Boyce Elementary Differential Equations. Solutions

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 C > œ "Þ&a b , 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 C > œ  "Þ&a b .

the equilibrium solution C > œ  "Î#a b . 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 C > œ  #a b .

8. For solutions to approach the equilibrium solution all C > œ #Î$a b , we must have C  !w for C  #Î$, and C  !w for C  #Î$. The required rates are satisfied by the differential equation C œ #  $Cw .

9. For solutions other than C > œ #a b 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 œ C  #w .

10. For solutions other than C > œ "Î$a b to diverge from C œ "Î$, we must have C  !w for C  "Î$, and C  !w for C  "Î$. The required rates are satisfied by the differential equation .C œ $C  "w

12.

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

solution C > œ !a b . For C  !, the slopes are all positive; thus solutions with initial values

less than approach the solution ! C > œ !a b . 14.

Observe that C œ !w for C œ ! and C œ #. The two equilibrium solutions are C > œ !a b and C > œ #a b . Based on the direction field, C  !w for C  #; thus solutions with initial values greater than diverge from # C > œ #a b . For !  C  #, the slopes are also

positive, and hence solutions with initial values between and all increase toward the! # solution

C > œ #a b . For C  !, the slopes are all negative; thus solutions with initial values less than diverge from the solution ! C > œ !a b .

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 > 2<=a b. The drug is administered into the body at a constant rate of &!! 71Î2<Þ

The rate at which the drug leaves the bloodstream is given by !Þ%Q > Þa b Hence the accumulation rate of the drug is described by the differential equation

.Q

.> œ &!!  !Þ% Q a71Î2< Þb a b,

7.@ œ 71  @ .> # # or equivalently, .@ .> œ 1  7@ Þ # _#

a b, After a long time, .@ ¸ ! Þ Hence the object attains a given by

.> terminal velocity

@ œ_{_ Ê}71 Þ #

a b- Using the relation #@ œ 71_{_}# , the required _{drag coefficient} is # œ !Þ!%!) 51Î=/- Þ a b.

19.

All solutions appear to approach a linear asymptote aA3>2 =69:/ /;?+6 >9 "b. It is easy to verify that C > œ >  $a b is a solution.

All solutions approach the equilibrium solution C > œ ! Þ_{a b} 23.

All solutions appear to diverge from the sinusoid C > œ a b _È$_# =38Ð>  Ñ  "_%1 , which is also a solution corresponding to the initial value C ! œ  &Î#a b . 25.

All solutions appear to converge to C > œ !a b . First, the rate of change is small. The slopes

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.

Section 1.2

1a b+ The differential equation can be rewritten as .C

&  C œ .> Þ

Integrating both sides of this equation results in  68 &  C œ >  -k k ", or equivalently, &  C œ - />. Applying the initial condition C ! œ Ca b ! results in the specification of

the constant as - œ &  C!. Hence the solution is C > œ &  C  & /a b a ! b >Þ

All solutions appear to converge to the equilibrium solution C > œ & Þa b 1_{a b}- Þ Rewrite the differential equation as

.C

"!  #C œ .> Þ

Integrating both sides of this equation results in  68 "!  #C œ >  -"_# k k ", or equivalently,

&  C œ - /#>_{. Applying the initial condition} C ! œ Ca b _{results in the specification of}

!

the constant as - œ &  C!. Hence the solution is C > œ &  C  & /a b a ! b #>Þ

All solutions appear to converge to the equilibrium solution C > œ &a b , but at a faster rate than in Problem 1a Þ

.C

C  & œ .> Þ

Integrating both sides of this equation results in 68 C  & œ >  -k k ", or equivalently, C  & œ - />_{. Applying the initial condition} C ! œ Ca b _{results in the specification of}

!

the constant as - œ C  &! . Hence the solution is C > œ &  C  & / Þa b a ! b >

All solutions appear to diverge from the equilibrium solution C > œ &a b . 2a b, Þ Rewrite the differential equation as

.C

#C  & œ .> Þ

Integrating both sides of this equation results in " , or equivalently,

#68 #C  & œ >  -k k "

#C  & œ - /#>. Applying the initial condition C ! œ Ca b ! results in the specification of

the constant as - œ #C  &! . Hence the solution is C > œ #Þ&  C  #Þ& / Þa b a ! b #>

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

.C

#C  "! œ .> Þ

Integrating both sides of this equation results in " , or equivalently,

#68 #C  "! œ >  -k k "

C  & œ - /#>. Applying the initial condition C ! œ Ca b ! results in the specification of

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

.C

,  +C œ .> ,

which is valid for C Á , Î+. Integrating both sides results in  68 ,  +C œ >  - , or

"

"

+ k k

equivalently, ,  +C œ - /+>. Hence the general solution is C > œ ,  - /a b a +>bÎ+ Þ Note that if C œ ,Î+, then .CÎ.> œ !, and C > œ ,Î+a b is an equilibrium solution. a b,

a b3 As increases, the equilibrium solution gets closer to + C > œ !a b , from above. Furthermore, the convergence rate of all solutions, that is, , also increases.+

a b33 As increases, then the equilibrium solution , C > œ ,Î+a b also becomes larger. In this case, the convergence rate remains the same.

_{a b}333 If and both increase + , _abut constant,Î+ œ _b, then the equilibrium solution C > œ ,Î+a b 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, re-write the equation as

.C

C œ  + .> Þ

" "

 +C  ! œ  + C  5  ," a " b .

That is,  +5  , œ !, and hence 5 œ ,Î+.

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

6a b+ . The general solution is : > œ *!!  - /a b >Î#, that is, : > œ *!!  :  *!! /a b a b >Î#.

!

With : œ )&!! , the specific solution becomes : > œ *!!  &!/a b >Î#. This solution is a decreasing exponential, and hence the time of extinction is equal to the number of months

it takes, say , for the population to reach >0 zero. Solving *!!  &!/>0Î# œ !, we find that > œ # 68 *!!Î&! œ &Þ()0 a b months.

a b, The solution, : > œ *!!  :  *!! /a b a ! b >Î#, is a decreasing exponential as long as :  *!!! . Hence *!!  :  *!! /a ! b >0Î#œ ! has only one root, given by

> œ # 68 *!! Þ *!!  :

0

!

Œ 

a b- . The answer in part a b, is a general equation relating time of extinction to the value of

the initial population. Setting > œ "#0 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 : > œ : /a b ! <>. Based on the discussion in the text, time is> measured in months. Assuming "monthœ $!days, the hypothesis can be expressed as : /! <†" œ #:!. Solving for the rate constant, < œ 68 #a b, with units of per month .

a b, . daysR œ R Î$! months . The hypothesis is stated mathematically as : /! <N/30 œ #:!

.

It follows that <R Î$! œ 68 #a b, and hence the rate constant is given by < œ $! 68 # ÎR Þa b The units are understood to be per month.

9_{a b}+ . Assuming no air resistance, with the positive direction taken as downward, Newton's

Second Law can be expressed as

7.@ œ 71 .>

written as .@Î.> œ 1, with solution @ > œ 1>  @ Þa b ! The object is released with an initial

velocity .@!

a b, . Suppose that the object is released from a height of 2units above the ground. Using the

fact that @ œ .BÎ.>, in which is the B downward displacement of the object, we obtain the

differential equation for the displacement as .BÎ.> œ 1>  @ Þ! With the origin placed at the point of release, direct integration results in B > œ 1> Î#  @ >a b # ! . Based on the chosen

coordinate system, the object reaches the ground when B > œ 2a b . Let > œ X be the time that it takes the object to reach the ground. Then 1X Î#  @ X œ 2# . Using the

!

quadratic

formula to solve for ,X

X œ  @ „ @  #12 Þ 1

! È !

The positive answer corresponds to the time it takes for the object to fall to the ground. The

negative answer represents a previous instant at which the object could have been launched

upward awith the same impact speedb, only to ultimately fall downward with speed ,@!

from a height of 2 units above the ground.

a b- . The impact speed is calculated by substituting > œ X into @ >a b in part a b+ Þ That is, @ X œa b È@  #12! .

10 , . The general solution of the differential equation is a b+b U > œ - /a b <>Þ Given that U ! œ "!!a b mg, the value of the constant is given by - œ "!!. Hence the amount of thorium-234 present at any time is given by U > œ "!! /a b <>. Furthermore, based on the hypothesis, setting > œ " results in )#Þ!% œ "!! / Þ< Solving for the rate constant, we find that < œ  68 )#Þ!%Î"!! œ Þ"*(*'a b /week /dayor < œ Þ!#)#) .

a b- . Let be the time that it takes the isotope to decay to X 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&".+Cs . 11. The general solution of the differential equation .UÎ.> œ  < U is U > œ U /a b ! <>, in which U œ U !a b is the initial amount of the substance. Let be the time that it takes7

12. The differential equation governing the amount of radium-226 is .UÎ.> œ  < U, with solution U > œ U ! /a b a b <>Þ Using the result in Problem 11, and the fact that the half-life 7 œ "'#!years, the decay rate is given by < œ 68 # Î"'#!a b per year. The amount of radium-226, after years, is therefore > U > œ U ! /a b a b !Þ!!!%#()'>Þ Let beX the time that it takes the isotope to decay to $Î% of its original amount. Then setting > œ X ,

and U X œ U !a b $_% a b, we obtain $_%U ! œ U ! /a b a b !Þ!!!%#()'XÞ Solving for the decay time, it follows that  !Þ!!!%#()' X œ 68 $Î%a b or X œ '(#Þ$'years .

13. The solution of the differential equation, with U ! œ !a b , is U > œ GZ "  /a b a  ÎGV> bÞ

As > p _, the exponential term vanishes, and hence the limiting value is U œ GZP . 14_{a b}+ . The accumulation rate of the chemical is Ð!Þ!"Ñ $!!_{a b}grams per hour. At any given time , the > concentration of the chemical in the pond is U > Î"!a b 'grams per gallon .

Consequently, the chemical leaves the pond at a rate of a$ ‚ "!%b a bU > grams per hour. Hence, the rate of change of the chemical is given by

.U

.> œ $  !Þ!!!$ U >a b gm/hr . Since the pond is initially free of the chemical, U ! œ !a b . a b, . The differential equation can be rewritten as

.U

"!!!!  U œ !Þ!!!$ .> Þ

Integrating both sides of the equation results in  68 "!!!!  U œ !Þ!!!$>  Gk k . Taking

the natural logarithm of both sides gives "!!!!  U œ - /!Þ!!!$>. Since U ! œ !a b , the value of the constant is - œ "!!!!. Hence the amount of chemical in the pond at any time

is U > œ "!!!! "  /a b a !Þ!!!$>b_grams. Note that "_{yearœ )('! hours}. Setting

> œ )('!, the amount of chemical present after one year is U )('! œ *#((Þ((a b grams, that is, *Þ#(((( kilograms.

a b- . With the accumulation zerorate now equal to , the governing equation becomes .UÎ.> œ  !Þ!!!$ U >a b gm/hr . Resetting the time variable, we now assign the new initial value as U ! œ *#((Þ((a b grams.

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

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 68 "!Î*#((Þ((a b or > œ ##ß (('hoursœ #Þ'years. a b0

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 #!! † ; > Î'!!!!c a b dkg/minœ #!! '!Î"!!! ; > Î'!a bc a b d gm per hour .

Hence the equation that governs the amount of dye in the pool is .;

.> œ  !Þ# ; agm/hr .b The initial amount of dye in the pool is ; ! œ &!!!a b grams.

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

a b- . The amount of dye in the pool after four hours is obtained by setting > œ %. That is, ; % œ &!!! /a b !Þ)œ ##%'Þ'% grams. Since size of the pool is '!ß !!!gallons, the concentration of the dye is !Þ!$(%grams/gallon.

a b. . Let be the time that it takes to reduce the concentration level of the dye toX !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 .; <

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.

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 and its derivatives.C

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 is an argument of the C sine function, which is not a linear function.

7. C > œ / Ê C > œ C > œ /"a b > "wa b "wwa b >. Hence C  C œ ! Þ"ww "

Also, C > œ -9=2 > Ê C > œ =382 >#a b "wa b and C > œ -9=2 >#wwa b . Thus C  C œ ! Þ#ww #

9. C > œ $>  > Ê C > œ $  #>a b # wa b . Substituting into the differential equation, we have > $  #>  $>  >a b a #bœ $>  #>  $>  > œ ># # #_{. Hence the given function is a solution.}

10. C > œ >Î$ Ê C > œ "Î$"a b "wa b and C > œ C"wwa b "wwwa b> œ C"wwwwa b> œ ! Þ Clearly, C >"a b is a solution. Likewise, C > œ /#a b >  >Î$ Ê C > œ  /#wa b > "Î$ C > œ /, #wwa b > , C_#wwwa b> œ  />, C_#wwwwa b> œ />. Substituting into the left hand side of the equation, we find that /> %  /a >b $ /a > >Î$ œ /b > %/>  $/> > œ >. Hence both functions are solutions of the differential equation.

11. C > œ >"a b "Î#Ê C > œ >"wa b "Î#Î# and C > œ  >"wwa b $Î#Î%. Substituting into the left hand side of the equation, we have

#>  > Î%  $> > Î#  > œ  > Î#  $ > Î#  > œ !

#ˆ $Î# ‰ ˆ "Î# ‰ "Î# "Î# "Î# "Î#

Likewise, C > œ >#a b " Ê C > œ  >#wa b # and C > œ # >#wwa b $. Substituting into the left hand side of the differential equation, we have #> # >#a $b $>  >a #b >" œ % >"

 $ >"  >" œ !_{. Hence both functions are solutions of the differential equation.} 12. C > œ >"a b # Ê C > œ  #>"wa b $ and C > œ ' >"wwa b %. Substituting into the left hand side of the differential equation, we have > ' >#a %b &>  #>a $b % ># œ ' >#

 & >2 "! > 68 >  % > 68 > œ ! Þ2 2 _{Hence both functions are solutions of the}

differential equation.

13. C > œ -9= > 68 -9= >  > =38 > Ê C > œ  =38 > 68 -9= >  > -9= >a b a b wa b a b and

C > œ  -9= > 68 -9= >  > =38 >  =/- >wwa b a b _{. Substituting into the left hand side of the}

differential equation, we have a -9= > 68 -9= >  > =38 >  =/- >  -9= > 68 -9= > a b b a b  > =38 > œ  -9= > 68 -9= >  > =38 >  =/- >  -9= > 68 -9= >  > =38 > œ =/- >a b a b . Hence the function C >a b is a solution of the differential equation.

15. Let C > œ /a b <>. Then C > œ < /wwa b # <>, and substitution into the differential equation results in < /  # / œ !# <> <> . Since / Á !<> , we obtain the algebraic equation <  # œ !Þ# The roots of this equation are < œ „ 3 # Þ"ß# È

17. C > œ / Ê C > œ < /a b <> wa b <> and C > œ < /wwa b # <>. Substituting into the differential equation, we have < /  </  ' / œ !# <> <> <> . Since / Á !<> , we obtain the algebraic equation <  <  ' œ !# , that is, a<  # <  $ œ !ba b . The roots are < œ  $ #, .

"ß#

18. Let C > œ /a b <>. Then C > œ </wa b <>, C > œ < /wwa b # <> and Cwwwa b> œ < /$ <>. Substituting the derivatives into the differential equation, we have < /  $< /  #</ œ !$ <> # <> <> . Since / Á !<> _{, we obtain the algebraic equation} <  $<  #< œ ! Þ$ # _{By inspection, it follows}

that < <  " <  # œ !a ba b . Clearly, the roots are < œ ! < œ "" , # and < œ # Þ$

20. C > œ > Ê C > œ < >a b < wa b <" and C > œ < <  " >wwa b a b <#. Substituting the derivatives into the differential equation, we have > < <  " >#c a b <#d %> < >a <"b % > œ !< . After some algebra, it follows that < <  " >  %< >  % > œ !a b < < < . For > Á !, we obtain the algebraic equation <  &<  % œ ! Þ# The roots of this equation are < œ " and < œ % Þ

" #

21. The order of the partial differential equation is two, since the highest derivative, in fact each one of the derivatives, is of second order. The equation is linear, since the left hand side is a linear function of the partial derivatives.

23. The partial differential equation is fourth order, since the highest derivative, and in fact each of the derivatives, is of order four. The equation is linear, since the left hand side is a linear function of the partial derivatives.

24. The partial differential equation is second order, since the highest derivative of the function ? Bß Ca b is of order two. The equation is nonlinear, due to the product ? † ?B on the left hand side of the equation.

25. ? Bß C œ -9= B -9=2 C Ê"a b ` ?<sub>`B</sub> œ  -9= B -9=2 C and ` ?<sub>`C</sub> œ -9= B -9=2 C Þ

# #

" "

# #

It is evident that ` ? ` ? Likewise, given , the second

`B `C # #

# #

" "

#  # œ ! Þ ? Bß C œ 68 B  C#a b a b

` ? # %B `B œ B  C  B  C ` ? # %C `C œ B  C  B  C # # # # # # # # # # # # # # 2 2 a b a b # #

Adding the partial derivatives,

` ? ` ? # %B # %C `B  `C œ B  C  B  C  B  C  B  C œ %  % B  C B  C œ ! # # # # # # # # # # # # # # # # # # 2 2 a b a b a b # # # # # aB  C b .

Hence ? Bß C#a b is also a solution of the differential equation.

27. Let ? Bß > œ =38 B =38 +>"a b - - . Then the second derivatives are ` ? `B œ  =38 B =38 +> ` ? `> œ  + =38 B =38 +> # # # # # # # " " - - -- -

-It is easy to see that +# œ . Likewise, given ? Bß > œ =38 B  +> , we have

# ` ? ` ? `B `> # # #" #" a b a b ` ? `B œ  =38 B  +> ` ? `> œ  + =38 B  +> # # # # # # # a b a b

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

28. Given the function ? Bß > œa b È1Î> /B Î%# _!#>, the partial derivatives are ? œ  Î> /  Î> B / # > % > ? œ  > /  #> B / % > > BB B Î% > # B Î% > # % # > B Î% > # # B Î% > # # È È È È È 1 1 ! ! 1 1 ! # # # # # # # # ! ! ! ! It follows that !#_{? œ ? œ } . BB > È ˆ1 #!_%!>B /_>ȉ_> # # B Î%# #> # # !

29a b+ .

a b, Þ 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 !Fœ 7 Þa 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 . Since the only force acting in the tangential direction is the component) Ñ 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 rod .Ñ

a b- . Rearranging the terms results in the differential equation

. 1 .>  P=38 œ ! Þ # # ) )

Chapter Two

Section 2.1 1a b+ Þ

a b, Þ Based on the direction field, all solutions seem to converge to a specific increasing function.

a b- Þ The integrating factor is .a b> œ /$>, and hence C > œ >Î$  "Î*  /a b #> - /$>Þ It follows that all solutions converge to the function C > œ >Î$  "Î* Þ"a b

2a b+ Þ

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

a b- Þ The integrating factor is .a b> œ /#>, and hence C > œ > / Î$  - / Þa b $ #> #> It is evident that all solutions increase at an exponential rate.

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

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

4_{a b}+ .

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

C > œ $-9= #>  =38 #> $ -%> # >

a b a b a b

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

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

a b- Þ The integrating factor is .a b> œ /B:  #.> œ /a ' b #>Þ The differential equation can

be written as /#> wC  #/#>C œ $/>, that is, a/#>C œ $/ Þbw > Integration of both sides of the equation results in the general solution C > œ  $/  - / Þa b > #> It follows that all solutions will increase exponentially.

6a b+

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

a b- Þ The integrating factor is .a b> œ >#, and hence the general solution is C > œ  -9= >  =38 #> 

-> > > a b a b a b_{# #}

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

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

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

8_{a b}+

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

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

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

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

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 b- Þ 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 Î>  CÎ> œ > /w # >, that is, aCÎ> œ > / Þbw > Integration leads to the general solution C > œ  >/a b > - > Þ For - Á !, solutions diverge, as implied by the direction field. For the case - œ !, the specific solution is C > œ  >/a b >_{, which evidently}

approaches zeroas .> p _ 11a b+ .

a b- Þ The integrating factor is .a b> œ />, and hence C > œ =38 #>  # -9= #>  - / Þa b a b a b >

It is evident that all solutions converge to the specific solution C > œ =38 #>  #!a b a b

-9= #>a b . 12a b+ Þ

a b, . All solutions eventually have positive slopes, and hence increase without bound. a b- Þ The integrating factor is .a b> œ /#>. The differential equation can be

written as />Î# wC  />Î#CÎ# œ $> Î## , that is, ˆ/>Î#CÎ# œ $> Î#Þ‰w # Integration of both sides of the equation results in the general solution C > œ $>  "#>  #%  - /a b # >Î#Þ It follows that all solutions converge to the specific solution C > œ $>  "#>  #%!a b # . 14. The integrating factor is .a b> œ /#>. After multiplying both sides by .a b> , the equation can be written as ˆ/ C2> ‰w œ > Þ Integrating both sides of the equation results in the general solution C > œ > /a b # #>Î#  - /#>Þ Invoking the specified condition, we require that / Î#  - /# # œ !. Hence - œ  "Î#, and the solution to the initial value problem is C > œ >  " /a b a # b #>Î# Þ

16. The integrating factor is .a b> œ /B:ˆ' #_>.> œ >‰ #. Multiplying both sides by .a b> , the equation can be written as a b> C œ -9= > Þ# w a b Integrating both sides of the equation results in the general solution C > œ =38 > Î>  - > Þa b a b # # Substituting > œ ₁ and setting the value equal to zero gives - œ !. Hence the specific solution is C >a b œ =38 > Î> Þa b # 17. The integrating factor is .a b> œ /#>, and the differential equation can be written as

Integrating, we obtain Invoking the specified initial ˆ/2>C‰w œ " Þ /2>C > œ >  - Þa b

condition results in the solution C > œ >  # / Þa b a b #>

19. After writing the equation in standard orm0 , 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 > œ  >  " /% a b 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 C > œ  >a b ˆ $ >%‰/ Þ>

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 b, Þ The integrating factor is .a b> œ / Î#> . The general solution of the differential equation is C > œ )=38 >  %-9= > Î&  - /a b a a b a bb >Î#. The solution is sinusoidal as long as - œ !. The initial value of this sinusoidal solution is

+ œ! a)=38 !  %-9= ! Î& œ  %Î& Þa b a bb

a b- Þ See part a b, . 22a b+ Þ

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 b, Þ The integrating factor is .a b> œ / Î#> , and the general solution of the differential equation is C > œ  $/a b >Î$ - />Î#Þ Invoking the initial condition C ! œ +a b , the solution

a b- . For + œ  "! , the solution is C > œ  $/a b >Î$ # />Î#, which for large isa >b dominated by the term containing / Þ>Î#

is .C > œ )=38 >  %-9= > Î&  - /a b a a b a bb >Î#

23a b+ Þ

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

a b, . The integrating factor is .a b> œ /B:ˆ' >"_> .> œ > / Þ‰ > The general solution of the differential equation is C > œ > /a b > - / Î>> . Invoking the specified value C " œ +a b , we have "  - œ + /. That is, - œ + /  ". Hence the solution can also be expressed as C > œ > /a b > + /  " / Î>a b > . 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 > œ > /a b >_{, which approaches as} ! > p ! .

24a b+ .

a b, . Given the initial condition, Ca Î#1 bœ +, the solution is C > œ +a b a 1#Î% -9= > Î>b Þ

Since lim , solutions increase without bound if , and solutions

>Ä

#

!-9= > œ " +  %Î1

decrease without bound if +  %Î Þ1# Hence the critical value is + œ %Î! 1# œ !Þ%&#)%(ÞÞÞ.

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

>Ä

lim

!

solution is bounded.

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

C > œ  %=38 >  )-9= >  %Þ& / Î& C > œ  %-9= >  )=38 >  #Þ#& / Î& w > ww > a b  a b a b ‘ a b  a b a b ‘  Î#  Î#

Setting C > œ !wa b , the first solution is > œ "Þ$'%$, which gives the location of the

" first

stationary point. Since C >wwa b ! . The

" , the first stationary point in a local maximum

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

be written as a/# Î$> C œ /bw # Î$>  > /# Î$> Î# Þ The general solution is C > œa b _{Ð#"  '>ÑÎ) }

- /# Î$ C > œÐ#"  '>ÑÎ)  C  #"Î) /# Î$

!

> _{. Imposing the initial condition, we have} a b _{a b} > _.

Since the solution is smooth, the desired intersection will be a point of tangency. Taking the derivative, C > œ  $Î%  #C  #"Î% /wa b a b > Î$ Þ Setting C > œ !wa b , the solution

! # Î$

is > œ 68 #"  )C Î* Þ" $_# ca !b d Substituting into the solution, the respective value at the stationary point is C >a b" œ $_#  68 $  68 #"  )C*_% *₎ a !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 a/>Î4C œ $ /bw >Î4 # / -9= #> Þ>Î4 a b _{The general solution is}

C > œ "#  )-9= #>  '%=38 #> Î'&  - /a b c a b a bd  Î>4Þ Invoking the initial condition, C ! œ !a b , the specific solution is

C > œ "#  )-9= #>  '%=38 #>  ()) /a b  a b a b  Î>4‘Î'& Þ

As > p _, the exponential term will decay, and the solution will oscillate about an average

29. The integrating factor is .a b> œ /$ Î#> , and the differential equation can be written

as a/$ Î#> C œ $> /bw $ Î#>  # / Î#> Þ The general solution is C > œ  #>  %Î$  % / a b >

 - /$ Î#Þ C > œ  #>  %Î$  % /  C  "'Î$ /$ Î#Þ

!

> _{Imposing the initial condition,} a b > a b >

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, C > œ  #>  %Î$  % /a b >, will also decrease without bound.

Note on Problems 31-34 :

Let 1 >a b be given, and consider the function C > œ C >  1 >a b "a b a b, in which C > p _"a b

as > p _. Differentiating, C > œ C >  1 >wa b wa b wa b. Letting be a +

" constant, it follows

that C >  +C > œ C >  +C >  1 >  +1 > Þwa b a b wa b a b wa b a b

" " Note that the hypothesis on the

function C >"a b will be satisfied, if C >  +C > œ !"wa b "a b . That is, C > œ - /"a b +>Þ Hence C > œ - /a b +> 1 >a b_{, which is a solution of the equation} C  +C œ 1 >  +1 > Þw wa b a b

For convenience, choose + œ ".

31. Here 1 > œ $a b , and we consider the linear equation C  C œ $ Þw The integrating factor is .a b> œ />, and the differential equation can be written as a/ C œ $/ Þ> bw > The general solution is C > œ $  - / Þa b >

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

34. 1 > œ %  > Þa b # Consider the linear equation C  C œ %  #>  > Þw # The integrating factor is .a b> œ />, and the equation can be written as a/ C œ %  #>  > / Þ> bw a #b >

Section 2.2

2. For B Á  ", the differential equation may be written as C .C œcB Î "  B# a $bd.B Þ Integrating both sides, with respect to the appropriate variables, we obtain the relation C Î# œ 68# "  - Þ C B œ „ #68  - Þ

$ k"  B$k That is, a b É$ k"  B$k

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= B C œ "b _{, in which is an arbitrary constant.}G Solving for the dependent variable, explicitly, C B œ "Î G  -9= Ba b a b .

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 œ B  /Cb a Bb.B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation

C  # / œ B  # /# C # 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 "  #B .B Þ# a b 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

C B œ "Îa b aB  B  '# bÞ a b,

10a b, Þ

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

a b, Þ

a b- Þ The function under the radical becomes negative near B œ  "Þ( and B œ !Þ(' Þ 11a b+ Þ Write the differential equation as < .< œ# ₎". Þ₎ Integrating both sides of the equation results in the relation  <" œ 68  - Þ₎ Imposing the condition < " œ #a b , we obtain . - œ  "Î# The explicit form of the solution is <a b) œ #Î "  # 68a )bÞ

a b, Þ

a b- Þ Clearly, the solution makes sense only if )  ! Þ Furthermore, the solution becomes singular when 68 œ "Î#) , that is, ) œÈ/ Þ

13a b a b+ Þ C B œ  È# 68 "  Ba #b % Þ a b, Þ

14a b+ . Write the differential equation as C .C œ B "  B$ a #b"Î#.B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation

 C Î# œ# È"  B  - Þ# _{Imposing the initial condition, we obtain} - œ  $Î# Þ Hence the specific solution can be expressed as C# œ $  #È"  B Þ# The _explicit form of the solution is C B œ "Îa b É$  #È"  B# Þ The positive _{sign is chosen to} satisfy the initial condition.

a b, Þ

a b- Þ The solution becomes singular when #È"  B œ $# _{. That is, at} B œ „È& Î# Þ 15a b a b+ Þ C B œ  "Î#  ÈB  "&Î% Þ#

a b, Þ

16a b+ . Rewrite the differential equation as %C .C œ B B  " .B Þ$ a # b Integrating both sides

of the equation results in C œ% aB  "# b#Î%  - Þ Imposing the initial condition, we obtain - œ ! Þ Hence the solution may be expressed as aB  "# b# %C œ ! Þ% The _explicit form of the solution is C B œ a b ÈaB  " Î## b Þ The sign is chosen based on C ! œa b  "ÎÈ#Þ

a b, Þ

a b- Þ The solution is valid for all B −‘. 17a b a b+ Þ C B œ  &Î#  ÈB  /  "$Î% Þ$ B

a b, Þ

a b- . The solution is valid for B   "Þ%& Þ This value is found by estimating the root of %B  %/  "$ œ ! Þ$ B

18a b+ . Write the differential equation as a$  %C .C œ /b a B / .B ÞBb Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation $C  #C œ  /  /# a B Bb - Þ _{Imposing the initial condition,} C ! œ "a b _{, we obtain} - œ (Þ Thus, the solution can be expressed as $C  #C œ  /  /# a B Bb (Þ Now by completing the square on the left hand side, # C  $Î% œa b#

 /  /a B Bb '&Î). Hence the explicit form of the solution is C B œ  $Î% a b È'&Î"'  -9=2 B Þ

a b, Þ

a b- . Note the '&  "' -9=2 B   !, as long as k kB  #Þ" Þ Hence the solution is valid on the interval  #Þ"  B  #Þ".

19a b+ Þ C B œ  Î$  =38 a b 1 "_$ "a$ -9= B Þ# b

a b, Þ

20a b+ Þ Rewrite the differential equation as C .C œ +<-=38 BÎ# È"  B .B Þ# Integrating both sides of the equation results in C Î$ œ +<-=38 B Î#  - Þ$ a b# Imposing the condition C ! œ !a b , we obtain - œ !Þ The explicit form of the solution is C B œa b É$ $a b Þ

# +<-=38 B

a b, .

a b- . Evidently, the solution is defined for  " Ÿ B Ÿ " Þ

22. The differential equation can be written as a$C  % .C œ $B .B Þ# b # Integrating both sides, we obtain C  %C œ B  - Þ$ $ Imposing the initial condition, the specific solution is C  %C œ B  " Þ$ $ Referring back to the differential equation, we find that C p _w as C p „#ÎÈ The respective values of the abscissas are $ Þ B œ  "Þ#(' "Þ&*) Þ,

Hence the solution is valid for  "Þ#('  B  "Þ&*) Þ

24. Write the differential equation as _a$  #C .C œ #  / .B Þ_{b a} B_b Integrating both sides, we obtain $C  C œ #B  /  - Þ# B Based on the specified initial condition, the solution can be written as $C  C œ #B  /  " Þ# B Completing the square, it follows that

C B œ  $Î# a b È#B  /  "$Î% ÞB _{The solution is defined if} #B  /  "$Î%   !B _,

that is,  "Þ& Ÿ B Ÿ # aapproximatelyb. In that interval, C œ !w , for B œ 68 # Þ It can be verified that C 68 #  !wwa b . In fact, C B  !wwa b on the interval of definition. Hence the solution attains a global maximum at B œ 68 # Þ

26. The differential equation can be written as _a"  C# "_{b .C œ # "  B .B Þ}a b Integrating both sides of the equation, we obtain +<->+8C œ #B  B  - Þ# Imposing the given initial

equation, the solution is stationary at B œ  "Þ Since C B  !wwa b on the entire interval of

definition, the solution attains a global minimum at B œ  " Þ

28a b+ . Write the differential equation as C"a%  Cb".C œ > "  >a b".> Þ Integrating both sides of the equation, we obtain 68 C  68 C  % œ %>  %68 "  >  - Þk k k k k k Taking the exponential of both sides, it follows that kCÎ C  % œ G / Î "  > Þa bk %> a b% It follows that as > p _ CÎ C  % œ "  %Î C  % p _, k a bk k a bk . That is, C > p % Þa b

a b, Þ Setting C ! œ #a b , we obtain that G œ ". Based on the initial condition, the solution may be expressed as CÎ C  % œ  / Î "  > Þa b %> a b% Note that CÎ C  %  !a b , for all >   !Þ Hence C  % for all >   !Þ Referring back to the differential equation, it follows that is always Cw positive. This means that the solution is monotone increasing. We find that the root of the equation / Î "  >%> a b% œ $** is near > œ #Þ)%% Þ

a b- Þ Note the C > œ %a b is an equilibrium solution. Examining the local direction field,

we see that if C !  !a b , then the corresponding solutions converge to C œ %. Referring back to part a b+ , we have CÎ C  % œ C Î C  % / Î "  >a b c ! a ! bd %> a b%, for C Á % Þ! Setting > œ #, we obtain C Î C  % œ $Î/! a ! b a # %b a b aC # Î C #  % Þa b b Now since the function 0 C œ CÎ C  %a b a b is monotone for C  % and C  %, we need only solve the equations C Î C  % œ! a ! b  $** $Î/a #b%and C Î C  % œ! a ! b %!" $Î/a #b%Þ The respective solutions are C œ $Þ''##! and C œ %Þ%!%# Þ!

31a b

-32a b+ . Observe that aB  $C Î#BC œ# #b _#"ˆ ‰C_B " $_{# B}C Þ Hence the differential equation is .homogeneous

a b, . The substitution C œ B @ results in @  B @ œ B  $B @ Î#B @w a # # #b # . The

transformed equation is @ œ "  @ Î#B@ Þw a #b 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 b- Þ

34a b+ Þ Observe that a%B  $C ÎÐ#B  CÑ œ  # b C_B#  _BC‘"Þ Hence the differential equation is homogeneous.

a b, . The substitution C œ B @ results in @  B @ œ  #  @Î #  @w a b. The transformed equation is @ œ  @  &@  % ÎÐ#  @ÑB Þw a # b This equation is _separable, with general solution a@%b k# @" œ GÎB Þk $ In terms of the original dependent variable, the solution is a%B  Cb k# BC œ GÞk

a b- Þ

35a b- .

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

a b, Þ Write the equation as a"  @b#.@ œ B .B Þ" Integrating both sides of the equation, we obtain the general solution  "Î "  @ œ 68 B  - Þa b k k In terms of the original

a b- Þ

37a b+ Þ The differential equation can be expressed as C œw "_#ˆ ‰_BC "  $_{# B}C. Hence the equation is homogeneous. The substitution C œ B @ results in B @ œ "  &@ Î#@w a #b . Separating variables, we have #@ "

"&@#.@ œ .B ÞB

a b, Þ Integrating both sides of the transformed equation yields  " &

68k"  &@#kœ 68 B  -k k ,

that is, "  &@ œ GÎ B Þ# k k& In terms of the original dependent variable, the general solution is &C œ B  GÎ B Þ# # _{k k}$

a b- Þ

38a b+ Þ The differential equation can be expressed as C œw $  "ˆ ‰ . Hence the

# B # B

C C "

equation is homogeneous. The substitution C œ B @ results in B @ œ @  " Î#@w a # b , that is, #@ "

@ "# .@ œ .B ÞB

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

Section 2.3

5a b+ . Let be the amount of salt in the tank. Salt enters the tank of water at a rate ofU #"_%ˆ"  =38 > œ"_# ‰ _#"  =38 >"_% 9DÎ738Þ 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 factor .a b> œ />Î&!Þ Write the equation as ˆ/>Î&!U œ /‰w >Î&!ˆ" " ‰Þ The

#  =38 >%

specific solution is U > œ #&  "#Þ&=38 >  '#&-9= >  '$"&! /a b  >Î&!‘Î#&!" 9D Þ a b, Þ

a b- Þ 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 > œ W / Þa b ! <> Setting W X œ #Wa b !, the required time is X œ 68 # Î< Þa b

a b, Þ For the case < œ (%œ Þ!( X ¸ *Þ* C<= Þ,

a b- Þ Referring to Parta b+ , < œ 68 # ÎXa b . Setting X œ ), the required interest rate is to be approximately < œ )Þ''%Þ

8a b+ . Based on the solution in Eq.a b"' , with W œ !! , the value of the investments with contributions is given by W > œ #&ß !!! /  " Þa b a <> b After _ten years, person A has W œ #&ß !!! "Þ##' œ $!ß '%! ÞE $ a b $ Beginning at age $&, the investments can now be analyzed using the equations W œ $!ß '%! /E Þ!)> and W œ #&ß !!! /F a Þ!)> " Þb

a b- .

a b. Þ The two balances can never be equal.

11a b+ . Let be the value of the mortgage. The debt accumulates at a rate of W <W, in which < œ Þ!*is the annual interest rate. Monthly payments of $)!! are equivalent to $*ß '!! per yearÞ The differential equation governing the value of the mortgage is .WÎ.> œ Þ!* W  *ß '!! Þ Given that is the original amount borrowed, the debt isW!

W > œ W /a b ! Þ!*> "!'ß ''( /a Þ!*> " Þb Setting W $! œ !a b , it follows that W œ **ß &!!! $ .

a b, Þ 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 > œ W / a b ! <> 5_<a/  " Þ<> b

a b, . The solution may also be expressed as W > œ ÐW  Ñ/  Þa b ! 5_< <> 5_< Note that if the withdrawal rate is 5 œ < W! !, the balance will remain at a constant level W Þ!

a b- . Assuming that 5  5!, W Xa b! œ ! for X œ 68! _<" ’_555 _!“Þ a b. . If < œ Þ!) and 5 œ #5!, then X œ )Þ''! years .

a b/ . Setting W > œ !a b and solving for in Part/<> a b, / œ, <> _5<W5 Þ Now setting > œ X

!

results in 5 œ <W / Î /  " Þ! <X a <X b

a b0 Þ In parta b/ , let 5 œ "#ß !!! < œ Þ!), , and X œ #!. The required investment becomes .W œ ""*ß ("&! $

14a b+ Þ Let U œ  < U Þw The general solution is U > œ U /a b <>Þ Based on the

 &($! < œ 68Ð"Î#Ñ, that is, < œ "Þ#!*( ‚ "!% per year.

a b, . Hence the amount of carbon-14 is given by U > œ U /a b ! "Þ#!*(‚"! >Þ

%

a b- Þ Given that U X œ U Î&a b ! , we have the equation "Î& œ /"Þ#!*(‚"! XÞ 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 <T Þ The population decreases by #!ß !!! per day. Hence the equation that models the population is given by .T Î.> œ <T  #!ß !!!. Note that the variable represents > days. The solution is T > œ T / a b ! <> #!ß!!!_< a/  " Þ<> b In the absence of predators, the governing equation is .T Î.> œ <T" ", with solution

T > œ T / Þ"a b ! <> Based on the data, set T ( œ #T"a b !, that is, #T œ T / Þ! ! (< The growth

rate is determined as < œ 68 # Î( œ Þ!**!#a b per dayÞ Therefore the population, including the predation by birds, is T > œ # ‚ "! /a b & Þ!**> #!"ß **( /a Þ!**> " œb

œ #!"ß **(Þ$  "*((Þ$ /Þ!**>Þ

16_{a b a b}+ . C > œ /B: #Î"!  >Î"!  #-9=Ð>ÑÎ"! Þ_{c d} The doubling-time is 7 ¸ #Þ*'$# Þ a b, Þ The differential equation is .CÎ.> œ CÎ"!, with solution C > œ C ! /a b a b >Î"!Þ The doubling-time is given by 7 œ "!68 # ¸ 'Þ*$"& Þa b

a b- . Consider the differential equation .CÎ.> œ !Þ&  =38Ð# >Ñ CÎ& Þa 1 b The equation is separable, with " " Integrating both sides, with respect to the

C.C œ !Þ"  =38Ð# >Ñ .> Þˆ & 1 ‰

appropriate variable, we obtain 68 C œa1>  -9=Ð# >Ñ Î"!  - Þ1 b 1 Invoking the initial condition, the solution is C > œ /B: "  >  -9=Ð# >Ñ Î"! Þa b ca 1 1 b 1d The doubling-time is 7 ¸ 'Þ$)!% Þ The doubling-time approaches the value found in parta b, .

sides yields the general solution C œ  5 ' a b. 7 .  C7 !.a b! Î >‘ .a b. In this problem, the

integrating factor is .a b> œ /B: -9= >  > Î& Þca b d

a b, Þ The population becomes extinct, if C >a b‡ œ !, for some > œ >‡. Referring to part_{a b}+ , we find that C > œ ! Êa b‡ ( ca b d ! > "Î& -‡ /B: -9= 7 7 Î& . œ & /7 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!)*$. Hence extinction can happen only if & /"Î&C  &Þ!)*$ C  !Þ)$$$ Þ

- , that is,

-a b- . Repeating the argument in parta b, , it follows that C >a b‡ œ ! Ê

( ca b d ! > "Î& -‡ /B: -9=  Î& . œ "/ C Þ 5 7 7 7

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

-

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

19a b+ . Let U >a b be the volume of carbon monoxide in the room. The rate of increase of CO is a ba bÞ!% !Þ" œ !Þ!!% 0 > Î738 Þ$ The amount of CO leaves the room at a rate of a b a b!Þ" U > Î"#!! œ U > Î"#!!! 0 > Î738 Þa b $ Hence the total rate of change is given by the differential equation .UÎ.> œ !Þ!!%  U > Î"#!!! Þa b This equation is linear and separable, with solution U > œ %)  %) /B:  >Î"#!!!a b a b 0> Þ$ Note that U œ !! 0 > Þ$ Hence the concentration at any time is given by B > œ U > Î"#!! œ U > Î"#a b a b a b .% a b, . The concentration of CO in the room is B > œ %  %/B:  >Î"#!!!a b a b%Þ A level of !Þ!!!"# corresponds to !Þ!"# . Setting % Ba b7 œ !Þ!"#, the solution of the equation

20a b+ Þ The concentration is - > œ 5  T Î<  -  5  T Î< /a b a ! b <>ÎZÞ It is easy to see that - >p_ œ 5  T Î< Þa b

a b a b, Þ - > œ - / ! <>ÎZ. The reduction times are X œ 68Ð#ÑZ Î<&! and X œ 68Ð"!ÑZ Î<Þ"!

a b- Þ The reduction times, in years, are X œ 68 "! '&Þ# Î"#ß #!! œ %$!Þ)&W a ba b

X œ 68 "! "&) Î%ß *!! œ ("Þ% àQ a ba b X œ 68 "! "(& Î%'! œ 'Þ!&I a ba b

X œ 68 "! #!* Î"'ß !!! œ "(Þ'$ ÞS a ba b

21a b- Þ

22a b+ . The differential equation for the motion is 7 .@Î.> œ  @Î$!  71 Þ Given the initial condition @ ! œ #!a b m/s , the solution is @ > œ  %%Þ"  '%Þ" /B:  >Î%Þ&a b a b. Setting @ >a b" œ !, the ball reaches the maximum height at > œ "Þ')$" sec. Integrating @ >a b, the position is given by B > œ $")Þ%&  %%Þ" >  #))Þ%& /B:  >Î%Þ& Þa b a b Hence the maximum heightis B >a b" œ %&Þ() m.

a b, Þ Setting B >a b# œ !, the ball hits the ground at > œ &Þ"#)# sec. a b- Þ

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

a b, Þ The differential equation for the downward motion is 7 .@Î.> œ  @  71 Þ. #

This equation is also separable, with 7 For convenience, set at

71 @. #.@ œ  .> Þ > œ !

the top of the trajectory. The new initial condition becomes @ ! œ !a b . Integrating both sides and invoking the initial condition, we obtain 68 %%Þ"$  @ Î %%Þ"$  @ œ >Î#Þ#&ca b a bd Þ

Solving for the velocity, @ > œ %%Þ"$ "  /a b a >Î#Þ#&b aÎ "  />Î#Þ#&bÞ Integrating @ >a b, the position is given by B > œ **Þ#* 68 /a b ’ >Î#Þ#&Î "  /a >Î#Þ#&b#“ ")'Þ# Þ To estimate the duration of the downward motion, set B >a b# œ !, resulting in > œ $Þ#('# sec. Hence the total time that the ball remains in the air is >  > œ &Þ"*#1 # sec.

a b- Þ

24_{a b}+ Þ Measure the positive direction of motion downward. Based on Newton's #nd law,

the equation of motion is given by

7.@ œ Þ

.>

 !Þ(& @  71 !  >  "!  "# @  71 >  "! œ _, ,

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 @ ! œ !a b fpsÞ This differential equation is separable and linear, with solution @ > œ #%! "  /a b a >Î(Þ&b_{. Hence} @ "! œ "('Þ(a b _fpsÞ

a b, . Integrating the velocity, with B > œ !a b , the distance fallen is given by B > œ #%! >  ")!! /a b >Î(Þ& ")!!_.

a b- Þ For computational purposes, reset time to > œ !. For the remainder of the motion, the initial value problem is .@Î.> œ  $#@Î"&  $#, with specified initial velocity @ ! œ "('Þ(a b fps Þ The solution is given by @ > œ "&  "'"Þ( /a b $# >Î"&Þ As > p _, @ > p @ œ "&a b P fps Þ Integrating the velocity, with B ! œ "!(%Þ&a b , the distance fallen after the parachute is open is given by B > œ "& >  (&Þ) /a b $# >Î"& ""&!Þ$ 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' _secÞ

a b. Þ

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 @ ! œ @a b !. The solution is @ > œ  71Î5  @  71Î5 /a b a ! b 5>Î7Þ Setting

@ >a b7 œ !, the maximum height is reached at time > œ 7Î5 68 71  5 @ Î71 Þ7 a b ca !b d

Integrating the velocity, the position of the body is

B > œ  71 >Î5  7 1  7 @ Ð"  / ÑÞ

5 5

a b _”Š ‹# !_• 5>Î7

Hence the maximum height reached is

B œ B > œ 7 @  1 7 68 71  5 @ Þ

5 5 71

7 7

! !

a b _{Š ‹ ”}# _•

a b, Þ Recall that for $ ¥ " 68 " , a $bœ $ "_#$# "_$$$ "_%$% á 26a b, . lim œlim  a5 @  71 /b œ  1> Þ 5Ä ! 5Ä ! 71 5 @ 71 / 5 7> 5>Î7 a ! b 5>Î7 !

a b- Þ lim ˆ @ /‰ ‘ œ !, since lim / œ ! Þ

7 Ä ! 7 Ä ! 71 71

5 5 ! 5>Î7 5>Î7

B @ œ  7@  7 168 71  5 @ Þ

5 5 71

a b #_{# º º}

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

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

29_{a b}+ Þ Let represent the height above the earth's surface. The equation of motion isB given by 7.@_.> œ  K Q 7 , in which is the universal gravitational constant. TheK

VB a b#

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

.B a_{V  B}b# .

This equation is separable, with @ .@ œ  KQ V  Ba b#.B Þ Integrating both sides, and

invoking the initial condition @ ! œa b È#1V , the solution is @ œ #KQ V  B# a b"  #1V  #KQ ÎV Þ From elementary physics, it follows that 1 œ KQ ÎV#. Therefore @ B œa b È ’#1 VÎÈV  B Þ“ aNote that 1 œ ()ß &%& mi/hr .#b

a b, Þ We now consider .BÎ.> œÈ ’#1 VÎÈV  B“. This equation is also separable, with ÈV  B .B œÈ#1 V .> Þ By definition of the variable , the initial condition isB B ! œ !Þa b Integrating both sides, we obtain B > œa b _#$ˆÈ#1 V >  V_$# $Î#‰‘#Î$ V Þ Setting the distance B X  V œ #%!ß !!!a b , and solving for , the duration of such aX flight would be X ¸ %* hours .

32a b+ Þ Both equations are linear and separable. The initial conditions are @ ! œ ? -9=a b E

and A ! œ ? =38 Ea b . The two solutions are @ > œ ? -9= E /a b <> and A > œ  1Î< a b  ? =38 E  1Î< /a b <>Þ

a b, Þ Integrating the solutions in parta b+ , and invoking the initial conditions, the coordinates are B > œ -9= E "  /a b ?_< a <>b and

C > œ  1>Î<  1  ?< =38 E  2< Î<  ?=38 E  1Î< / Þ <

a b ˆ #‰ # Š #‹ <>

a b- Þ

a b. Þ Let be the time that it takes the ball to go X $&! horizontally. Then from above,ft /X Î& œ ? -9= E  (! Î? -9= E Þa b At the same time, the height of the ball is given by C X œ  "'! X  #'(  "#&?=38 E  Ð)!!  &? =38 EÑ ? -9= E  (! Î? -9= E Þa b ca b d Hence and must satisfy the inequalityE ?

)!!68 ? -9= E  (!  #'(  "#&?=38 E  Ð)!!  &? =38 EÑ ? -9= E  (! Î? -9= E   "! Þ ? -9= E

” • ca b d

33a b+ Þ Solving equation a b3 , C B œ 5  C ÎCwa b ca # b d"Î#. The positive answer is chosen, since is an C increasing function of .B

a b, . Let C œ 5 =38 ># # . Then .C œ #5 =38 > -9= > .> Þ# Substituting into the equation in parta b+ , we find that

#5 =38 > -9= > .> -9= > .B œ =38 > Þ

#

Hence #5 =38 > .> œ .B Þ# #

a b- Þ Letting ) œ #>, we further obtain 5 =38# #_#). œ .B Þ) Integrating both sides of the equation and noting that > œ) œ ! corresponds to the origin, we obtain the solutions Ba b) œ 5#a) =38)bÎ# _{and from part}c a b, Cd a b) œ 5 "  -9=#a )bÎ# Þ