2 W = x^^v{x) — bui{x) log x, w i t h v{x) a power series. T h e change of variable, SUBCASE 3b.
y = y*
-
am.MlW log X
-
OmiUzix)
log X
-
iflms^UiW {log x ) ^
then yields for^-* a differential equation w i t h an integral of form 11.22. Example 8.
T o find a particular integral of the differentigd equation 2x^>" + xy' + xy = sin X + 2x^ cos x
about the point XQ = 0, T h e indices are i n this instance ^-2, and 0, a n d fix) = x i l + 2x - i x ^ - x^ - h • • -1.
This is therefore an instance of Case 1.
Example 9.
T h e integral is found to be
T o find a particular integral of the differential equation xy
+ 3 x V - 2> = x'^{2 -
7 x + 4x^1
about the point XQ = 0. T h e indices are k\ = 2 and = — 1 . Since A = 2, this example is a n instance of Case 2. F o r the reduced equation we may find the integrzil «l(x) = X 2 { 1 - | ; c + |Jx^ +
• • -1.
T h e change of variable 11.23, w i t h m = 0, transforms the given equation into x ^ " +
3x2,*'
_
= ¡2 +
3(2olx2 -
¡7
+ fflol^' + !4 + f ^ a o ) * * + • • • ,
which has the particular integral = x'^{—% -\- •rkjix'^ +•••). integral of the given differential equation is thus :v
- ; c 2 { - | +
Example 10. ^xY
T^i7x'+
• • •) + V { 1 - | ; c + K x 2 +
The
• - .}logx.
T o find a particular integral of the differential equation -
x[\
+ x\y' -
about the point xo = 0.
I I + |xl> = x^^{4 + 2x + x^)
215
The Regular Singular Point at oo
T h e indices are ¿ 1 = 4 and kz = —h T h i s example is therefore a n instance of Case 2, w i t h m = 1. T h e reduced equation has the integral
«iW
= x^l\
S
+
^2
^
f T h e change of variable 11.23 transforms the given differential equation into 3xW
- '{1 +
{I + ixljK* = 4x^ + {2 + SmW'
-
which is found to have the integral = — 2 + ix^ + this instance it happens that ai = 0; hence, by (11.23), y integral of the given differential equation is thus
+ • •). I n y*. The
I, and does not involve log x. W e shall not consider any instances of the more intricate Case 3. PROBLEMS Find a particular integral of each of the following differential equations about the point *o = 0.
2 37. x M l + 4x-\y" + xy 38. 5 x V ' + (ixy' -
39. 2 x V '
-ly^lx
4> =
+*}/
3x{l + x | l
+ U
40. Axy" - lOxV -Sxy^ 41. x|2
42.
x V y
1-^ 4
+ 3{I - x } /
+x»}> = x^^{2 - 14x|
x^4{3 - 5x + x^}. - x>Ml
\x-k-i^x^\.
+ / sin X - ¡1 + x)> = .r^
43. x^\l -
x]y" +x)r'
+ xy = l + 2x -
3x1
44. x ^ ; - " + 4x^'y' + 2> = ^ ! 1 + - x= X 3 45. x V + x'-^^ =. X - M 6 + 2x + 3x2 46. 3x'>" -
llxy' -
j^3j
5;F cos X = x-^^|38 + 4x + 6x2|.
11.8. The regular singular point at oo W h e n the coefficients of differential equation 11.19 are representable by power series i n l/x, thus:
216
Solutions in Power Series poix)
q\{x) =
?i,o H
?2W
?2,0
fix)
^
= pO,(i +
=
+
X
+
X
1—2~
H
• ••,
with/Jo.o 5^ 0,
+
• +
+ f:
=
*r"
[
X
the differential equation has a regular singular point at infinity reduced equation, the substitution 1
u —
_L '^l _ L ^2 ,
Co H
h
+
with Co
For the
0,
leads to an indicial equation for A, and yields for each index a system of relations i n the coefficients CQ, CU C2, ' ' ' • F o r the complete equation the substitution 1 ' = = 7 r ^
X
leads to a system of relations i n the coefficients aoj oi, 02, ' ' ' • I n general, two integrals of the reduced equation and a particular integrzil of the complete equation are so determinable. I n the exceptional cases i n which these integrals are not obtainable, the given differential equation may be transformed by the change of variable x = \/z. T h e transformed equation has a singular point at z = 0, and for this transformed equation the integrals are obtainable by the methods of Sections 11.6 and 11.7. PROBLEMS For each of the following differential equations find an integral in powers of \/x. 47. 2x\"
+
1 +
X
1 u' -\--u=
1
X
48. Ix'^u" + 5xu' -
1 + -
2
u = 0,
X
49. 9x-y" + \Sxy' + 2> = 1 +
-+ 4 X
50. x'y 2..H
2
1 X
0,
y
8 +
X
2 X
1 ^ X
3
2
X
X*
Answers to Problems 51. xY
1 -
5 -
+x
f y
X
7 -
+
8
4
2
X
1 ll •+-2
52. x^
2 -
i
1
2
1
ANSWERS TO ODD-NUMBERED PROBLEMS 1. u =* 2 -
3x
3. a = - 1 +
+
X*
- x« + 4x* -
U + 1) +
2^5 + • •
jx + I p -
U + ll'*+
•• •.
5. u =• 2x -j-x"^ + X* + Ox'' -\- • • • 7. „ = , o l l + | : c 2 - f x ' + ¥ ^ * - K x ^ +
- • •!.
9. y U.y^x'-W-ih'^hUx'-^-
n.
l
3
1 4
5
1
> = Ifx — Y^x^ — YTUX +
15. u = CO 1
17. u
CO
-
1 1 +-5
3A:^
3
2
5
x^ '
X*
;
+ C1
+
1 2 12-13
+
1 -
• •
^
1 + 1 + — 4 +
2 U y
23.
1
6x
1
1
1
2x^
3x'
x^
-
••• •
ui (1 + 2 x -
«2
3*2 +
-
25. a i i.
27.
«1
-
fl + T ( * + 1) +
u2-(x-\-
D^Mi -
29. 01 - * ^ { 1 U2
31. ui
x'(l i l
• • •}>
+ 1) + ^ ( *
+ 2 * +4*'' +
X-^{1
n
+
+
+
• •
+Jt2 + * 4 4x + 9*2 + 32
1.
13
3 I
1 + T r « i W log X.
33. ui = *(1 + x + i*2 4- • • "2
"Ml
1.
+ x + 3*2 +
4 *
} - 3m(x) log X.
37^ 37^
Solutions in Power Series 1, +
U - *+
•••}-
+ • • -1 - {1 - * + i ' ' +
- + 2 + 3x -
3*« +
X
2x
2
X
5x^
Sx^
V**
#
X
X*
Ki(jf) log X.
2x'
•
*
-
2
1
2
+ 41
CHAPTER
12 Some Differential Equations Involving Parameters 12.1. The separation of variables in a partial differential equation T h e mathematical expression of a physical law of nature generally requires the use of a partial differential equation. T h e course of any chain of events which is subject to that law is then governed by an integral of that equation, that is, by a particular integral that is characterized by certain conditions which the chain of events fulfills. A n important method for determining the requisite integral depends upon ordinary differential equations. W e shall indicate some elements of that method. T h e partial differential equation
a^r
d^T
a^r
governs a variety of physical phenomena. It is known as Laplace's equo' tion, i n honor of the French mathematician P . S. Laplace (1749-1827). A product of a function of x by a function of > by a function of namely, u{x)v{y)w{z)y is a n integral of (12.1) if u" {x)v(y)w{z) + u{x)v" {y)w{z) + u{x)v{y)w" {z) = 0,
T h a t is, if v"{y)
^ w"{z)
v{y)
w(z)
_
_
u"{x) u(x)
I n this equation the left-hand member does not vary w i t h x. T h e righthand member does not vary w i t h y or z. T h e i r common value is therefore a constant. O n designating this constant by A, we have
u{x)
'
v(y)
w{z)
I n the second of these equations the members are again a constant, say /. W e see thus that u{x)v(y)w{z) is a n integral of the partial differential 219
220
Differential Equations with Parameters
equation 12.1, if and only if its factors are integrals of the respective o r d i nary differential equations u'\x)
+
huix) =
v"(y)
-
h{y)
(12.2) w"{z)
+
{h-
0,
= 0,
l\w{z) =
0.
I n these differential equations k and / do not depend upon x, y, or T h e y are therefore constants. But their values are not specified. T h e y can be assigned various values. They are therefore called parameters. This method of referring a partial differential equation to a set of ordinary
F i o . 31.
differential equations involving parameters is called the separation of variables. W e have applied it to differential equation 12.1, namely, to the Laplace equation i n the rectangular coordinates x, y, z. It can be applied also i n other coordinate systems, as we shall show. T h e cylindrical space coordinates (r, tp, z) indicated i n F i g . 31 are related to x, y, z by the formulas X
—
r cos ip^
y
r sm (p.
z =
z.
These define a change of variables which transforms differential equation 12.1 into d^T 1 dT 1 a^r d^T dr
r dr
r^ d m-O
^ 1)"*A!
2fc-2m
m\lk-m}\
k times, to get [k/2\
. , rfx* '
V '
(-l)"*|2A-2m}!
.k-2m
¿1/ ml\k - m]\{k - 2m\\
m-0
A comparison of the last equation with equation 12.24 proves the formula
230
Differential Equations with Parameters
T h e Legendre polynomials are interconnected by the relations xPk'ix) Pk'(x) -
(12.27)
Pk+x'{x) {x^ {k +
\]Pk+iix)
xPk-i'{x)
-
Pk-i'ix)
\\Pk'{x) -
-
{2k +
l]xPk{x} +
These are called recurrence formulas. that an equivalent form of it is
kPkix) =
0,
=
0,
kPt-i(x)
kxPkix) +
{2k +
-
Pk-i'ix)
\]Pk{x) = 0, =
0,
APA_I(X) =
0.
kPk-dx)
T o prove the first one we observe
B y substituting for Pk{x) and Pk_i{x) their forms 12.24, and performing the indicated differentiations, the formula is proved. T h e second formula 12.27 can be proved in the same way, and the remaining ones can then be inferred from the first two. Some important quadratures of Legendre polynomials can be evaluated. Thus, to begin with, (12.28)
2, 0,
=
j',Pk{x)dx
if if
>t = 0, A > 0.
This is obvious if A = 0, since Po{x) = 1. W h e n A > 0, a quadrature of the first relation 12.27 from — 1 to 1, w i t h an integration by parts of the terms i n Pk{x) and Pk-i{x), yields the equation [xPkix) -
Pj^i{x)]l,
-
{/: +
1 j / _ \ Pkix) dx =
0.
N o w , by (12.25), the integrated part of the last equation is zero. result (12.28) then follows. A more general evaluation is [0, (12.29)
\\x^Pkix)
dx =
The
if ffl < i ,
I 2^+M/:!P
.
This also is obvious when k = 0. T o prove it for other values of A, let the first formula 12.27 be multiplied by x"^ and then integrated. T h a t yields the equation [x^+'Pk(x) -
x"^Pk-i{x)]l,
-
{A +
m +
1 i / _ \ x'-Pkix) dx
+
m f\x"'''Pk-i{x)dx
= 0,
The Legendre Polynomicds wherein the integrated part vanishes, by (12.25).
-1
x^Pkix) dx ==
I
k -\- m -\- \ J ~\
231
Hence
x - - i P f c _ i ( x ) dx.
A n m-fold iteration of this relation yields - 1 x'^Pkix) dx = -r { A - m + 3 l { / t - m + 5 j - - - l y t + m-|-lj 1 Pk-m{x) dx. - 1
Evaluation 12.29 then follows, by equations 12.28 and 12.13 W e may now show finally that [0, (12.30)
|_\ PH{x)Pk{x) dx =
2 2k + 1
i{
k ^ k,
if
k = k.
T o do so we replace the polynomial Ph{x) or Pk{x) that is of the lower degree, or one of them if h = k, by its form (12.24) i n powers of x. Quadrature 12.30 then appears as a sum of quadratures of type 12.29, a n d evaluation 12.30 follows. Result 12.30 is applicable to the discussion of Section 12.2. T h e function there denoted by vic($) is Pk(x) when x = cos 6. Thus relation 12.8 is, effectively,! CO
f{x) =
^
cyz^Pk^x),
f{x) being an arbitrary function. T o determine the coefficients Ck we multiply the above relation by Ph{x), with any A, and integrate it term by term from — 1 to 1. Every resulting quadrature on the right except that for which k = k then vanishes, by quadrature 12.30. T h e integrated relation therefore reduces to ^
f{x)PH(x)dx
=
2 CKa^——-,
and this determines Ch* ASSIGNMENTS 1. By the use of formula 12.24, show that
ax and thus derive the second of the recurrence formulas 12.27,
232
Differential Equations with Parameters
2. From the first and second of the recurrence formulas 12.27, derive the remaining ones, 3, Show, by the use of formulas 12.16, that ( — 1)*^*
; '
^ uo.iW)
when
k is even,
_r 2
Pkix)
«
n«o,jW,
when A U o d d .
4. Show that when k is zero or a positive integer the Laguerre equation 12,20 has the polynomial solution k
(,2.M)
^'•W =
I*^I^X^'^= n =0
(This is called the kxh Laguerre polynomial.) 5. Show that
(12.32)
iiW
= X -
Liix)
-
1, -
4x + 2,
Lz{x) « x ' -
U{x)
= X* -
+ 18x -
6,
16x' + 72x* -
96x + 24.
6. Show that when k is zero or a positive integer, the Hermitc equation 12.21 has the polynomial solution f*/2I (12.33)
2^
H.(x)
-
c-ir
^
j
^
^
(This is called the i t h Hermite polynomial.) 7. Show that //o(^) = 1, Hi{x) - 2x, Ha(x) = 4x= ff3(^)
= 8x' -
HA{X)
-
i/6(:f) -
2, 12x,
16X* - 48x2 _j. 32x^ -
Hiiix) = 64x" -
160x» + 120x, 480x* + 720x -
120,
The Hypergeometric Equation
233
8. Show that d dx and. from this, that 0, 9. By the use of formulas 12.32, show that
when A -
0, 1, 2, 3, 4.
10. Show that
0
0, 1,
e-'Lk{x) dx =
if k if
0, A = 0.
12.6. The hypergeometric equation T h e differential equation (12.34)
x{\ - x]u" + (T -
(a + /3 + 1)^1«' - a0u = 0,
i n w h i c h a , |9, and y are parameters, is called the hypergeometric equation. Its singular points are at x = 0, x = 1, and x = oc, and are a l l regular. T h e indices at x = 0 are zero a n d ¡1 — y\. O n substituting the form as
u — X * ^ r„x" into differential equation 12.34, the relations n-O (12.35)
¡7 + i + «1 ¡1 + A +
-
{a + A + «| (i3 + A + n]c, = 0, n = 0, 1, 2, • • • ,
are obtained. W h e n k — 0, these can be fulfilled, except possibly when ¡7 — 1) is a negative integer, and with that possible exception, (12.36)
Cn+i
=
, w , 1 Cn. {1 + nj{7 + n]
T h e power series w i t h these coefficients, when CQ is assigned the value 1, is denoted by F ( a , /3; 7; x). It is called the hypergeometric function a n d has the formula (12.37)
F(a,P;y;x) go
= 1 +
a j g + l i • • • ( a + M g ( g + l ! ' ' ' !j3 + M „ + , l " + l } ! 7 l 7 + l | • ' ' {y + n\
234
Differential Equations with Parameters
T h e series is convergent Vk^hen — 1 < x < 1, and is unchanged when a a n d 0 are interchanged. Differential equation 12.34 thus has the integral (12.38)
«o.i(x) -
F{a,p;y;x).
W h e n i = 1 — 7, equations 12.35 can be fulfilled, except possibly when ¡7 — 1) is a positive integer, and with that possible exception, (12.39)
r ^ ¡1 + « 1 1 2 - 7 + «l
Cn+l =
:
C
« = 0, 1, 2, • T h e series w h i c h results from this can be found i n the usual way. It c a n , however, also be obtained by observing that relation 12.39 is derivable from 12.36 merely by making the replacements ( a — 7 + l j for a; 1/3 — 7 - f 1! for j3; and ¡2 — 7) for 7. T h e second integral i n powers of X is thus (12.40)
uo,2{x) = x'-^Fia
- 7 + 1, /3 - 7 + 1; 2 - 7; ;c).
W h e n {7 — 1 j is an integer, the system of equations 12.35 may include a n equation that is not fulfilled. Whether it does so or not depends upon the values of a and /3. W h e n it does so, there is only one integral i n powers of X, the second one then involving log x. T o obtain the integrals i n powers of {A: — 1), we transform differential equation 12.34 by the change of variable \ — x = z into d^u
^¡1 - ^1 ^
+ {(a + ^ - 7 + 1) -
du
(a + i3 + 1)^) ^ - a^M = 0.
This is recognizably again a hypergeometric equation, i n fact one that is obtainable from equation 12.34 by making the replacements z for x\ a n d {a + /3 — 7 + 1) for 7. T h e integrals, i n terms of the original variable, are therefore, by equations 12.38 and 12.40, (12.41)
ui,x{x) -
F ( a , /3; a + /3 -
ui.2{x) = ¡1 -
x]''-''-'F{y
7 + 1; 1 -
^, 7 -
x\
«;7 + 1 -
a -
^; 1 -
^)-
T o obtain the integrals in powers of 1 /x, we make the change of variables u — x~*'w,
X — \/Zj
which transforms differential equation 12.34 into -
^1 ^
+ {(a - /3 + 1) -
(2a - 7 + 2)z\ J
-
235
The Hypergeometric Equation
T h i s is again a hypergeometric equation, and is seen to be obtainable from equation 12.34 by making the replacements w for «; z for x] (a — 7 + 1} for j3; and {a — j8 + 1) for 7. Formulas 12.38 and 12.40 may thus be d r a w n upon for the integrals w, and from these formulas it follows that (12.42)
««.i(^) = ! i M ' " ^ ( a , a - 7 =
u^Ax)
13 -
ir/x]^F{^,
+ i;a-)9 + i ; lA),
7 +
1; ^ -
a +
1;
M a n y familiar functions are of the hypergeometric type. by the following list. ¡1 -
= P{-a,
x]"
= -i^(l, l;2;x).
-log{l - * | X
1 2x
1
log
1; l ; x ) .
+x
1 -
X
1 - sm 1* X X
1 _ i - t a n ^X
2)H ¡1 - x M
\2
cos \k sin ^ x|
=
2' 2'
^ V i ' T ' 2 '
It can also be shown that e' =
lim F
0—* 00
COS X
- sin X = l i m F X
ergeometric equation. 18. By use of the result of assignment 17, show that F"{OL — 2, /3 — 2; 7 2; JV) is a linear combination of the functions uo,iW ^ " d uo.2(^), when 7 is not an integer. Then, by comparing coefficients of like powers of show that
-
2. , -
2, . -
2; .) = l ^ - 2 | | a - » H . - 2 H , - l ) \y ~ 2\\y - \ ]
12.7. Equations that are solvable by hyper geometric functions Some differential equations are transformable into the hypergeometric type by suitable changes of variables. S u c h equations are therefore solvable by hypergeometric functions. A n y differential equation of the f o r m (12.43)
a{x -
n l \x ~
r2l«" +
b{x -
r^W + c« -
i n w h i c h a, b, c, r i , r 2 , and rs are constants, and this way. For the change of variable (12.44)
x =
(r2-ri}z
+
0,
9^ r j , can be solved i n
r1
transforms (12.43) into differential equation 12.34 with values a, /3, and 7*, obtainable from the relations a + )3 + 1 = b/a,
a& = c/a,
7 = ^-j^^^'^* a\ri - raí
Equations Solvable by Hypergeometric Functions T h e singular points x = ri and x = respectively. Example 1.
237
then appear as z = 0 and z
T o find the integrals of the differential equation 4 ) u " H- [Ix + l\u' -
{2x^ - 2 x -
3u = 0
about the singular point x = 2. I n factored form the coefficient of u" is 2[x — 2)|x + 1). W i t h T\ — 2 and rz — — 1 , the change of variable 12.44 is x= — 3 ^ + 2, which transforms the given equation into d^
1-1 2~ 2^\
dz
, ^
+ - « = 0.
This is the hypergeometric equation w i t h a = 3, j8 therefore has the integrals «0.1 =
FO,
- h h
7
—
5
It
z),
-3;
uo.2 = z-^F{h
—
z).
-I;
I n terms of the original variable x, these integrals are
"o.iW ^f(3
-
V 2-
W0.2W
Example 2.
'-I-
2-
X
2'2'
x\
=
V2'
" 2' ~ 3 ~
T o find the integrals of the differentiail equation {3^2
-
15x + 1 8 i « " -
-
6x]u'
+ 1« = 0,
i n forms that are appropriate to large values of x. T h e change of variable 12.44 is x = z + 2, and the transformed equation is hypergeometric with a = i, 0 — ^, y = — ^. Large values of z correspond to large values of x; therefore w^,i and « ^ , 2 are appropriate. These integrals are 1
c-/^ 2
X -
Example 3.
-
1
\
^V3'2'5^"^/
1 X
^ 2
/2 2
\3
n
1
4
\
6'3'x-2)
T o find the integrals of the diflferential equation ¡1 - e']u"
-\-W
by making use of the change of variable
+
e'u =
= z.
0
238
Differential Equations with Parameters
This change of variable transforms the equation into the hypergeometric type, and makes x = 0 correspond to z = 1. F r o m formulas 12.41, therefore,
PROBLEMS 9, Find the integrals of the differential equation {x^ -
\ \u" +
|5JC +
4]u' + 3u = 0
about the point x = — 1, 10, Find the integrals of the differential equation
about the point x = — ^ , 11, T h e Legendrc equation 12.14 is of the hypergeometric form.
Assuming that
2k is not an integer, find for it a pair of integrals as hypergeometric functions that arc appropriate to large values of jr, 12, Find an integral of the differential equation -
2W'
+
{1 -
+ u -
0
about the point JC = — \ / 2 . 13, Find two integrals of the differential equation
about the point JT = 0, by using the change of variable x =
"sfz.
14, F i n d two integrals of the differential equation {1 -
*-2^Ju" -
3u' + Ar'^'u = 0,
in a form that is suitable for use with large positive values of x, by making the change of variable e~^' = z.
12.8. The Bessel equation T h e differential equation (12.45)
x\" + xu' +
- jt^jw = 0
is called the Bessel equation, i n honor of the G e r m a n mathematician a n d astronomer F . W . Bessel (1784-1846). It has singular points at = 0 a n d X = CO, but only .v = 0 of these is regular. W e shall therefore consider only the solutions i n powers of x.
The Bessel Equation
239
T h e indices at x = 0 and k and ~k. F o r the integral associated w i t h the index A, the relations giving the coefficients are ¡1 •^2k\ci = 0, n{n + 2k]c„ — Cn~i = 0,
n = 2, 3, 4, •
W h e n n is even, say n — 2m, these relations are -1
2'^m{m + A} and yield the evaluations m
"^"^
(-1) 22"»m!{A + l ) { A + 2 l • • • jA + m l " * ' '
A l l the coefficients w i t h odd subscripts may be taken to be zero. CQ is assigned the value l / [ 2 * r ( A + 1)], the series obtained is _
(12.46)
Jk{x)
When
fc+2m
'
^
'
T h i s is called the Bessel function of the order k, of the first kind. T h e integral associated with the index — A is similarly obtainable. W e find thus that (12.47)
«i(x) = /*(*),
U2{x) =
J_j,(x).
T h e integrals 12.47 are linearly independent unless A is an integer. W h e n A is a n integer, a second integral, linearly independent of MI(X), involves log X. T h e values of the Bessel functions are obtainable, for many values of A, from tables that are i n common use; hence they are known i n the same sense as the values of log x or cos x. W e may therefore regard a differential equation as solved if its integrals are expressed i n Bessel functions. Example 4.
T o find a pair of integrals of the differential equation « " -1- 9x^u = 0.
T h e change of variables u = 's/x w,
X
transforms the differential equation into
=
A/IZ
240
Differential Equations with Parameters
This is the Bessel equation w i t h k — \. Thus wi = I n terms of the original variables, therefore,
J^(z),
and
1V2
PROBLEMS 15. F i n d the general integral of the differential equation 1
-
4 9*'
a = 0
by making the change of variable u = "x/x w. 16. Find the general integral of the differential equation
x'u" - xu' +
{x2 + | ) u
= 0
by making the change of variable u ^ xw. 17. Find the general integral of the differential equation
by making the change of variable 4 i ' = z^. 18. Find a pair of integrals of the differential equation 4 u " + 2Sx\ = 0 by making the change of variables u = \/x w. 19. Find a pair of integrals of the differential equation
as power scries in jr, assuming that 2k is not an integer. 20. Find a pair of integrals of the differential equation
as power series in x, assuming that k is not an integer.
12.9. Bessel function graphs and inter-relationships Formula 12.46 shows that every* Bessel function Jjt(x) of a real order k is real for positive values of x. W e shall show that it is an oscillatory function for such J , that is, that its graph has the shape of a wave. T h e change of variable u = w/'\/x transforms the Bessel equation i n t o (12.48)
w"
+
1 -
k^-V
.
w ^ 0
241
Bessel Function Graphs
L e t xi be any value greater than and set a = v ' l — k^/xi^. T h e function V = sin a(x — xi) is an integral of the differential equation v" + a^v = 0. It follows from this equation, and equation 12.48, that {vw'
~
v'wY
=
-
x^ ^
4x^
vw.
and hence, by a quadrature from x\ to x\ -f- ir/a, that aw
4- aw(j:i) —
x^
Ax
w{x) sin
— x\) dx.
T h i s woxild be impossible if w{x) maintained its sign over the whole interval from x\ to x\ + v/a, for then the two members of the equation would have opposite signs. Therefore w{x) changes its sign on any interval of the length ir/a to the right of the point x i , and the function Jk{x) — w{x)/'\/lc is accordingly oscillatory. T h e graphs of JQ{X), J\{X)J and Jzix), are shown i n F i g . 33.
Fio.
33.
Let the zeros of Jk{x) with any chosen real value k be a i , « 2 , « 3 , • • • T h e functions v(x) = Jkiocnx) and w{x) = Jkiax), in which a is a parameter, are integrals of the respective differential equations .2
xv" +
xw"
-\-w'
v'-\-
-\-
Oln X — k^
V = 0,
w; = 0.
N o w the result of multiplying these equations respectively by w and by
242
Differential Equations with Parameters
—Df and adding them, is the relation {x{v'w -
vw')]' - h {cxn^ -
a^}xvw = 0.
A quadrature from 0 to 1 gives this relation the form »'(1M1)
+ {a„2 -
j^xvwdx
= 0,
or, more explicitly, (12.49)
a„7jb'(a„)7fc(a) = (a^ - a^^j
xJk{ar,x)Jk{ax) dx.
T h e derivative of this w i t h respect to a is the relation (12.50)
a„Jfc'(an)^* (a) = 2 a \^ xJ^MJ^iwc) -
dx + x'JkMJk'iax)
«n')
If a is now assigned any one of the values reduces to
dx.
other than a „ , relation 12.49
^ = fo xJk{anx)Jk(a^)
dx.
O n the other hand, as a —> an, the limiting form of relation 12,50 is anJk\<Xn)
= 2a„
xJk\otnX ) dx.
W e have thus found that (12.51)
xJk{anx)Jk{a„x)
dx =
0,
if a „ 5^ any
iA'\an)y
if
a „ = a„.
Evcduations 12.51 are applicable to the discussion of Section 12.2. T h e first of equations 12.3 is transformable into the Bessel type. If a function f{x) is representable as a series of Bessel functions, thus, (12.52)
f(x) =
2 Cm-Jk{ctmX), m-1
a determination of the coefficients Cm is made possible by evciluations 12.51. For, if relation 12.52 is multiplied by xJk{anx)y w i t h any n, a n d is then integrated term by term, the resulting quadratiu"cs, except one, vanish, by (12.51).
T h e result is the relation 0
xf{x)Jk(anX) dx =
CnWk^(0Cn\
and thereby the coefficient c„ is determined.
243
Answers to Problems ASSIGNMENTS 19. Show, by use of formula 12.46 and the evaluation r ( ^ ) = -yjr, that /2 •
/2 J—}^{x) — ^ /
COS X.
20. Show, by use of formula 12.46, that
dx that is, that Jk\x) "-Jkix)
(12.53)
=
-Jjt+iW-
X
21. Show, by use of formula 12.46, that
dx that is, that Jk'ix) -\--Jk(x)
(12.54)
-
Jk-i{x).
x 22- From relations 12.53 and 12.54, derive the further recurrence formulas
Jk-iix)
+ Jk+i(x) -
2k - Jkix), X
Jk^iix)
-
Jk+lix) = 2Jk'{x).
23, Show that in the product of the two series ^ 2 ^2=2! ^ 2 ' 3 ! .-x/2t
„ 1 _
£
I
2z^2W
^ ^
I
.
*
2^3
the total term in z° is Joix), and the total term in z is zJi(x). 24. Show that f^^ xJk{fimx)Jk{M
dx = 0.
if /3i, /32, 03, • • • , are the values of ;c for which Jk'ix) = 0, and /3m 7* p„.
ANSWERS TO ODD-NUMBERED PROBLEMS 6[-t n - i
2H* -
3} . . . {A -
n-
1}
244
Differential Equations with Parameters m
l 2 m } ! r C i * - m + 1) m>0
( - i r 2 ^ " ' r ( i A + |)
U2
|2m +
m-0
li!r(iA+i-m)
.Sftt+l
m-l
+
«2
(-l)n2'"2-5-8
+
i3m + l i !
00
7. Ul -
1
1 n-0
I2n + l l !
1
U2
X
n-0
9. Ul =• F
"2
11.
j
2
Ul
\2'2'2'
2
/
F l A + l , A + l ; 2 A + 2;
13. U l
''F(i,i;hx\
15. a = ^/x
UlJ^iix) +
f2-/-4a(*)i-
17. a = fiJ^,(2
