Student Solutions Manual to accompany Boyce Elementary Differential Equations and Boundary Value Problems

Math 219, Homework 2 Due date: 23.11.2005, Wednesday This homework concerns two (fictitious) design problems about the solar car “MES¸e” of the METU Robotics Society, which won the Formula-G trophy in September 2005. Just for the purposes of this homework, assume that they want to modify the car, and they are asking for your help on two issues. 1. The first problem is about the shock absorbing system of the car. We may model the shock absorber as a single linear spring. This question concerns how to adjust the damping coefficient in order to meet certain requirements. (a) It is known that when the pilot, weighing 80kg, gets into the car seat, the shock absorber is compressed by 5cm. From this data, compute the spring constant k (in kg/sec2 ).

(b) The car (without the pilot) weighs 240kg. Write a differential equation which governs the vertical motion of the car (this could for instance describe the vertical

displacement when the car goes over a speed bump). (Hint: Check “Damped free vibrations” from Boyce,Di Prima, section 3.8). (c) It is required that, when the car goes over a speed bump of 5cm high, the vertical displacement x(t) should approach the equilibrium point 0 in a way that for t ≥ 1sec, |x(t)| ≤ 1cm. For several values of the damping coefficient and for this initial condition, sketch the graph of the solution curve for 0 ≤ t ≤ 2sec. Finally, decide which values of the damping coefficient are allowable in order to meet this requirement. 2. The second problem is about the power supply of the motor. The panels convert solar energy to electrical energy and store it in the accumulator. Assume that the accumulator provides a voltage of E(t) = 48 cos(ωt) Volts to the system. The frequency ω is adjusted by the gas pedal. The circuit can be modeled as a series L − C circuit. Take L = 0.05Henry and C = 10−6 Farad. (a) Write a differential equation for the current I(t) through the voltage source (Hint: Check “Electric Circuits” from Boyce,Di Prima, section 3.8. We are assuming R = 0). (b) Solve this differential equation for I(t) in terms of ω using the method of undetermined coefficients. (c) Currents over 20 Amperes may harm the accumulator. Using your result from part (b), find out which frequencies (in Hertz) should not be allowed. For several values of ω graph I(t) for 0 ≤ t ≤ 0.2sec using ODE Architect, and denote which of these are potentially harmful and which are not.

Math 219, Homework 3 Due date: 9.12.2005, Friday 1. Consider the initial value problem d2 x dx + + x = u4 (t), dt2 dt

y(0) = y 0 (0) = 0

(a) Solve this initial value problem using the Laplace transform. dx dt with respect to t (You can use the function Step(t, 4) to create a unit step function with discontinuity at t = 4). (b) Use ODE Architect to solve the equation, and graph the solution. Also graph

(c) Discuss how the graphs agree with the solutions in (a): in particular determine dx (if any) all the points where x(t) and are discontinuous, behavior of these two dt functions for t → ∞, their maxima and minima. 2. Write each of the following systems of differential equations in matrix form, find the eigenvalues and eigenvectors of the coefficient matrices, and using these, find all solutions of each system. Also, graph the phase portraits (x − y graph) using ODE Architect. Please use a scale which includes the point (0, 0), and graph several solutions in order to clearly observe the behavior around (0, 0). Also, place arrows on the solution curves which indicate the direction of increasing t, and make sure that solution curves along the eigenvector directions are graphed if there are any real eigenvectors. (a) dx = 2x − y dt dy = 3x + 3y dt

(b) dx = −x + y dt dy = 3x − 4y dt (c) dx = 2x + 3y dt dy = 5x + 5y dt (d) dx = −4x + 3y dt dy = −3x + 2y dt (e) dx = −x − 3y dt dy = 2x + y dt

Math 219, Homework 4 Due date: 30.12.2005, Wednesday Suppose that K > 0, and f (t) is defined as ( f (t) =

1 0

if 4n ≤ t < 4n + 1 otherwise

where n runs through the set of integers. (a) Determine the Fourier series for f (t). (b) Consider the differential equation d2 x + x = f (t). dt2 By using ODE Architect, solve (and graph the solutions of) this equation for as many values of K as possible between 0.5 and 10 for 0 ≤ t ≤ 100.(You can enter f (t) in ODE Architect using the built in command SqW ave(t, L, K) and then set L = 4 ∗ K, and assign K definite values on the lines below). Record the maximum values of x(t) for each of these K’s, and plot a K vs. max(x(t)) graph by hand. (c) Which values of K result in a resonance in the system? (Hint: you should find 5 such values). Can you relate these values to the Fourier series terms? Please print the resonance graphs. (d) What do the Fourier coefficients correspond to on the resonance graphs?

We wish you a happy new year and success on your finals!

—————————————————————————— CHAPTER 1. ——

Chapter One Section 1.1 1.

For C  "Þ& , the slopes are negative, and hence the solutions decrease. For C  "Þ& , the slopes are positive, and hence the solutions increase. The equilibrium solution appears to be Ca>b œ "Þ& , to which all other solutions converge. 3.

For C   "Þ& , the slopes are :9=3tive, and hence the solutions increase. For C   "Þ& , the slopes are negative, and hence the solutions decrease. All solutions appear to diverge away from the equilibrium solution Ca>b œ  "Þ& . 5.

For C   "Î# , the slopes are :9=3tive, and hence the solutions increase. For C   "Î# , the slopes are negative, and hence the solutions decrease. All solutions diverge away from ________________________________________________________________________ page 1

—————————————————————————— CHAPTER 1. —— the equilibrium solution Ca>b œ  "Î# . 6.

For C   # , the slopes are :9=3tive, and hence the solutions increase. For C   # , the slopes are negative, and hence the solutions decrease. All solutions diverge away from the equilibrium solution Ca>b œ  # . 8. For all solutions to approach the equilibrium solution Ca>b œ #Î$ , we must have C w  ! for C  #Î$ , and C w  ! for C  #Î$ . The required rates are satisfied by the differential equation C w œ #  $C .

9. For solutions other than Ca>b œ # to diverge from C œ # , C a>b must be an increasing function for C  # , and a decreasing function for C  # . The simplest differential equation whose solutions satisfy these criteria is C w œ C  # . 10. For solutions other than Ca>b œ "Î$ to diverge from C œ "Î$ , we must have C w  ! for C  "Î$ , and C w  ! for C  "Î$ . The required rates are satisfied by the differential equation C w œ $C  " . 12.

Note that C w œ ! for C œ ! and C œ & . The two equilibrium solutions are C a>b œ ! and Ca>b œ & . Based on the direction field, C w  ! for C  & ; thus solutions with initial values greater than & diverge from the solution Ca>b œ & . For !  C  &, the slopes are negative, and hence solutions with initial values between ! and & all decrease toward the

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—————————————————————————— CHAPTER 1. —— solution Ca>b œ ! . For C  ! , the slopes are all positive; thus solutions with initial values less than ! approach the solution Ca>b œ ! . 14.

Observe that C w œ ! for C œ ! and C œ # . The two equilibrium solutions are C a>b œ ! and Ca>b œ # . Based on the direction field, C w  ! for C  # ; thus solutions with initial values greater than # diverge from Ca>b œ # . For !  C  #, the slopes are also positive, and hence solutions with initial values between ! and # all increase toward the solution Ca>b œ # . For C  ! , the slopes are all negative; thus solutions with initial values less than ! diverge from the solution Ca>b œ ! .

16. a+b Let Q a>b be the total amount of the drug ain milligramsb in the patient's body at any given time > a2

a71Î2b œ >  $ is a solution. 20.

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—————————————————————————— CHAPTER 1. ——

All solutions approach the equilibrium solution Ca>b œ ! Þ 23.

All solutions appear to diverge from the sinusoid Ca>b œ 

$ È#

=38Ð>  1% Ñ  " ,

which is also a solution corresponding to the initial value Ca!b œ  &Î# . 25.

All solutions appear to converge to Ca>b œ ! . First, the rate of change is small. The slopes eventually increase very rapidly in magnitude. 26.

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—————————————————————————— CHAPTER 1. ——

The direction field is rather complicated. Nevertheless, the collection of points at which the slope field is zero, is given by the implicit equation C$  'C œ #># Þ The graph of these points is shown below:

The y-intercepts of these curves are at C œ ! , „È' . It follows that for solutions with initial values C  È' , all solutions increase without bound. For solutions with initial values in the range C   È' and !  C  È' , the slopes remain negative, and hence these solutions decrease without bound. Solutions with initial conditions in the range  È'  C  ! initially increase. Once the solutions reach the critical value, given by the equation C$  'C œ #># , the slopes become negative and remain negative. These solutions eventually decrease without bound.

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—————————————————————————— CHAPTER 1. —— Section 1.2 1a+b The differential equation can be rewritten as .C œ .> Þ &C

Integrating both sides of this equation results in  68k&  C k œ >  -" , or equivalently, &  C œ - /> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ &  C! . Hence the solution is C a>b œ &  aC!  &b/> Þ

All solutions appear to converge to the equilibrium solution Ca>b œ & Þ 1a- bÞ Rewrite the differential equation as

.C œ .> Þ "!  #C

Integrating both sides of this equation results in  "# 68k"!  #C k œ >  -" , or equivalently, &  C œ - /#> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ &  C! . Hence the solution is C a>b œ &  aC!  &b/#> Þ

All solutions appear to converge to the equilibrium solution Ca>b œ & , but at a faster rate than in Problem 1a Þ 2a+bÞ The differential equation can be rewritten as

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—————————————————————————— CHAPTER 1. ——

.C œ .> Þ C&

Integrating both sides of this equation results in 68kC  &k œ >  -" , or equivalently, C  & œ - /> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ C!  & . Hence the solution is C a>b œ &  aC!  &b/> Þ

All solutions appear to diverge from the equilibrium solution Ca>b œ & . 2a,bÞ

Rewrite the differential equation as .C œ .> Þ #C  &

Integrating both sides of this equation results in "# 68k#C  &k œ >  -" , or equivalently, #C  & œ - /#> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ #C!  & . Hence the solution is C a>b œ #Þ&  aC!  #Þ&b/#> Þ

All solutions appear to diverge from the equilibrium solution Ca>b œ #Þ& . 2a- b. The differential equation can be rewritten as

.C œ .> Þ #C  "!

Integrating both sides of this equation results in "# 68k#C  "!k œ >  -" , or equivalently, C  & œ - /#> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ C!  & . Hence the solution is C a>b œ &  aC!  &b/#> Þ

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—————————————————————————— CHAPTER 1. ——

All solutions appear to diverge from the equilibrium solution Ca>b œ & . 3a+b. Rewrite the differential equation as

.C œ .> , ,  +C

which is valid for C Á , Î+. Integrating both sides results in +" 68k,  +C k œ >  -" , or equivalently, ,  +C œ - /+> . Hence the general solution is C a>b œ a,  - /+> bÎ+ Þ Note that if C œ ,Î+ , then .CÎ.> œ ! , and C a>b œ ,Î+ is an equilibrium solution. a, b

a3b As + increases, the equilibrium solution gets closer to Ca>b œ ! , from above. Furthermore, the convergence rate of all solutions, that is, + , also increases. a33b As , increases, then the equilibrium solution C a>b œ ,Î+ also becomes larger. In this case, the convergence rate remains the same. a333b If + and , both increase abut ,Î+ œ constantb, then the equilibrium solution Ca>b œ ,Î+ remains the same, but the convergence rate of all solutions increases.

5a+b. Consider the simpler equation .C" Î.> œ  +C" . As in the previous solutions, rewrite the equation as .C" œ  + .> Þ C"

Integrating both sides results in C" a>b œ - /+> Þ a,bÞ Now set Ca>b œ C" a>b  5 , and substitute into the original differential equation. We find that

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—————————————————————————— CHAPTER 1. ——  +C"  ! œ  +aC"  5 b  , . That is,  +5  , œ ! , and hence 5 œ ,Î+ . a- b. The general solution of the differential equation is Ca>b œ - /+>  ,Î+ Þ This is exactly the form given by Eq. a"(b in the text. Invoking an initial condition Ca!b œ C! , the solution may also be expressed as Ca>b œ ,Î+  aC!  ,Î+b/+> Þ

6a+b. The general solution is :a>b œ *!!  - />Î# , that is, :a>b œ *!!  a:!  *!!b/>Î# . With :! œ )&! , the specific solution becomes :a>b œ *!!  &!/>Î# . This solution is a decreasing exponential, and hence the time of extinction is equal to the number of months it takes, say >0 , for the population to reach zero. Solving *!!  &!/>0 Î# œ ! , we find that >0 œ # 68a*!!Î&!b œ &Þ() months. a,b The solution, :a>b œ *!!  a:!  *!!b/>Î# , is a decreasing exponential as long as :!  *!! . Hence *!!  a:!  *!!b/>0 Î# œ ! has only one root, given by >0 œ # 68Œ

*!! Þ *!!  :!

a- b. The answer in part a,b is a general equation relating time of extinction to the value of the initial population. Setting >0 œ "# months , the equation may be written as *!! œ /' , *!!  :! which has solution :! œ )*(Þ('*" . Since :! is the initial population, the appropriate answer is :! œ )*) mice .

7a+b. The general solution is :a>b œ :! / . Based on the discussion in the text, time > is measured in months . Assuming " month œ $! days , the hypothesis can be expressed as :! / œ " results in )#Þ!% œ "!! /< Þ Solving for the rate constant, we find that < œ  68a)#Þ!%Î"!!b œ Þ"*(*'/week or < œ Þ!#)#)/day . a- b. Let X be the time that it takes the isotope to decay to one-half of its original amount. From part a+b, it follows that &! œ "!! /<X , in which < œ Þ"*(*'/week . Taking the natural logarithm of both sides, we find that X œ $Þ&!"% weeks or X œ #%Þ&" .+C s .

11. The general solution of the differential equation .UÎ.> œ  < U is Ua>b œ U! / , in which U! œ Ua!b is the initial amount of the substance. Let 7 be the time that it takes the substance to decay to one-half of its original amount , U! . Setting > œ 7 in the solution, we have !Þ& U! œ U! / Þ Let X be the time that it takes the isotope to decay to $Î% of its original amount. Then setting > œ X, and UaX b œ $% Ua!b , we obtain $% Ua!b œ Ua!b/!Þ!!!%#()'X Þ Solving for the decay time, it follows that  !Þ!!!%#()' X œ 68a$Î%b or X œ '(#Þ$' years . 13. The solution of the differential equation, with Ua!b œ !, is Ua>b œ GZ a"  />ÎGV bÞ As > p _ , the exponential term vanishes, and hence the limiting value is UP œ GZ .

14a+b. The accumulation rate of the chemical is Ð!Þ!"Ña$!!b grams per hour . At any given time > , the concentration of the chemical in the pond is Ua>bÎ"!' grams per gallon . Consequently, the chemical leaves the pond at a rate of a$ ‚ "!% bUa>b grams per hour . Hence, the rate of change of the chemical is given by .U œ $  !Þ!!!$ Ua>b gm/hr . .>

Since the pond is initially free of the chemical, Ua!b œ ! .

a,b. The differential equation can be rewritten as

.U œ !Þ!!!$ .> Þ "!!!!  U

Integrating both sides of the equation results in  68k"!!!!  Uk œ !Þ!!!$>  G . Taking the natural logarithm of both sides gives "!!!!  U œ - /!Þ!!!$> . Since Ua!b œ ! , the value of the constant is - œ "!!!! . Hence the amount of chemical in the pond at any time is Ua>b œ "!!!!a"  /!Þ!!!$> b grams . Note that " year œ )('! hours . Setting > œ )('! , the amount of chemical present after one year is Ua)('!b œ *#((Þ(( grams , that is, *Þ#(((( kilograms . a- b. With the accumulation rate now equal to zero, the governing equation becomes .UÎ.> œ  !Þ!!!$ Ua>b gm/hr . Resetting the time variable, we now assign the new initial value as Ua!b œ *#((Þ(( grams .

a. b. The solution of the differential equation in Part a- b is Ua>b œ *#((Þ(( /!Þ!!!$> Þ Hence, one year after the source is removed, the amount of chemical in the pond is Ua)('!b œ '(!Þ" grams .

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—————————————————————————— CHAPTER 1. —— a/b. Letting > be the amount of time after the source is removed, we obtain the equation "! œ *#((Þ(( /!Þ!!!$> Þ Taking the natural logarithm of both sides,  !Þ!!!$ > œ œ 68a"!Î*#((Þ((b or > œ ##ß ((' hours œ #Þ' years .

a0 b

15a+b. It is assumed that dye is no longer entering the pool. In fact, the rate at which the dye leaves the pool is #!! † c; a>bÎ'!!!!d kg/min œ #!!a'!Î"!!!bc; a>bÎ'!d gm per hour . Hence the equation that governs the amount of dye in the pool is .; œ  !Þ# ; .>

a gm/hrb .

The initial amount of dye in the pool is ; a!b œ &!!! grams .

a,b. The solution of the governing differential equation, with the specified initial value, is ; a>b œ &!!! /!Þ# > Þ

a- b. The amount of dye in the pool after four hours is obtained by setting > œ % . That is, ; a%b œ &!!! /!Þ) œ ##%'Þ'% grams . Since size of the pool is '!ß !!! gallons , the concentration of the dye is !Þ!$(% grams/gallon . a. b. Let X be the time that it takes to reduce the concentration level of the dye to !Þ!# grams/gallon . At that time, the amount of dye in the pool is "ß #!! grams . Using the answer in part a,b, we have &!!! /!Þ# X œ "#!! . Taking the natural logarithm of both sides of the equation results in the required time X œ (Þ"% hours . a/b. Note that !Þ# œ #!!Î"!!! . Consider the differential equation .; < œ  ;. .> "!!! Here the parameter < corresponds to the flow rate, measured in gallons per minute . Using the same initial value, the solution is given by ; a>b œ &!!! /< >Î"!!! Þ In order to determine the appropriate flow rate, set > œ % and ; œ "#!! . (Recall that "#!! gm of ________________________________________________________________________ page 13

—————————————————————————— CHAPTER 1. —— dye has a concentration of !Þ!# gm/gal ). We obtain the equation "#!! œ &!!! /< Î#&! Þ Taking the natural logarithm of both sides of the equation results in the required flow rate < œ $&( gallons per minute .

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—————————————————————————— CHAPTER 1. —— Section 1.3 1. The differential equation is second order, since the highest derivative in the equation is of order two. The equation is linear, since the left hand side is a linear function of C and its derivatives. 3. The differential equation is fourth order, since the highest derivative of the function C is of order four. The equation is also linear, since the terms containing the dependent variable is linear in C and its derivatives. 4. The differential equation is first order, since the only derivative is of order one. The dependent variable is squared, hence the equation is nonlinear. 5. The differential equation is second order. Furthermore, the equation is nonlinear, since the dependent variable C is an argument of the sine function, which is not a linear function. 7. C" a>b œ /> Ê C"w a>b œ C"ww a>b œ /> . Hence C"ww  C" œ ! Þ Also, C# a>b œ -9=2 > Ê C"w a>b œ =382 > and C#ww a>b œ -9=2 > . Thus C#ww  C# œ ! Þ

9. Ca>b œ $>  ># Ê C w a>b œ $  #> . Substituting into the differential equation, we have >a$  #>b  a$>  ># b œ $>  #>#  $>  ># œ ># . Hence the given function is a solution.

10. C" a>b œ >Î$ Ê C"w a>b œ "Î$ and C"ww a>b œ C"www a>b œ C"wwww a>b œ ! Þ Clearly, C" a>b is a solution. Likewise, C# a>b œ />  >Î$ Ê C#w a>b œ  />  "Î$ , C#ww a>b œ /> , C#www a>b œ  /> , C#wwww a>b œ /> . Substituting into the left hand side of the equation, we find that />  %a  /> b  $a/>  >Î$b œ />  %/>  $/>  > œ > . Hence both functions are solutions of the differential equation. 11. C" a>b œ >"Î# Ê C"w a>b œ >"Î# Î# and C"ww a>b œ  >$Î# Î% . Substituting into the left hand side of the equation, we have #># ˆ  >$Î# Î% ‰  $>ˆ>"Î# Î# ‰  >"Î# œ  >"Î# Î#  $ >"Î# Î#  >"Î# œ!

Likewise, C# a>b œ >" Ê C#w a>b œ  ># and C#ww a>b œ # >$ . Substituting into the left hand side of the differential equation, we have #># a# >$ b  $>a  ># b  >" œ % >"   $ >"  >" œ ! . Hence both functions are solutions of the differential equation. 12. C" a>b œ ># Ê C"w a>b œ  #>$ and C"ww a>b œ ' >% . Substituting into the left hand side of the differential equation, we have ># a' >% b  &>a  #>$ b  % ># œ ' >#   "! >#  % ># œ ! . Likewise, C# a>b œ >2 68 > Ê C#w a>b œ >$  #>$ 68 > and C#ww a>b œ  & >%  ' >% 68 >. Substituting into the left hand side of the equation, we have ># a  & >%  ' >% 68 >b  &>a>$  #>$ 68 >b  %a>2 68 >b œ  & >2  ' >2 68 >  ________________________________________________________________________ page 15

—————————————————————————— CHAPTER 1. ——  & >2  "! >2 68 >  % >2 68 > œ ! Þ Hence both functions are solutions of the differential equation. 13. Ca>b œ a-9= >b68 -9= >  > =38 > Ê C w a>b œ  a=38 >b68 -9= >  > -9= > and C ww a>b œ  a-9= >b68 -9= >  > =38 >  =/- > . Substituting into the left hand side of the differential equation, we have a  a-9= >b68 -9= >  > =38 >  =/- >b  a-9= >b68 -9= >   > =38 > œ  a-9= >b68 -9= >  > =38 >  =/- >  a-9= >b68 -9= >  > =38 > œ =/- > . Hence the function Ca>b is a solution of the differential equation.

15. Let Ca>b œ / . Then C ww a>b œ b œ / . Then C w a>b œ b œ b œ <  %< ><  % >< œ ! . For > Á ! , we obtain the algebraic equation b œ =38 -B =38 -+> . Then the second derivatives are ` # ?" œ  -# =38 -B =38 -+> # `B ` # ?" œ  -# +# =38 -B =38 -+> `># #

It is easy to see that +# ``B?#" œ

` # ?" `>#

. Likewise, given ?# aBß >b œ =38aB  +>b , we have

` # ?# œ  =38aB  +>b `B# ` # ?# œ  +# =38aB  +>b # `>

Clearly, ?# aBß >b is also a solution of the partial differential equation.

28. Given the function ?aBß >b œ È1Î> /B Î%! > , the partial derivatives are #

#

È1Î> /B# Î%!# > È1Î> B# /B# Î%!# > ?BB œ   #! # > %! % > # # # # ! B Î% > È1> / È1 B /B# Î%!# > ?> œ   #># %!# ># È>

It follows that !# ?BB œ ?> œ 

È1 ˆ#!# >B# ‰/B# Î%!# > %! # > # È >

.

Hence ?aBß >b is a solution of the partial differential equation.

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—————————————————————————— CHAPTER 1. —— 29a+b.

a,bÞ The path of the particle is a circle, therefore polar coordinates are intrinsic to the problem. The variable < is radial distance and the angle ) is measured from the vertical. Newton's Second Law states that ! F œ 7a Þ In the tangential direction, the equation of motion may be expressed as ! J) œ 7 +) , in which the tangential acceleration, that is,

the linear acceleration along the path is +) œ P . # )Î.># Þ Ð +) is positive in the direction of increasing ) Ñ. Since the only force acting in the tangential direction is the component of weight, the equation of motion is  71 =38 ) œ 7P

.# ) Þ .>#

ÐNote that the equation of motion in the radial direction will include the tension in the rodÑ. a- b. Rearranging the terms results in the differential equation .# ) 1  =38 ) œ ! Þ # .> P

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—————————————————————————— CHAPTER 2. ——

Chapter Two Section 2.1 1a+bÞ

a,bÞ Based on the direction field, all solutions seem to converge to a specific increasing function. a- bÞ The integrating factor is .a>b œ /$> , and hence Ca>b œ >Î$  "Î*  /#>  - /$> Þ It follows that all solutions converge to the function C" a>b œ >Î$  "Î* Þ 2a+bÞ

a,b. All slopes eventually become positive, hence all solutions will increase without bound. a- bÞ The integrating factor is .a>b œ /#> , and hence Ca>b œ >$ /#> Î$  - /#> Þ It is evident that all solutions increase at an exponential rate. 3a+b

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—————————————————————————— CHAPTER 2. ——

a,b. All solutions seem to converge to the function C! a>b œ " Þ

a- bÞ The integrating factor is .a>b œ /#> , and hence Ca>b œ ># /> Î#  "  - /> Þ It is clear that all solutions converge to the specific solution C! a>b œ " .

4a+b.

a,b. Based on the direction field, the solutions eventually become oscillatory. a- bÞ The integrating factor is .a>b œ > , and hence the general solution is C a >b œ

$-9=a#>b $  =38a#>b  %> # >

in which - is an arbitrary constant. As > becomes large, all solutions converge to the function C" a>b œ $=38a#>bÎ# Þ

5a+b.

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—————————————————————————— CHAPTER 2. ——

a,b. All slopes eventually become positive, hence all solutions will increase without bound.

a- bÞ The integrating factor is .a>b œ /B:a  ' #.>b œ /#> Þ The differential equation can w be written as /#> C w  #/#> C œ $/> , that is, a/#> C b œ $/> Þ Integration of both sides of the equation results in the general solution Ca>b œ  $/>  - /#> Þ It follows that all solutions will increase exponentially. 6a+b

a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ

a- bÞ The integrating factor is .a>b œ ># , and hence the general solution is Ca>b œ 

-9=a>b =38a#>b   # # > > >

in which - is an arbitrary constant. As > becomes large, all solutions converge to the function C! a>b œ ! Þ 7a+b.

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—————————————————————————— CHAPTER 2. ——

a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ

a- bÞ The integrating factor is .a>b œ /B:a># b, and hence C a>b œ ># />  - /> Þ It is clear that all solutions converge to the function C! a>b œ ! . #

#

8a+b

a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ

a- bÞ Since .a>b œ a"  ># b# , the general solution is Ca>b œ c>+8" a>b  G dÎa"  ># b# Þ It follows that all solutions converge to the function C! a>b œ ! . 9a+bÞ

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—————————————————————————— CHAPTER 2. —— a,b. All slopes eventually become positive, hence all solutions will increase without bound.

a- bÞ The integrating factor is .a>b œ /B:ˆ' "# .>‰ œ />Î# . The differential equation can w be written as />Î# C w  />Î# CÎ# œ $> />Î# Î# , that is, ˆ/>Î# CÎ#‰ œ $> />Î# Î#Þ Integration of both sides of the equation results in the general solution Ca>b œ $>  '  - />Î# Þ All solutions approach the specific solution C! a>b œ $>  ' Þ

10a+b.

a,b. For C  ! , the slopes are all positive, and hence the corresponding solutions increase without bound. For C  ! , almost all solutions have negative slopes, and hence solutions tend to decrease without bound. a- bÞ First divide both sides of the equation by > . From the resulting standard form, the integrating factor is .a>b œ /B:ˆ  ' "> .>‰ œ "Î> . The differential equation can be written as C w Î>  CÎ># œ > /> , that is, a CÎ>bw œ > /> Þ Integration leads to the general solution Ca>b œ  >/>  - > Þ For - Á ! , solutions diverge, as implied by the direction field. For the case - œ ! , the specific solution is Ca>b œ  >/> , which evidently approaches zero as > p _ . 11a+b.

a,bÞ The solutions appear to be oscillatory. ________________________________________________________________________ page 22

—————————————————————————— CHAPTER 2. —— a- bÞ The integrating factor is .a>b œ /> , and hence Ca>b œ =38a#>b  # -9=a#>b  - /> Þ It is evident that all solutions converge to the specific solution C! a>b œ =38a#>b  # -9=a#>b . 12a+bÞ

a,b. All solutions eventually have positive slopes, and hence increase without bound.

a- bÞ The integrating factor is .a>b œ /#> . The differential equation can be w written as />Î# C w  />Î# CÎ# œ $># Î# , that is, ˆ/>Î# CÎ#‰ œ $># Î#Þ Integration of both sides of the equation results in the general solution Ca>b œ $>#  "#>  #%  - />Î# Þ It follows that all solutions converge to the specific solution C! a>b œ $>#  "#>  #% .

14. The integrating factor is .a>b œ /#> . After multiplying both sides by .a>b, the w equation can be written as ˆ/2> C‰ œ > Þ Integrating both sides of the equation results in the general solution Ca>b œ ># /#> Î#  - /#> Þ Invoking the specified condition, we require that /# Î#  - /# œ ! . Hence - œ  "Î# , and the solution to the initial value problem is Ca>b œ a>#  "b/#> Î# Þ 16. The integrating factor is .a>b œ /B:ˆ' #> .>‰ œ ># . Multiplying both sides by .a>b, w the equation can be written as a># Cb œ -9=a>b Þ Integrating both sides of the equation results in the general solution Ca>b œ =38a>bÎ>#  - ># Þ Substituting > œ 1 and setting the value equal to zero gives - œ ! . Hence the specific solution is Ca>b œ =38a>bÎ># Þ

17. The integrating factor is .a>b œ /#> , and the differential equation can be written as ˆ/2> C‰w œ " Þ Integrating, we obtain /2> C a>b œ >  - Þ Invoking the specified initial condition results in the solution Ca>b œ a>  #b/#> Þ 19. After writing the equation in standard 0 orm, we find that the integrating factor is .a>b œ /B:ˆ' %> .>‰ œ >% . Multiplying both sides by .a>b, the equation can be written as ˆ>% C‰w œ > /> Þ Integrating both sides results in >% C a>b œ  a>  "b/>  - Þ Letting > œ  " and setting the value equal to zero gives - œ ! Þ Hence the specific solution of the initial value problem is Ca>b œ  ˆ>$  >% ‰/> Þ 21a+b.

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—————————————————————————— CHAPTER 2. ——

The solutions appear to diverge from an apparent oscillatory solution. From the direction field, the critical value of the initial condition seems to be +! œ  " . For +   " , the solutions increase without bound. For +   " , solutions decrease without bound. a,bÞ The integrating factor is .a>b œ />Î# . The general solution of the differential equation is Ca>b œ a)=38a>b  %-9=a>bbÎ&  - />Î# . The solution is sinusoidal as long as - œ ! . The initial value of this sinusoidal solution is +! œ a)=38a!b  %-9=a!bbÎ& œ  %Î& Þ a- bÞ See part a,b. 22a+bÞ

All solutions appear to eventually increase without bound. The solutions initially increase or decrease, depending on the initial value + . The critical value seems to be +! œ  " Þ a,bÞ The integrating factor is .a>b œ />Î# , and the general solution of the differential equation is Ca>b œ  $/>Î$  - />Î# Þ Invoking the initial condition C a!b œ + , the solution may also be expressed as Ca>b œ  $/>Î$  a+  $b />Î# Þ Differentiating, follows that C w a!b œ  "  a+  $bÎ# œ a+  "bÎ# Þ The critical value is evidently +! œ  " Þ

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—————————————————————————— CHAPTER 2. —— a- b. For +! œ  " , the solution is Ca>b œ  $/>Î$  # />Î# , which afor large >b is dominated by the term containing />Î# Þ is Ca>b œ a)=38a>b  %-9=a>bbÎ&  - />Î# . 23a+bÞ

As > p ! , solutions increase without bound if C a"b œ +  Þ% , and solutions decrease without bound if Ca"b œ +  Þ% Þ

> ‰ a,b. The integrating factor is .a>b œ /B:ˆ' >" > .> œ > / Þ The general solution of the differential equation is Ca>b œ > />  - /> Î> . Invoking the specified value C a"b œ + , we have "  - œ + / . That is, - œ + /  " . Hence the solution can also be expressed as Ca>b œ > />  a+ /  "b /> Î> . For small values of > , the second term is dominant. Setting + /  " œ ! , critical value of the parameter is +! œ "Î/ Þ

a- b. For +  "Î/ , solutions increase without bound. For +  "Î/ , solutions decrease without bound. When + œ "Î/ , the solution is C a>b œ > /> , which approaches ! as > p ! . 24a+b.

As > p ! , solutions increase without bound if C a"b œ +  Þ% , and solutions decrease without bound if Ca"b œ +  Þ% Þ ________________________________________________________________________ page 25

—————————————————————————— CHAPTER 2. —— a,b. Given the initial condition, Ca  1Î#b œ + , the solution is Ca>b œ a+ 1# Î%  -9= >bÎ> Þ Since lim -9= > œ " , solutions increase without bound if +  %Î1# , and solutions >Ä!

decrease without bound if +  %Î1# Þ Hence the critical value is +! œ %Î1# œ !Þ%&#)%(ÞÞÞ.

a- bÞ For + œ %Î1# , the solution is C a>b œ a"  -9= >bÎ> , and lim C a>b œ "Î# . Hence the solution is bounded.

>Ä!

25. The integrating factor is .a>b œ /B:ˆ' "# .>‰ œ />Î# Þ Therefore general solution is Ca>b œ c%-9=a>b  )=38a>bdÎ&  - />Î# Þ Invoking the initial condition, the specific solution is Ca>b œ c%-9=a>b  )=38a>b  * />Î# dÎ& . Differentiating, it follows that C w a>b œ   %=38a>b  )-9=a>b  %Þ& />Î# ‘Î& C ww a>b œ   %-9=a>b  )=38a>b  #Þ#& />Î# ‘Î&

Setting C w a>b œ ! , the first solution is >" œ "Þ$'%$ , which gives the location of the first stationary point. Since C ww a>" b  ! , the first stationary point in a local maximum. The coordinates of the point are a"Þ$'%$ ß Þ)#!!)b.

26. The integrating factor is .a>b œ /B:ˆ' #$ .>‰ œ /#>Î$ , and the differential equation can w be written as a/#>Î$ Cb œ /#>Î$  > /#>Î$ Î# Þ The general solution is C a>b œ Ð#"  '>ÑÎ)   - /#>Î$ . Imposing the initial condition, we have C a>b œ Ð#"  '>ÑÎ)  aC!  #"Î)b/#>Î$ . Since the solution is smooth, the desired intersection will be a point of tangency. Taking the derivative, C w a>b œ  $Î%  a#C!  #"Î%b/#>Î$ Î$ Þ Setting C w a>b œ ! , the solution is >" œ $# 68ca#"  )C! bÎ*dÞ Substituting into the solution, the respective value at the stationary point is Ca>" b œ $#  *% 68 $  *) 68a#"  )C! b. Setting this result equal to zero, we obtain the required initial value C! œ a#"  * /%Î$ bÎ) œ  "Þ'%$ Þ 27. The integrating factor is .a>b œ />Î4 , and the differential equation can be written as w a/>Î4 Cb œ $ />Î4  # />Î4 -9=a#>bÞ The general solution is Ca>b œ "#  c)-9=a#>b  '%=38a#>bdÎ'&  - />Î4 Þ

Invoking the initial condition, Ca!b œ ! , the specific solution is

Ca>b œ "#  )-9=a#>b  '%=38a#>b  ()) />Î4 ‘Î'& Þ

As > p _ , the exponential term will decay, and the solution will oscillate about an average value of "# , with an amplitude of )ÎÈ'& Þ

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—————————————————————————— CHAPTER 2. —— 29. The integrating factor is .a>b œ /$>Î# , and the differential equation can be written w as a/$>Î# Cb œ $> /$>Î#  # />Î# Þ The general solution is C a>b œ  #>  %Î$  % />   - /$>Î# Þ Imposing the initial condition, C a>b œ  #>  %Î$  % />  aC!  "'Î$b /$>Î# Þ As > p _ , the term containing /$>Î# will dominate the solution. Its sign will determine the divergence properties. Hence the critical value of the initial condition is C! œ  "'Î$Þ The corresponding solution, Ca>b œ  #>  %Î$  % /> , will also decrease without bound. Note on Problems 31-34 : Let 1a>b be given, and consider the function Ca>b œ C" a>b  1a>b , in which C" a>b p _ as > p _ . Differentiating, C w a>b œ C"w a>b  1 w a>b . Letting + be a constant, it follows that C w a>b  +C a>b œ C"w a>b  +C" a>b  1 w a>b  +1a>bÞ Note that the hypothesis on the function C" a>b will be satisfied, if C"w a>b  +C" a>b œ ! . That is, C" a>b œ - /+> Þ Hence Ca>b œ - /+>  1a>b, which is a solution of the equation C w  +C œ 1 w a>b  +1a>bÞ For convenience, choose + œ " . 31. Here 1a>b œ $ , and we consider the linear equation C w  C œ $ Þ The integrating w factor is .a>b œ /> , and the differential equation can be written as a/> C b œ $/> Þ The general solution is Ca>b œ $  - /> Þ

33. 1a>b œ $  > Þ Consider the linear equation C w  C œ  "  $  > ÞThe integrating w factor is .a>b œ /> , and the differential equation can be written as a/> C b œ a#  >b/> Þ The general solution is Ca>b œ $  >  - /> Þ 34. 1a>b œ %  ># Þ Consider the linear equation C w  C œ %  #>  ># ÞThe integrating w factor is .a>b œ /> , and the equation can be written as a/> C b œ a%  #>  ># b/> Þ The general solution is Ca>b œ %  >#  - /> Þ

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—————————————————————————— CHAPTER 2. —— Section 2.2 2. For B Á  " , the differential equation may be written as C .C œ cB# Îa"  B$ bd.B Þ Integrating both sides, with respect to the appropriate variables, we obtain the relation C# Î# œ "$ 68k"  B$ k  - Þ That is, C aBb œ „É #$ 68k"  B$ k  - Þ

3. The differential equation may be written as C# .C œ  =38 B .B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation  C " œ -9= B  - Þ That is, aG  -9= BbC œ ", in which G is an arbitrary constant. Solving for the dependent variable, explicitly, CaBb œ "ÎaG  -9= Bb . 5. Write the differential equation as -9=# #C .C œ -9=# B .B, or =/- # #C .C œ -9=# B .BÞ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation >+8 #C œ =38 B -9= B  B  - Þ 7. The differential equation may be written as aC  /C b.C œ aB  /B b.B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C #  # / C œ B #  # / B  - Þ 8. Write the differential equation as a"  C# b.C œ B# .B Þ Integrating both sides of the equation, we obtain the relation C  C $ Î$ œ B$ Î$  - , that is, $C  C $ œ B$  GÞ 9a+b. The differential equation is separable, with C# .C œ a"  #Bb.B Þ Integration yields  C" œ B  B#  - Þ Substituting B œ ! and C œ  "Î' , we find that - œ ' Þ Hence the specific solution is C" œ B#  B  ' . The explicit form is CaBb œ "ÎaB#  B  'bÞ a, b

a- b. Note that B#  B  ' œ aB  #baB  $b . Hence the solution becomes singular at B œ  # and B œ $ Þ 10a+bÞ CaBb œ  È#B  #B#  % Þ

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—————————————————————————— CHAPTER 2. —— 10a,bÞ

11a+bÞ Rewrite the differential equation as B /B .B œ  C .C Þ Integrating both sides of the equation results in B /B  /B œ  C # Î#  - Þ Invoking the initial condition, we obtain - œ  "Î# Þ Hence C # œ #/B  #B /B  "Þ The explicit form of the solution is CaBb œ È#/B  #B /B  " Þ The positive sign is chosen, since C a!b œ "Þ a, bÞ

a- bÞ The function under the radical becomes negative near B œ  "Þ( and B œ !Þ(' Þ

11a+bÞ Write the differential equation as a"  >b" .> Þ Integrating both sides of the equation, we obtain 68 kCk  68kC  %k œ %>  %68k"  >k  - Þ Taking the exponential of both sides, it follows that kCÎaC  %bk œ G /%> Îa"  >b% Þ It follows that as > p _ , kCÎaC  %bk œ k"  %ÎaC  %bkp _ . That is, C a>b p % Þ

a,bÞ Setting Ca!b œ # , we obtain that G œ " . Based on the initial condition, the solution may be expressed as CÎaC  %b œ  /%> Îa"  >b% Þ Note that CÎaC  %b  ! , for all >   !Þ Hence C  % for all >   !Þ Referring back to the differential equation, it follows that C w is always positive. This means that the solution is monotone increasing. We find that the root of the equation /%> Îa"  >b% œ $** is near > œ #Þ)%% Þ a- bÞ Note the Ca>b œ % is an equilibrium solution. Examining the local direction field,

we see that if Ca!b  ! , then the corresponding solutions converge to C œ % . Referring back to part a+b, we have CÎaC  %b œ cC! ÎaC!  %bd/%> Îa"  >b% , for C! Á % Þ Setting % > œ # , we obtain C! ÎaC!  %b œ a$Î/# b C a#bÎaC a#b  %bÞ Now since the function 0 aCb œ CÎaC  %b is monotone for C  % and C  % , we need only solve the equations C! ÎaC!  %b œ  $**a$Î/# b% and C! ÎaC!  %b œ %!"a$Î/# b% Þ The respective solutions are C! œ $Þ''## and C! œ %Þ%!%# Þ 30a0 bÞ

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—————————————————————————— CHAPTER 2. —— 31a- b

" 32a+b. Observe that aB#  $C # bÎ#BC œ "# ˆ BC ‰  is homogeneous.

$C #B

Þ Hence the differential equation

a,b. The substitution C œ B @ results in @  B @ w œ aB#  $B# @# bÎ#B# @ . The transformed equation is @ w œ a"  @# bÎ#B@ Þ This equation is separable, with general solution @#  " œ - B Þ In terms of the original dependent variable, the solution is B#  C # œ - B$ Þ a- bÞ

33a- bÞ

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—————————————————————————— CHAPTER 2. —— " 34a+bÞ Observe that  a%B  $C bÎÐ#B  CÑ œ  #  BC #  BC ‘ Þ Hence the differential equation is homogeneous.

a,b. The substitution C œ B @ results in @  B @ w œ  #  @Îa#  @b. The transformed equation is @ w œ  a@#  &@  %bÎÐ#  @ÑB Þ This equation is separable, with general solution a@%b# k@"k œ GÎB$ Þ In terms of the original dependent variable, the solution is a%B  Cb# kBCk œ GÞ a- bÞ

35a- b.

36a+bÞ Divide by B# to see that the equation is homogeneous. Substituting C œ B @ , we obtain B @ w œ a"  @b# Þ The resulting differential equation is separable.

a,bÞ Write the equation as a"  @b# .@ œ B" .B Þ Integrating both sides of the equation, we obtain the general solution  "Îa"  @b œ 68kBk  - Þ In terms of the original dependent variable, the solution is C œ B cG  68kBkd"  B Þ

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—————————————————————————— CHAPTER 2. —— a- bÞ

" 37a+bÞ The differential equation can be expressed as C w œ "# ˆ BC ‰  #$ BC . Hence the equation is homogeneous. The substitution C œ B @ results in B @ w œ a"  &@# bÎ#@ . #@ " Separating variables, we have "&@ # .@ œ B .B Þ

a,bÞ Integrating both sides of the transformed equation yields  "& 68k"  &@# k œ 68kBk  - , that is, "  &@# œ GÎkBk& Þ In terms of the original dependent variable, the general solution is &C# œ B#  GÎkBk$ Þ a- bÞ

" 38a+bÞ The differential equation can be expressed as C w œ $# BC  #" ˆ BC ‰ . Hence the equation is homogeneous. The substitution C œ B @ results in B @ w œ a@#  "bÎ#@, that is, @##@" .@ œ B" .B Þ

a,bÞ Integrating both sides of the transformed equation yields 68k@#  "k œ 68kBk  - , that is, @#  " œ G kBkÞ In terms of the original dependent variable, the general solution is C# œ G B# kBk  B# Þ

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—————————————————————————— CHAPTER 2. —— a- bÞ

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—————————————————————————— CHAPTER 2. —— Section 2.3 5a+b. Let U be the amount of salt in the tank. Salt enters the tank of water at a rate of # "% ˆ"  "# =38 >‰ œ "#  "% =38 > 9DÎ738 Þ It leaves the tank at a rate of # UÎ"!! 9DÎ738Þ Hence the differential equation governing the amount of salt at any time is .U " " œ  =38 >  UÎ&! Þ .> # % The initial amount of salt is U! œ &! 9D Þ The governing ODE is linear, with integrating w factor .a>b œ />Î&! Þ Write the equation as ˆ/>Î&! U‰ œ />Î&! ˆ "#  "% =38 >‰Þ The specific solution is Ua>b œ #&  "#Þ&=38 >  '#&-9= >  '$"&! />Î&! ‘Î#&!" 9D Þ a, bÞ

a- bÞ The amount of salt approaches a steady state, which is an oscillation of amplitude "Î% about a level of #& 9D Þ 6a+b. The equation governing the value of the investment is .WÎ.> œ < W . The value of the investment, at any time, is given by W a>b œ W! / Þ Setting W aX b œ #W! , the required time is X œ 68a#bÎ< Þ

a,bÞ For the case < œ (% œ Þ!( , X ¸ *Þ* C œ Þ!* W  *ß '!! Þ Given that W! is the original amount borrowed, the debt is W a>b œ W! /Þ!*>  "!'ß ''(a/Þ!*>  "bÞ Setting W a$!b œ ! , it follows that W! œ $ **ß &!! . a,bÞ The total payment, over $! years, becomes $ #))ß !!! . The interest paid on this purchase is $ "))ß &!! .

13a+b. The balance increases at a rate of < W $/yr, and decreases at a constant rate of 5 $ per year. Hence the balance is modeled by the differential equation .WÎ.> œ <W  5 . The balance at any time is given by W a>b œ W! /  5< a/  "bÞ a,b. The solution may also be expressed as W a>b œ ÐW!  5< Ñ/  5< Þ Note that if the withdrawal rate is 5! œ < W! , the balance will remain at a constant level W! Þ 5 a- b. Assuming that 5  5! , W aX! b œ ! for X! œ "< 68’ 55 “Þ !

a. b. If < œ Þ!) and 5 œ #5! , then X! œ )Þ'' years .

a/b. Setting W a>b œ ! and solving for / in Parta,b, / œ results in 5 œ <W! /<X Îa/<X  "bÞ

5 5<W! Þ

Now setting > œ X

a0 bÞ In parta/b, let 5 œ "#ß !!! , < œ Þ!) , and X œ #! . The required investment becomes W! œ $ ""*ß ("& . 14a+bÞ Let U w œ  < U Þ The general solution is Ua>b œ U! / Þ Based on the definition of half-life, consider the equation U! Î# œ U! /&($! < Þ It follows that

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—————————————————————————— CHAPTER 2. ——  &($! < œ 68Ð"Î#Ñ, that is, < œ "Þ#!*( ‚ "!% per year. a,b. Hence the amount of carbon-14 is given by Ua>b œ U! /"Þ#!*(‚"! > Þ %

a- bÞ Given that UaX b œ U! Î& , we have the equation "Î& œ /"Þ#!*(‚"! X Þ Solving for the decay time, the apparent age of the remains is approximately X œ "$ß $!%Þ'& years . %

15. Let T a>b be the population of mosquitoes at any time > . The rate of increase of the mosquito population is œ represents days . The solution is T a>b œ T! /  #!ß!!! < a/  "bÞ In the absence of predators, the governing equation is .T" Î.> œ b œ T! / Þ Based on the data, set T" a(b œ #T! , that is, #T! œ T! /(< Þ The growth rate is determined as < œ 68a#bÎ( œ Þ!**!# per dayÞ Therefore the population, including the predation by birds, is T a>b œ # ‚ "!& /Þ!**>  #!"ß **(a/Þ!**>  "b œ œ #!"ß **(Þ$  "*((Þ$ /Þ!**> Þ 16a+b. Ca>b œ /B:c#Î"!  >Î"!  #-9=Ð>ÑÎ"!dÞ The doubling-time is 7 ¸ #Þ*'$# Þ

a,bÞ The differential equation is .CÎ.> œ CÎ"! , with solution C a>b œ C a!b/>Î"! Þ The doubling-time is given by 7 œ "!68a#b ¸ 'Þ*$"& Þ

a- b. Consider the differential equation .CÎ.> œ a!Þ&  =38Ð#1>Ñb CÎ& Þ The equation is separable, with C" .C œ ˆ!Þ"  &" =38Ð#1>щ.> Þ Integrating both sides, with respect to the appropriate variable, we obtain 68 C œ a1>  -9=Ð#1>ÑbÎ"!1  - Þ Invoking the initial condition, the solution is Ca>b œ /B:ca"  1>  -9=Ð#1>ÑbÎ"!1dÞ The doubling-time is 7 ¸ 'Þ$)!% Þ The doubling-time approaches the value found in parta,b. a. b .

17a+b. The differential equation .CÎ.> œ b C  5 is linear, with integrating factor .a>b œ /B:  ' b.>‘Þ Write the equation as a. C bw œ  5 .a>b Þ Integration of both

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—————————————————————————— CHAPTER 2. —— sides yields the general solution C œ   5 ' .a7 b. 7  C! .a!b‘Î.a>b . In this problem, the integrating factor is .a>b œ /B:ca-9= >  >bÎ&dÞ

a,bÞ The population becomes extinct, if Ca>‡ b œ ! , for some > œ >‡ . Referring to parta+b, we find that Ca>‡ b œ ! Ê (

>‡ !

/B:ca-9= 7  7 bÎ&d. 7 œ & /"Î& C- Þ

It can be shown that the integral on the left hand side increases monotonically, from zero to a limiting value of approximately &Þ!)*$ . Hence extinction can happen only if & /"Î& C-  &Þ!)*$ , that is, C-  !Þ)$$$ Þ a- b. Repeating the argument in parta,b, it follows that Ca>‡ b œ ! Ê (

!

>‡

/B:ca-9= 7  7 bÎ&d. 7 œ

" "Î& / C- Þ 5

Hence extinction can happen only if /"Î& C- Î5  &Þ!)*$ , that is, C-  %Þ"''( 5 Þ

a. b. Evidently, C- is a linear function of the parameter 5 .

19a+b. Let Ua>b be the volume of carbon monoxide in the room. The rate of increase of CO is aÞ!%ba!Þ"b œ !Þ!!% 0 >$ Î738 Þ The amount of CO leaves the room at a rate of a!Þ"bUa>bÎ"#!! œ Ua>bÎ"#!!! 0 >$ Î738 Þ Hence the total rate of change is given by the differential equation .UÎ.> œ !Þ!!%  Ua>bÎ"#!!! Þ This equation is linear and separable, with solution Ua>b œ %)  %) /B:a  >Î"#!!!b 0 >$ Þ Note that U! œ ! 0 >$ Þ Hence the concentration at any time is given by Ba>b œ Ua>bÎ"#!! œ Ua>bÎ"# % . a,b. The concentration of CO in the room is Ba>b œ %  %/B:a  >Î"#!!!b %Þ A level of !Þ!!!"# corresponds to !Þ!"# % . Setting Ba7 b œ !Þ!"# , the solution of the equation %  %/B:a  >Î"#!!!b œ !Þ!"# is 7 ¸ $' minutes .

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—————————————————————————— CHAPTER 2. —— 20a+bÞ The concentration is - a>b œ 5  T Î<  a-!  5  T Îp_b œ 5  T Î< Þ a,bÞ - a>b œ -! /ÎZ . The reduction times are X&! œ 68Ð#ÑZ Î< and X"! œ 68Ð"!ÑZ Î" œ "Þ')$ sec . Integrating @a>b , the position is given by Ba>b œ $")Þ%&  %%Þ" >  #))Þ%& /B:a  >Î%Þ&bÞ Hence the maximum height is Ba>" b œ %&Þ() m . a,bÞ Setting Ba># b œ ! , the ball hits the ground at ># œ &Þ"#) sec .

a- bÞ

23a+bÞ The differential equation for the upward motion is 7 .@Î.> œ  . @#  71 , in which . œ "Î"$#& . This equation is separable, with . @#7 71 .@ œ  .> Þ Integrating

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—————————————————————————— CHAPTER 2. —— both sides and invoking the initial condition, @a>b œ %%Þ"$$ >+8aÞ%#&  Þ### >bÞ Setting @a>" b œ ! , the ball reaches the maximum height at >" œ "Þ*"' sec . Integrating @a>b , the position is given by Ba>b œ "*)Þ(& 68c-9=a!Þ### >  !Þ%#&bd  %)Þ&( Þ Therefore the maximum height is Ba>" b œ %)Þ&' m .

a,bÞ The differential equation for the downward motion is 7 .@Î.> œ  .@#  71 Þ This equation is also separable, with 717. @# .@ œ  .> Þ For convenience, set > œ ! at the top of the trajectory. The new initial condition becomes @a!b œ ! . Integrating both sides and invoking the initial condition, we obtain 68ca%%Þ"$  @bÎa%%Þ"$  @bd œ >Î#Þ#& Þ Solving for the velocity, @a>b œ %%Þ"$a"  />Î#Þ#& bÎa"  />Î#Þ#& b Þ Integrating @a>b , the # position is given by Ba>b œ **Þ#* 68’/>Î#Þ#& Îa"  />Î#Þ#& b “  ")'Þ# Þ To estimate the

duration of the downward motion, set Ba># b œ ! , resulting in ># œ $Þ#(' sec . Hence the total time that the ball remains in the air is >1  ># œ &Þ"*# sec . a- bÞ

24a+bÞ Measure the positive direction of motion downward . Based on Newton's #nd law, the equation of motion is given by 7

.@  !Þ(& @  71 , !  >  "! œœ  "# @  71 , >  "! .>

Þ

Note that gravity acts in the positive direction, and the drag force is resistive. During the first ten seconds of fall, the initial value problem is .@Î.> œ  @Î(Þ&  $# , with initial velocity @a!b œ ! fps Þ This differential equation is separable and linear, with solution @a>b œ #%!a"  />Î(Þ& b. Hence @a"!b œ "('Þ( fps Þ a,b. Integrating the velocity, with Ba>b œ ! , the distance fallen is given by Hence Ba"!b œ "!(%Þ& ft .

Ba>b œ #%! >  ")!! />Î(Þ&  ")!! .

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—————————————————————————— CHAPTER 2. —— a- bÞ For computational purposes, reset time to > œ ! . For the remainder of the motion, the initial value problem is .@Î.> œ  $#@Î"&  $# , with specified initial velocity @a!b œ "('Þ( fps Þ The solution is given by @a>b œ "&  "'"Þ( /$# >Î"& Þ As > p _ , @a>b p @P œ "& fps Þ Integrating the velocity, with Ba!b œ "!(%Þ& , the distance fallen after the parachute is open is given by Ba>b œ "& >  (&Þ) /$# >Î"&  ""&!Þ$ Þ To find the duration of the second part of the motion, estimate the root of the transcendental equation "& X  (&Þ) /$# X Î"&  ""&!Þ$ œ &!!! Þ The result is X œ #&'Þ' sec Þ a. bÞ

25a+b. Measure the positive direction of motion upward . The equation of motion is given by 7.@Î.> œ  5 @  71 . The initial value problem is .@Î.> œ  5@Î7  1 , with @a!b œ @! . The solution is @a>b œ  71Î5  a@!  71Î5 b/5>Î7 Þ Setting @a>7 b œ !, the maximum height is reached at time >7 œ a7Î5 b68ca71  5 @! bÎ71dÞ Integrating the velocity, the position of the body is Ba>b œ  71 >Î5  ”Š

7 # 7 @! 5>Î7 ÑÞ ‹ 1 •Ð"  / 5 5

Hence the maximum height reached is B7 œ Ba>7 b œ

7 @! 7 # 71  5 @!  1Š ‹ 68” •Þ 5 5 71

a,bÞ Recall that for $ ¥ " , 68a"  $ b œ $  "# $ #  "$ $ $  "% $ %  á 26a,b. a- bÞ

lim 71a5 @!571b/

5Ä !

lim  

7Ä!

71 5

5>Î7

œ lim  5Ä !

> 7 a5 @ !

 71b/5>Î7 œ  1> Þ

‰ 5>Î7 ‘ œ ! , since  ˆ 71 5 @! /

lim /5>Î7 œ ! Þ

7Ä!

28a+b. In terms of displacement, the differential equation is 7@ .@Î.B œ  5 @  71 Þ .@ .B .@ This follows from the chain rule : .@ .> œ .B .> œ @ .> . The differential equation is separable, with

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—————————————————————————— CHAPTER 2. —— Ba@b œ 

7@ 7# 1 71  5 @  # 68º ºÞ 5 5 71

The inverse exists, since both B and @ are monotone increasing. In terms of the given parameters, Ba@b œ  "Þ#& @  "&Þ$" 68k!Þ!)"' @  "kÞ

a,bÞ Ba"!b œ "$Þ%& meters . The required value is 5 œ !Þ#% . a- bÞ In parta+b, set @ œ "! m/s and B œ "! meters .

29a+bÞ Let B represent the height above the earth's surface. The equation of motion is Q7 given by 7 .@ .> œ  K aVBb# , in which K is the universal gravitational constant. The

symbols Q and V are the mass and radius of the earth, respectively. By the chain rule, 7@

.@ Q7 œ K . .B aV  Bb#

This equation is separable, with @ .@ œ  KQ aV  Bb# .B Þ Integrating both sides, and invoking the initial condition @a!b œ È#1V , the solution is @# œ #KQ aV  Bb"   #1V  #KQ ÎV Þ From elementary physics, it follows that 1 œ KQ ÎV # . Therefore @aBb œ È#1 ’VÎÈV  B “Þ a Note that 1 œ ()ß &%& mi/hr# .b a,bÞ We now consider .BÎ.> œ È#1 ’VÎÈV  B “. This equation is also separable, with ÈV  B .B œ È#1 V .> Þ By definition of the variable B, the initial condition is #Î$ Ba!b œ !Þ Integrating both sides, we obtain Ba>b œ  $# ˆÈ#1 V >  #$ V $Î# ‰‘  V Þ Setting the distance BaX b  V œ #%!ß !!! , and solving for X , the duration of such a flight would be X ¸ %* hours .

32a+bÞ Both equations are linear and separable. The initial conditions are @a!b œ ? -9= E and Aa!b œ ? =38 E . The two solutions are @a>b œ ? -9= E / and Aa>b œ  1Î<   a? =38 E  1Îb œ ?< -9= Ea"  / b and a- bÞ

? Ca>b œ  1>Î<  ˆ1  ?< =38 E  2b

.a=b1a=b .=.

a,b. By definition, .a">b œ /B:ˆ  ' :a>b.>‰. Hence C"w œ  :a>b That is, C"w  :a>bC" œ !Þ a- bÞ C#w œ Š  :a>b

" '> .a>b ‹ ! .a=b1a=b .=

That is, C#w  :a>bC# œ 1a>bÞ

" .a>b

œ  :a>bC" Þ

 Š .a">b ‹.a>b1a>b œ  :a>bC#  1a>bÞ $

.C C .@ $ .C 30. Since 8 œ $, set @ œ C # . It follows that .@ .> œ  #C .> and .> œ  # .> Þ $ $ Substitution into the differential equation yields  C# .@ .>  &C œ  5 C , which further results in @ w  #&@ œ #5 Þ The latter differential equation is linear, and can be written as w a/#&> b œ #5 Þ The solution is given by @a>b œ #5> /#&>  -/#&> Þ Converting back to the original dependent variable, C œ „@"Î# Þ $

.C C .@ $ .C 31. Since 8 œ $, set @ œ C # . It follows that .@ .> œ  #C .> and .> œ  # .> Þ $ $ The differential equation is written as  C# .@ .>  a>-9= >  X bC œ 5 C , which upon further substitution is @ w  #a>-9= >  X b@ œ #Þ This ODE is linear, with integrating factor .a>b œ /B:ˆ#' a>-9= >  X b.>‰ œ /B:a  #>=38 >  #X >bÞ The solution is

@a>b œ #/B:a#>=38 >  #X >b( /B:a  #>=38 7  #X 7 b. 7  - /B:a  #>=38 >  #X >bÞ >

!

Converting back to the original dependent variable, C œ „@"Î# Þ 33. The solution of the initial value problem C"w  #C" œ !, C" a!b œ " is C" a>b œ /#> Þ Therefore ya" b œ C" a"b œ /# Þ On the interval a"ß _bß the differential equation is C#w  C# œ !, with C# a>b œ -/> Þ Therefore C a" b œ C# a"b œ -/" Þ Equating the limits Ca" b œ Ca" b, we require that - œ /" Þ Hence the global solution of the initial value problem is Ca>b œ œ

/#> , ! Ÿ > Ÿ " Þ /"> , >  "

Note the discontinuity of the derivative C a >b œ œ

 #/#> , !  >  " Þ  /"> , >  "

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—————————————————————————— CHAPTER 2. —— Section 2.5 1.

For C!   ! , the only equilibrium point is C ‡ œ ! . 0 w a!b œ +  ! , hence the equilibrium solution 9a>b œ ! is unstable. 2.

The equilibrium points are C‡ œ  +Î, and C ‡ œ ! . 0 w a  +Î, b  ! , therefore the equilibrium solution 9a>b œ  +Î, is asymptotically stable. 3.

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—————————————————————————— CHAPTER 2. —— 4.

The only equilibrium point is C‡ œ ! . 0 w a!b  ! , hence the equilibrium solution 9a>b œ ! is unstable. 5.

The only equilibrium point is C‡ œ ! . 0 w a!b  ! , hence the equilibrium solution 9a>b œ ! is +=C7:>9>3-+66C stable. 6.

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—————————————————————————— CHAPTER 2. —— 7a,b.

8.

The only equilibrium point is C‡ œ " . Note that 0 w a"b œ ! , and that C w  ! for C Á " . As long as C! Á " , the corresponding solution is monotone decreasing. Hence the equilibrium solution 9a>b œ " is semistable. 9.

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—————————————————————————— CHAPTER 2. —— 10.

The equilibrium points are C‡ œ ! , „" . 0 w aC b œ "  $C # . The equilibrium solution 9a>b œ ! is unstable, and the remaining two are asymptotically stable. 11.

12.

The equilibrium points are C‡ œ ! , „# . 0 w aC b œ )C  %C $ . The equilibrium solutions 9a>b œ  # and 9a>b œ  # are unstable and asymptotically stable, respectively. The equilibrium solution 9a>b œ ! is semistable.

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—————————————————————————— CHAPTER 2. —— 13.

The equilibrium points are C‡ œ ! and " . 0 w aC b œ #C  'C #  %C $ . Both equilibrium solutions are semistable. 15a+b. Inverting the Solution a""b, Eq. a"$b shows > as a function of the population C and the carrying capacity O . With C! œ OÎ$, " a"Î$bc"  aCÎO bd > œ  68º ºÞ < aCÎO bc"  a"Î$bd

Setting C œ #C! ,

" a"Î$bc"  a#Î$bd 7 œ  68º ºÞ < a#Î$bc"  a"Î$bd

That is, 7 œ "< 68 % Þ If < œ !Þ!#& per year, 7 œ &&Þ%& years.

a,bÞ In Eq. a"$b, set C! ÎO œ ! and CÎO œ " . As a result, we obtain " ! c"  " d X œ  68º ºÞ < " c"  ! d

Given ! œ !Þ", " œ !Þ* and < œ !Þ!#& per year, 7 œ "(&Þ() years. 16a+b.

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—————————————————————————— CHAPTER 2. —— 17. Consider the change of variable ? œ 68aCÎO bÞ Differentiating both sides with respect to >, ? w œ C w ÎCÞ Substitution into the Gompertz equation yields ? w œ 