Introductor soution manual elemenrary differential equations with boundary value probolems

Chapter 10: Series Solutions of Linear 268Differential Equations Chapter 11: Second Order Partial Differential ??. Equations and Fourier Series Chapter 12: First Order Partial Differenti

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Chapter 10: Series Solutions of Linear 268

Differential Equations Chapter 11: Second Order Partial Differential ??

Equations and Fourier Series Chapter 12: First Order Partial Differential

Equations and the Method of Characteristics Chapter 13: Linear Two-Point Boundary Value Problems

Introduction to Differential Equations

5 (b) Substituting into the differential equation yields y( )1 =Ce12 =Ce Using the initial condition,

y( )1 =2=Ce Solving for C , we find C=2e-1

Order = 3 3 arbitrary constants

7 (a) y=C1sin2t C+ 2cos2t Differentiating gives us y¢ =2C1cos2t-2C2sin2t and

11 y=t r Differentiating gives us y¢ =rt r 1- and ¢¢ = -

13 From (12), y=C e1 2t +C e2 -2t, which we differentiate to get y¢ =2C e1 2t-2C e2 -2t Using the

initial conditions, y( )0 =2 and y ( )¢ 0 =0, we have two equations containing C1 and C2:

C1+C2 =2 and 2C1-2C2 =0 Solving these simultaneous equations gives us C1=C2=1.Thus, the solution to the initial value problem is y=e2t +e-2t =2cosh( ).2t

14 y( )0 =c1+c2=1 2, c1-2c2=2 \ c1=1, c2=0 ( )y t =e2t

15 From (12), y t( ) =C e t +C e-t

1 2 2 2

Using the initial condition y( )0 =3, we find that C1+C2 =3.From the initial condition lim ( )

From graph, only at

Also c= -1 From graph ( )y1 = -0 5 \ -1= m- fi m=

cos The object is

dropped from rest, so y¢( )0 =0= -4 +C

pe Solving for C yields C = 4

pe , and putting this

value back into the equation for y¢ and simplifying gives us ¢ = - + - Ê

Integrating again gives us y= - t + t-Ê t C

Ë

Á ˆ

¯

˜ ÊË

from a height of 252 ft (at t = 0), y( )0 =C¢ =252 and thus

1 (a) The equation is autonomous because y¢ depends only on y.

1 (b) Setting y¢ =0, we have 0= - +y 1 Solving this for y yields the equilibrium solution: y = 1

2 (a) not autonomous

2 (b) no equilibrium solutions, isoclines are t = constant.

3 (a) The equation is autonomous because y¢ depends only on y.

3 (b) Setting y¢ =0, we have 0 = siny Solving this for y yields the equilibrium solutions: y= ±np

4 (a) autonomous

4 (b) y y( -1)=0, y=0 1, .

5 (a) The equation is autonomous because y¢ does not depend explicitly on t.

5 (b) There are no equilibrium solutions because there are no points at which y¢ =0

6 (a) not autonomous

6 (b) y = 0 is equilibrium solution, isoclines are hyperbolas

7 (a) c = -1: Setting c = -1 gives us - + = -y 1 1 which, solved for y, reads y = 2 This is the

11 One example that would fit these criteria is y¢ = -(y-1)2 For this autonomous D.E., y¢ =0 at

y = 1 and y¢ <0 for -• < y 1< and 1 < y< •

First Order Linear Differential Equations

Section 2.1

1 This equation is linear because it can be written in the form y¢ + p t y( ) = g t( ) It is

nonhomogeneous because when it is put in this form, g t( ) π 0

9 This equation is linear because it cannot be written in the form y¢ + p t y( ) =g t( ) It is

nonhomogeneous because when it is put in this form, g t( ) π 0

10 linear, homogeneous

11 (a) Theorem 2.1 guarantees a unique solution for the interval (-• •, ), since t

t2 +1 and sin( )t areboth continuous for all t and -2 is on this interval.

11 (b) Theorem 2.1 guarantees a unique solution for the interval (-• •, ), since t

t2

1+ and sin( )t areboth continuous for all t and 0 is on this interval.

11 (c) Theorem 2.1 guarantees a unique solution for the interval (-• •, ), since t

t2

1+ and sin( )t areboth continuous for all t and p is on this interval.

12 (a) 2 < < •t

12 (b) - < <2 t 2

12 (c) - < <2 t 2

12 (d) -• < < -t 2

13 (a) For this equation, p t( ) is continuous for all t π -2 2, and g t( ) is continuous for all t π 3.

Therefore, Theorem 2.1 guarantees a unique solution for ( , )3 • , the largest interval that

includes t = 5.

13 (b) For this equation, p t( ) is continuous for all t π -2 2, and g t( ) is continuous for all t π 3.

Therefore, Theorem 2.1 guarantees a unique solution for (-2 2, ), the largest interval that

includes t = -3

2.

13 (c) For this equation, p t( ) is continuous for all t π -2 2, and g t( ) is continuous for all t π 3.

Therefore, Theorem 2.1 guarantees a unique solution for (-2 2, ), the largest interval that

includes t = 0.

13 (d) For this equation, p t( ) is continuous for all t π -2 2, and g t( ) is continuous for all t π 3.

Therefore, Theorem 2.1 guarantees a unique solution for (-• -2, ), the largest interval thatincludes t = -5.

13 (e) For this equation, p t( ) is continuous for all t π -2 2, and g t( ) is continuous for all t π 3.

Therefore, Theorem 2.1 guarantees a unique solution for (-2 2, ), the largest interval that

15 y t( ) = 3e t2 Differentiating gives us ¢ =y 3e t2( )2t =2ty Substituting these values into the given

equation yields 2ty+p t y( ) =0 Solving this for p t( ), we find that p t( ) = -2t Putting t = 0

into the equation for y gives us y0=3

17 y t( ) = 0 satisfies all of these conditions.

3 (a) We can rewrite this equation into the conventional form: y¢ -2ty=0 Then we will integrate

p t( ) = -2t to find P t( ) = -t2 The general solution, then, is y t( )=Ce-P t( )=Ce t2

3 (b) y( )1 =Ce=3 Solving for C yields C= e

-3 1 Therefore, the solution to the initial valueproblem is y t( )=3e e-1 t2 =3e(t2-1)

5 (b) y( )1 =C=1 Therefore, the solution to the initial value problem is y t( ) =t-4

6 (a) m =exp( -cos ) \ ( )= - - ( cos )

2

1.

9 (a) We can rewrite this equation into the conventional form: y¢ -3(t2 +1)y=0 Then we will

integrate p t( )= -3(t2 +1) to find P t( ) = -t3-3 The general solution, then, ist

2 a We can then substitute this value for C

into the general solution at the point ( , )4 4 : y( ) / /

15 Putting this equation into a form more like #14, we have y¢ = -2ty+6t= -2t y( -3) We will

then let z= y- 3 (and z¢ = ¢y , accordingly) Substituting into our modified original equationyields an equation for z t( ): z¢ = -2 , or put in a more conventional form, tz z¢ +2tz=0 Usingthe same substitution for the initial condition yields z( )0 =4- =3 1 Integrating p t( ) = 2t gives

us P t( ) = t2 The general solution is then z t( ) =Ce-t2 Our initial condition requires that C = 1,

so the solution for z t( ) is z t( ) =e-t2 In terms of y t( ), this solution reads y- =e-t

3 2 Solvedfor y t( ), this solution is y t( ) =e-t2 +3

17 Solving the equation y¢ +cy=0 with our method yields the general solution y t( ) = y e0 -ct

Looking at the graph, we can see that y( )0 =2=y0 and y(-0 4 )=3=y e0 -c(-0 4 )=2e0 4 c

Solving for c gives us c = Ê

Ë

Á ˆ

¯

˜ ª5

19 (a) The general solution to this D.E is y t( ) = y e0 -t, which can be rewritten as ln( )y = - +t c

Thus, this D.E corresponds to graph #2 with y0 y e y0 e2

0

19 (b) The general solution to this D.E is y t( )= y e0 tsin4t, which can be rewritten as

ln( )y =tsin4t c+ Thus, this D.E corresponds to graph #1 with y y e y

, which can be rewritten as

ln( )y = -t sin4t c+ Thus, this D.E corresponds to graph #3 with y y e y

1.

Substituting values from the table gives us the necessary equations to solve for y0 and n First,

2 11

14

Section 2.3

1 For this D.E., p t( ) = 2 Integrating gives us P t( ) = 2t An integrating factor is, then, m( )t =e2t

Multiplying the D.E by m( )t , we obtain e y2t e y2t e y2t e2t

¢ + =( )¢ = Integrating both sides yields

e y2t = +t C Therefore, the general solution is y t( ) =te-2t +Ce-2t

4 y¢ +2ty=t fi e y t ¢ =te t fi e y t = 1e t +C fi y= +Ce-t

2

12

7 For this D.E., p t( ) = 1 Integrating gives us P t( ) =t An integrating factor is, then, m( )t =e t

Multiplying the D.E by m( )t , we obtain e y t ¢ +e y t =(e y t )¢ =te t Integrating both sides yields

e y t =te t -e t +C Therefore, the general solution is y t( ) = - +t 1 Ce-t

9 For this D.E., p t( ) = -3 Integrating gives us P t( ) = -3t An integrating factor is, then,

m( )t =e-3t Multiplying the D.E by m( )t , we obtain e y- t e y-t e y-t e-t

¢ - = ¢ =

3 ( ) 6 Integrating both sides yields e y- t e- t C

3 2

3 2

5 2

32

12

¢ + =( )¢ = Integrating both sides yields

3

2

5 2

15

= + Solving for y gives us y= 1e t +Ce- t

= = + Solving for C yields C = -1

5, and thus our final solution is y= 1e t- e- t

5

15

3

2 Integrating gives us P t( )= sin( )t

2 An integrating factor is, then, m( )

1

14

= + - + + fi = - + + +

-y(-1)= -e 1+ +Ce =e fi C= e

-2

14

14

\ y=e-t + t + + e-(t+)

2

14

14

= + + Solving for y gives us y= t + +Ct

-4

13

1

3 Solving for C yields C = -1

4, and thus our final solution

is y= t + - t

-4

13

14

3

The t-interval on which this solution exists is -• < < t 0

18 2tCe t2 + pCe t2 =0fi p t( )= -2t Substituting, (Ce t2 +2)¢ -2t Ce( t2 +2)= -4tfig t( )= -4t.

19 Multiplying both sides of the equation by the integrating factor, m( )t =t, we have

ty=t Ct( -1+1)= +t C Differentiating gives us ( )ty ¢ = 1 Therefore,

y (1 cos )t y 1 cos , ( )t y 0 3, m e t sint

(e t+sint y) ( cos )t e t+sint (e t+sint) e t+sint y e t+sint C y Ce- +(t sin )t

23 Putting this D.E in the conventional form, we have y¢ +2y=e-t-2 For this D.E., p t( ) = 2

An integrating factor is, then, m( )t =e2t Multiplying the D.E by m( )t , we obtain

e y2t e y2t e y2t e t e2t

¢ + =( )¢ = - Integrating both sides yields e y2t =e t-e2t +C Solving for

y gives us y=e-t- +1 Ce-2t, and with our initial condition, y( )0 = - = - +2 1 1 C Solving for

C yields C = -2 , and thus our final solution is y=e-t- -1 2e-2t Therefore, lim ( )

Æ• = -1

(sin ) ( ) (sin ) Integrating both sides yields

e- t y e- t C

Solving for y gives us y= +1 Cecost, and with our initial condition,

y( )0 =3 1= +CefiC=2e-1 Therefore, the solution for 0 £ £t p is y= +1 2ecost-1 and

(sin ) ( ) ( sin ) Integrating both sides yields

e- t y e- t C

= - +

cos cos Solving for y gives us y= - +1 Cecost, and with our initial condition,

y( )p = +1 2e-2= - +1 Ce-1fiC=2e1+2e-1 Therefore, the solution for p £ £t 2 isp

t y¢ - t y=(t y)¢ = Integrating both sides yields 1

t y=C Solving for y gives us

y=Ct, and with our initial condition, y( )3 =3=C( )3 fiC=1 Therefore, the solution for

30 1 025 5000( )=( ) ◊

2 0

1

2 1 03( )=( + ) =( ) Setting P t2( ) =2A0 yields 2= 1 032t, and solving for t gives us

\ 8 =0 5 fi = 1 ª0 0625 ( 6 25%)

16

5 (a) P B¢ =( 0 04+0 004 t P) B; P B( )0 = A0.

5 (b) P B = A e0 .04t+.002t2 This can be verified easily through differentiation

5 (c) For Plan A, P t A( )= A e0 06t To find the time t at which Plan B “catches up” with Plan A, let us

set P t A( )=P t B( ): A e t A e t t

0 06 0

7 We can simplify this problem by considering the two deposits separately and then adding the

principals of each deposit together at a time of twelve years We have, then,

Solving this with the quadratic formula yields one positive value of x: x ª1 5616 =e6r

Solving for r yields r ª 0 0743

8 11 000 000 10 000 000, , = , , e5k Solving for k yields k = Ê

1110

11 80 000 100 000, = , e6k Solving for k yields k = 1

6ln( ) Using this value for 8 k , we have

0 and must be nonnegative fi - ≥0 If net immigration rate M > 0,

net growth rate k < 0 and vice versa.

13 (a) For Strategy I, we have M I =kP0 For Strategy II, we have M II =P e0( k-1).

13 (b) The net profit for each strategy would equal (M profit fish)( ), and so the profit for Strategy I

is, then: PrI =500 000 3172 75, ( )( )=118 950, , and the profit for Strategy II

14 (c) If k > 0, introduce the immigrants as early as possible If k < 0, introduce as late as possible.

15 (a) From the general solution of the radioactive decay equation, Q t( ) =Ce-kt, we can use the data

given to find C and k Q( )1 =Ce-k =100

10

ln Using this value of k

with the t = 1 data, we find that C=Q0 =149 4 mg C=Q0, since the exponential falls off theexpression for Q at t = 0.

ln

ln( )ln

10 3

10 3

10 3

10 3

17 Q¢ = -kQ+M Writing this D.E in the conventional form, we have Q¢ +kQ= M For this

D.E., p t( ) =k and P t( ) = kt, which yields an integrating factor of m( )t =e kt Thus,

e Q kt ¢ +ke Q kt =(e Q kt )¢ =e M kt Integrating both sides gives us e Q e M

- = for T and for T

To determine the angle a , part of the diagram is shown here with the radii of the circles

The corresponding contact angles and belt tensions are:

1 (a) To begin, Q( )0 =0 and Q¢ =( )( )0 2 3 - Q ( )

100 3 Putting the second equation in the conventionalform, we have Q¢ +0 03 Q=0 6 Multiplying both sides of this equation by the integratingfactor m( )t =e0 03. t gives us (e0 03. t Q) e0 03. t

Q( )0 =0=20+C, so C = -20 With this value for C , our final equation for Q is

form, we have Q¢ +0 005 rQ=0 25 r Multiplying both sides of this equation by the integratingfactor m( )t =e0 005. rt gives us (e0 005. rt Q) re0 005. rt

= = - - , and solving for r yields

5 (a) To begin, Q( )0 =10, V ( )0 =100, and V t( ) =100+t Since the tank has a capacity of 700

gallons, 100+ =t 700 Solving for t yields t = 600 minutes.

2.Multiplying both sides of the equation by the integrating factor m( )t =e2ln(100+t)=(100+t)2

3 1000

Ê Ë ˆ

20003

= + , and

solving for Q B, Q B e Ce

= 2000 - + 3

Á ˆ

¯

˜Ê Ë

7 (c) Setting Q B¢ = 0, we have 0 2000

3

1500

1200

=ÊË

Á ˆ

¯

˜ -Ê +Ë

3

5

ln hours

7 (d) Here, we want to determine t A such that Q t A( ) =A 1

2 lb and t B such that Q t B( )£ 0 2 lb where

t£t B This can be solved via plotting: t A ª 3800 hours and t B ª 4056 hours Therefore,

( sin ) ( ( cos ) ) ( sin ) ( cos )

Qe( 3t cos )t e t cost C Q Ce ( tcos )t

Substituting this back into our equation for Q yields Q=100 90- e-200

0 6 200

3 200 3 200 2

+

ÏÌÓ

¸

˝

˛+

3

200 =40+-0 6 +0 009 +

-1 000225

3 200

cos sin

3 200

Q =Q e- k Solving for k , we have k = ln2

18 Thus for the decay of the radioactive material

alone, we have Q t( ) e t

ln

=5

-2

18 with t measured in hours Now, for the lake, we know that Q

varies both with decay and with the water flow Accordingly, we will begin with the

relationship Q t t Q t kQ t t Q t

V r t

( +D )- ( )ª - ( )D - ( ) D

Using a form of the definition of the derivative and solving for Q©, we have

¯

˜ ª

-218

10

278218

ln

( ) min.

¯

˜ ª

110

350 120

ln The temperature of the food after 20 minutes in the oven is,

then,q(20)=350+(40-350)e-20k=350-(310 0 550)( )ª179 5 degrees Finally, the food iscooled at room temperature, so q( )t = = +( - )e-. t

110 72 179 5 72 0 02985 Solving for t yields

-ÊË

¯

˜ ª1

0 02985

110 72

179 5 72 34 8 ln . minutes.

5

14042

15

103

ˆ

¯

˜

ÏÌÔÓ

¸

˝Ô

¯

˜

ÏÌÓ

100 10

21214

3227ln

12

228150

First Order Nonlinear Differential Equations

Section 3.1

1 (a) Solving for y¢, we have y¢ = 1 - t y

3(1 2 cos ) Thus, f t y( , )= ( - tcos )y

( sin ) sin f and ∂

∂

f

y are continuous in the entire ty plane.

1 (c) The largest open rectangle is the entire ty plane, since f and ∂

y are continuous in the entire ty plane.

3 (c) The largest open rectangle is the entire ty plane, since f and ∂

61

t y

y t

¢ = , so - = From the initial condition, we have 8

3 1

53

14

114

t

6 (a) e y-y t t t t C

¢ +( -sin )=0, cosso e- -y+(22 + )= From the initial condition, we have

- + =1 1 0=C Then we have e-y= t22 +costfi y= -ln(t22 +cos t)

8 (a) (cos )y y¢ +t-2= , siny t1=C

0 so - - From the initial condition, we have 0- -( 1)= =1 C Then

we have siny= +t- fiy=sin-( +t- )

1

1 1

121

121

ÊË

11

, and solving for y yields y e

e

t t

= +

12 (b) -• < < •t

13 (a) sec2y y( )¢ +e-t =0, so tany-e-t =C From the initial condition, we have C = - =1 1 0 Then

we have tany e t, tany e t

y- y y( )¢ + t- t = so y2+ y+ t + t=C From the initial condition, we

have 0 1 0 1+ + + =2= C Then we have y2+cosy= -2 t -cost

1

2 0 0 4

12

-ÊË

246

23

18 y3+t2 +siny=4fi3y y2 ¢ +2t+(cos )y y¢ =0fi(3y2+cos )y y¢ +2t=0

When t=2, siny03+4+ y0 =4fi y03 +siny0 =0fiy0=0fi y( )2 =0

19 First, y e¢ y +ye y y ¢ +2t=cos Then 1t ( +y e y) y ¢ +(2t-cost)=0 At t0 =0, we have

0 1

21 (b) When t = 0, S( )0 =S0=1, so C=K◊ + =0 1 1 Then we have Kln +S S= -at+1 From the

other conditions, we have Kln 3 ln K

4

3

34

14

ÊË

C C, , and tan u t In terms of y, this reads y= - + Êt - p

Ë

¯

˜2

32

32

32

-2 From the implicit initial condition, we have

Q0-2= -C, so Q-2=2kt+Q0-2 Solved for Q, we have Q t

Thus 1

2 0 1 2

0 0 2

kQ

=+ t , where t is the half-life of the reactant Therefore,

2= 1 2+ kQ02t, which, solved for t, gives t = 3

2kQ02 Thus the half-life depends upon Q0.

0 1

0 0

0

0 Thus

dy y

ÏÌÓ

<

ÏÌ

-ln ,

( ) , ( ) ,

00

28 y¢ = -y2 is graph c y¢ = y3 is graph a y¢ = y(4-y) is graph b.

29 This is a translation three units to the right of graph (a) in problem 28

3 From the initial condition, we know that H( )0 = +1 CfiC=H( )0 -1, and

H( )0 1 C

3 2

= + Thus C2 C 2

3

= + Now we have H y( ) = +t C and ysin -y 3t+ =3 0, so

ysiny=3(t-1)fi 1ysiny= + -t ( 1)fiH y( )= ysiny C= -1

3

1

3 , and

Therefore, C2 C 2

3 1

23

13

= + = - + = - and the implicit solution of the initial value problem is

2 + -2 =3fi = - 6+4 -2 (the minus sign can be checked by the initial condition)

2 M= y+t3, , , so the equation is exact.N = +t y3 M y=N t =1

Ë

¯

˜tan 3

4 .

4 M=3t y N2 , , =6t+y3 M y=3t2, Nt=6 Therefore, the differential equation is not exact

5 M=e t y+ +3t2, , , so the equation is exact.N =e t y+ +2y M y=N t =e t y+

cos( ) cos( ) 2 2 2 2, and so h=e y2 From the initial

condition, we have 0+p + =1 C, and thus sin( ) ty + +t e y2 =p +1

7 M y=cos(t+y)-ysin(t+y)+1, cos(N t = t+y)-ysin(t+y)+1, so the equation is exact

y + y¢ +t+ = filn(y + )+ln(t+ )=C From the initial condition, we have ln2 = C

Thus (y2 +1) (t+1)=2, and solving for y yields y t

t

-+1

1 (a) The equation is both separable and exact.

1 (b) (i) y¢ = y(2-y)fi ¢ -y 2y= -y2fi 1-n= - =1 m v, = y-1fi y=v-1, thus y¢ = -v v2 ¢ and

-v v-2 ¢ =2v-1-v-2 or v¢ +2v=1, ( )v 0 =1

(ii) e v2t e2t e v2t 1e2t C v Ce 2t

2

12

1

1

2

12

( )¢ = fi = + or = + From the initial condition, 1

12+C= fiC= ,

1 3

32

94

32

or From the initial

condition, -3+ = fi =

112

112

¯

˜

112

32

2

3 2

6 (c) -• < < •t

7 First, let u=e y Then y u y u

Solving for u gives us

u= - +t Ct2 From the initial condition, we have y( )1 =0fiu( )1 =1, and so

1

1 3

ÊËÁ

ˆ

¯

˜ = - +

-

t t

t t

e e rt