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Solution Manual for A First Course in Differential Equations with Modeling Applications 11th Edition by Zill

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미분적분학 (BNMU0233)

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Chapter 1

Introduction to Differential Equations

1 Definitions and Terminology

1 order; linear

2 order; nonlinear because of (dy/dx) 4

3 order; linear

4 order; nonlinear because of cos(r+u)

5 order; nonlinear because of (dy/dx) 2 or

√

1 + (dy/dx) 2

6 order; nonlinear because ofR 2

7 order; linear

8 order; nonlinear because of ̇x 2

9 the differential equation in the formx(dy/dx) +y 2 = 1, we see that it is nonlinear inybecause ofy 2. However, writing it in the form (y 2 −1)(dx/dy) +x= 0, we see that it is linear inx.

10 the differential equation in the formu(dv/du)+(1+u)v=ueuwe see that it is linear inv. However, writing it in the form (v+uv−ueu)(du/dv)+u= 0, we see that it is nonlinear inu.

11=e−x/ 2 we obtainy′=− 12 e−x/ 2. Then 2y′+y=−e−x/ 2 +e−x/ 2 = 0.

12= 65 − 65 e− 20 twe obtaindy/dt= 24e− 20 t, so that

dy dt+ 20y= 24e

− 20 t+ 20

(

6 5 −

6

5 e

− 20 t

)

= 24.

13=e 3 xcos 2xwe obtainy′= 3e 3 xcos 2x− 2 e 3 xsin 2xandy′′= 5e 3 xcos 2x− 12 e 3 xsin 2x, so thaty′′− 6 y′+ 13y= 0.

1

Full file at TestbankDirect/

2 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

14=−cosxln(secx+ tanx) we obtainy′=−1 + sinxln(secx+ tanx) and

y′′= tanx+ cosxln(secx+ tanx). Theny′′+y= tanx.

15 domain of the function, found by solvingx+2≥0, is [− 2 ,∞). Fromy′= 1+2(x+2)− 1 / 2 we have

(y−x)y′= (y−x)[1 + (2(x+ 2)− 1 / 2 ]

=y−x+ 2(y−x)(x+ 2)− 1 / 2

=y−x+ 2[x+ 4(x+ 2) 1 / 2 −x](x+ 2)− 1 / 2

=y−x+ 8(x+ 2) 1 / 2 (x+ 2)− 1 / 2 =y−x+ 8.

An interval of definition for the solution of the differential equation is (− 2 ,∞) becausey′is not defined atx=−2.

16 tanxis not defined forx=π/2 +nπ,nan integer, the domain ofy = 5 tan 5xis {x

∣∣

5 x 6 =π/2 +nπ} or{x

∣∣

x 6 =π/10 +nπ/ 5 }. Fromy′= 25 sec 25 xwe have

y′= 25(1 + tan 25 x) = 25 + 25 tan 25 x= 25 +y 2.

An interval of definition for the solution of the differential equation is (−π/ 10 , π/10). Another interval is (π/ 10 , 3 π/10), and so on.

17 domain of the function is{x

∣∣

4 −x 26 = 0}or{x

∣∣

x 6 =−2 andx 6 = 2}. From y′= 2x/(4−x 2 ) 2 we have

y′= 2x

(

1

4 −x 2

) 2

= 2xy 2.

An interval of definition for the solution of the differential equation is (− 2 ,2). Other intervals are (−∞,−2) and (2,∞).

18 function isy= 1/

√

1 −sinx, whose domain is obtained from 1−sinx 6 = 0 or sinx 6 = 1. Thus, the domain is{x

∣∣

x 6 =π/2 + 2nπ}. Fromy′=− 12 (1−sinx)− 3 / 2 (−cosx) we have

2 y′= (1−sinx)− 3 / 2 cosx= [(1−sinx)− 1 / 2 ] 3 cosx=y 3 cosx.

An interval of definition for the solution of the differential equation is (π/ 2 , 5 π/2). Another one is (5π/ 2 , 9 π/2), and so on.

Full file at TestbankDirect/

4 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

21=c 1 et/

(

1 +c 1 et

)

we obtain dP dt =

(

1 +c 1 et

)

c 1 et−c 1 et·c 1 et (1 +c 1 et) 2

=

c 1 et 1 +c 1 et

[(

1 +c 1 et

)

−c 1 et

]

1 +c 1 et

=

c 1 et 1 +c 1 et

[

1 −

c 1 et 1 +c 1 et

]

=P(1−P).

22= 2x 2 −1 +c 1 e− 2 x 2 we obtaindy dx

= 4x− 4 xc 1 e− 2 x 2 , so that

dy dx+ 4xy= 4x− 4 xc 1 e

− 2 x 2 + 8x 3 − 4 x+ 4c 1 xe−x 2 = 8x 3

23=c 1 e 2 x+c 2 xe 2 xwe obtaindydx= (2c 1 +c 2 )e 2 x+ 2c 2 xe 2 xand d

2 y dx 2 = (4c 1 + 4c 2 )e

2 x+

4 c 2 xe 2 x, so that d 2 y dx 2 − 4

dy dx+ 4y= (4c 1 + 4c 2 − 8 c 1 − 4 c 2 + 4c 1 )e

2 x+ (4c 2 − 8 c 2 + 4c 2 )xe 2 x= 0.

24=c 1 x− 1 +c 2 x+c 3 xlnx+ 4x 2 we obtain dy dx

=−c 1 x− 2 +c 2 +c 3 +c 3 lnx+ 8x,

d 2 y dx 2

= 2c 1 x− 3 +c 3 x− 1 + 8,

and d 3 y dx 3

=− 6 c 1 x− 4 −c 3 x− 2 ,

so that

x 3 d

3 y dx 3

2x 2 d

2 y dx 2

−xdy dx

+y= (− 6 c 1 + 4c 1 +c 1 +c 1 )x− 1 + (−c 3 + 2c 3 −c 2 −c 3 +c 2 )x

(−c 3 +c 3 )xlnx+ (16−8 + 4)x 2

= 12x 2

In Problems25–28, we use the Product Rule and the derivative of an integral (( 12 ) of this section): d dx

ˆx

g(t)dt=g(x).

25=e 3 x

ˆx

e− 3 t t

dtwe obtain dy dx

= 3e 3 x

ˆx

e− 3 t t

dt+e

− 3 x x

·e 3 xor dy dx= 3e

3 x

ˆx

e− 3 t t dt+

1

x, so that

xdy dx

− 3 xy=x

(

3 e 3 x

ˆx

e− 3 t t

dt+ 1 x

)

− 3 x

(

e 3 x

ˆx

e− 3 t t

)

= 3xe 3 x

ˆx

e− 3 t t dt+ 1− 3 xe

3 x

ˆx

e− 3 t t dt= 1

Full file at TestbankDirect/

1 Definitions and Terminology 5

26=

√

ˆx

cos√t t

dtwe obtaindy dx

= 1

2 √

ˆx

cos√t t

dt+cos√x x

· √

xor dy dx=

1

2 √

ˆx

cos√t t

dt+ cosx, so that

2 x

dy dx−y= 2x

(

1

2 √

ˆx

cos√t t

dt+ cosx

)

−

√

ˆx

cos√t t

=

√

ˆx

cos√t t

dt+ 2xcosx−

√

ˆx

cos√t t

dt= 2xcosx

27=

5

10

ˆx

sint t dtwe obtain

dy dx=−

5

x 2 −

10

x 2

ˆx

sint t dt+

sinx x ·

10

x or dy dx=−

5

x 2 −

10

x 2

ˆx

sint t dt+

10 sinx x 2 , so that

x 2 dy dx

+xy=x 2

(

− 5

x 2

− 10

x 2

ˆx

sint t

dt+10 sinx x 2

)

(

5 + 10

ˆx

sint t

)

=− 5 − 10

ˆx

sint t dt+ 10 sinx+ 5 + 10

ˆx

sint t dt= 10 sinx

28=e−x 2 +e−x 2

ˆx

et 2 dtwe obtaindy dx

=− 2 xe−x 2 − 2 xe−x 2

ˆx

et 2 dt+ex 2 ·e−x 2

ordy dx

=− 2 xe−x 2 − 2 xe−x 2

ˆx

et 2 dt+ 1, so that

dy dx+ 2xy=

(

− 2 xe−x

2 − 2 xe−x

2 ˆx 0

2 dt+ 1

)

(

e−x

2 +e−x

2 ˆx 0

2 dt

)

=− 2 xe−x 2 − 2 xe−x 2

ˆx

et 2 dt+ 1 + 2xe−x 2 + 2xe−x 2

ˆx

et 2 dt= 1

29 y=

{

−x 2 , x < 0 x 2 , x≥ 0 we obtain y′=

{

− 2 x, x < 0 2 x, x≥ 0 so thatxy′− 2 y= 0.

30 functiony(x) is not continuous atx= 0 since lim x→ 0 −

y(x) = 5 and lim x→ 0 +

y(x) =−5. Thus, y′(x) does not exist atx= 0.

31 the functiony=emxinto the equationy′+ 2y= 0 to get (emx)′+ 2(emx) = 0

memx+ 2emx= 0

emx(m+ 2) = 0 Now sinceemx>0 for all values ofx, we must havem=−2 and soy=e− 2 xis a solution.

Full file at TestbankDirect/

1 Definitions and Terminology 7

36 the functiony=xminto the equationx 2 y′′− 7 xy′+ 15y= 0 to get

x 2 ·(xm)′′− 7 x·(xm)′+ 15(xm) = 0

x 2 ·m(m−1)xm− 2 − 7 x·mxm− 1 + 15xm= 0

(m 2 −m)xm− 7 mxm+ 15xm= 0

xm[m 2 − 8 m+ 15] = 0

xm[(m−3)(m−5)] = 0

The last line implies thatm= 3 orm= 5 thereforey=x 3 andy=x 5 are solutions.

In Problems37–40, we substitutey=cinto the differential equations and usey′= 0andy′′= 0

37 5c= 10 we see thaty= 2 is a constant solution.

38 2 + 2c−3 = (c+ 3)(c−1) = 0 we see thaty=−3 andy= 1 are constant solutions.

39 1/(c−1) = 0 has no solutions, the differential equation has no constant solutions.

40 6c= 10 we see thaty= 5/3 is a constant solution.

41=e− 2 t+ 3e 6 tandy=−e− 2 t+ 5e 6 twe obtain

dx dt

=− 2 e− 2 t+ 18e 6 t and dy dt

= 2e− 2 t+ 30e 6 t.

Then x+ 3y= (e− 2 t+ 3e 6 t) + 3(−e− 2 t+ 5e 6 t) =− 2 e− 2 t+ 18e 6 t=dx dt and 5 x+ 3y= 5(e− 2 t+ 3e 6 t) + 3(−e− 2 t+ 5e 6 t) = 2e− 2 t+ 30e 6 t=dy dt

.

42= cos 2t+ sin 2t+ 15 etandy=−cos 2t−sin 2t− 15 etwe obtain

dx dt=−2 sin 2t+ 2 cos 2t+

1

5 e

t and dy dt= 2 sin 2t−2 cos 2t−

1

5 e

and d 2 x dt 2

=−4 cos 2t−4 sin 2t+ 1 5

et and d

2 y dt 2

= 4 cos 2t+ 4 sin 2t− 1 5

et.

Then

4 y+et= 4(−cos 2t−sin 2t−

1

5 e

t) +et=−4 cos 2t−4 sin 2t+ 1 5 e

t=d 2 x dt 2 and 4 x−et= 4(cos 2t+ sin 2t+ 1 5

et)−et= 4 cos 2t+ 4 sin 2t− 1 5

et=d

2 y dt 2

.

Full file at TestbankDirect/

8 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

43.(y′) 2 + 1 = 0 has no real solutions because (y′) 2 + 1 is positive for all differentiable functions y=φ(x).

44 only solution of (y′) 2 +y 2 = 0 isy= 0, since ify 6 = 0,y 2 >0 and (y′) 2 +y 2 ≥y 2 >0.

45 first derivative off(x) =ex isex. The first derivative off(x) = ekx iskekx. The differential equations arey′=yandy′=ky, respectively.

46 function of the formy=cexory=ce−xis its own second derivative. The corresponding differential equation isy′′−y= 0. Functions of the formy=csinxory=ccosxhave second derivatives that are the negatives of themselves. The differential equation isy′′+y= 0.

47 first note that

√

1 −y 2 =

√

1 −sin 2 x=

√

cos 2 x=|cosx|. This prompts us to consider values ofxfor which cosx <0, such asx=π. In this case

dy dx

∣∣

∣x=π=

d dx

(sinx)

∣∣

∣x=π= cosx

∣∣

x=π= cosπ=− 1 ,

but √ 1 −y 2 |x=π=

√

1 −sin 2 π=

√

1 = 1.

Thus,y = sinxwill only be a solution ofy′ =

√

1 −y 2 when cosx >0. An interval of definition is then (−π/ 2 , π/2). Other intervals are (3π/ 2 , 5 π/2), (7π/ 2 , 9 π/2), and so on.

48 the first and second derivatives of sintand costinvolve sintand cost, it is plausible that a linear combination of these functions,Asint+Bcost, could be a solution of the differential equation. Usingy′=Acost−Bsintandy′′=−Asint−Bcostand substituting into the differential equation we get

y′′+ 2y′+ 4y=−Asint−Bcost+ 2Acost− 2 Bsint+ 4Asint+ 4Bcost

= (3A− 2 B) sint+ (2A+ 3B) cost= 5 sint

Thus 3A− 2 B= 5 and 2A+ 3B= 0. Solving these simultaneous equations we findA= 1513 andB=− 1013. A particular solution isy= 1513 sint− 1013 cost.

49 solution is given by the upper portion of the graph with domain approximately (0, 2 .6). The other solution is given by the lower portion of the graph,also with domain approximately (0, 2 .6).

50 solution, with domain approximately (−∞, 1 .6) is the portion of the graph in the second quadrant together with the lower part of the graph in the firstquadrant. A second solution, with domain approximately (0, 1 .6) is the upper part of the graph in the first quadrant. The third solution, with domain (0,∞), is the part of the graph in the fourth quadrant.

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10 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

57 differential equationyy′−xy= 0 has normal formdy/dx=x. These are not equivalent becausey= 0 is a solution of the first differential equation but not a solution of the second.

58 we gety′=c 1 + 2c 2 xandy′′= 2c 2. Thenc 2 =y′′/2 andc 1 =y′−xy′′, so

(

y′−xy′′

)

(

y′′ 2

)

x 2 =xy′−

1

2 x

2 y′′

and the differential equation isx 2 y′′− 2 xy′+ 2y= 0.

(a)Sincee−x 2 is positive for all values ofx,dy/dx >0 for allx, and a solution,y(x), of the differential equation must be increasing on any interval.

(b)x→−∞lim

dy dx= limx→−∞e

−x 2 = 0 and lim x→∞

dy dx= limx→∞e

−x 2 = 0. Sincedy/dxapproaches 0 as xapproaches−∞and∞, the solution curve has horizontal asymptotes to the left and to the right.

d 2 y dx 2 =

d dx

(

dy dx

)

=

d dx

(

e−x

2 )

=− 2 xe−x

2 .

Since the second derivative is positive forx <0 and negative forx >0, the solution curve is concave up on (−∞,0) and concave down on (0,∞).

(d)

(a)The derivative of a constant solutiony=cis 0, so solving 5−c= 0 we see thatc= 5 and soy= 5 is a constant solution.

(b)A solution is increasing wheredy/dx= 5−y >0 ory <5. A solution is decreasing wheredy/dx= 5−y <0 ory >5.

(a)The derivative of a constant solution is 0, so solvingy(a−by) = 0 we see thaty= 0 and y=a/bare constant solutions.

(b)A solution is increasing wheredy/dx=y(a−by) =by(a/b−y)>0 or 0< y < a/b. A solution is decreasing wheredy/dx=by(a/b−y)<0 ory <0 ory > a/b.

Full file at TestbankDirect/

1 Definitions and Terminology 11

d 2 y dx 2 =y(−by

′) +y′(a−by) =y′(a− 2 by).

Thusd 2 y/dx 2 = 0 wheny=a/ 2 b. Sinced 2 y/dx 2 >0 for 0< y < a/ 2 bandd 2 y/dx 2 < 0 fora/ 2 b < y < a/b, the graph ofy=φ(x) has a point of inflection aty=a/ 2 b.

(d)

y = a / b

y = 0 x

(a)Ify=cis a constant solution theny′= 0, butc 2 + 4 is never 0 for any real value ofc.

(b)Sincey′=y 2 + 4>0 for allxwhere a solutiony=φ(x) is defined, any solution must be increasing on any interval on which it is defined. Thus it cannot have any relative extrema.

(c) Using implicit differentiation we compute d 2 y/dx 2 = 2yy′ = 2y(y 2 + 4). Setting d 2 y/dx 2 = 0 we see that y= 0 corresponds to the only possible point of inflection. Sinced 2 y/dx 2 <0 fory <0 andd 2 y/dx 2 >0 fory >0, there is a point of inflection wherey= 0.

(d)

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1 Initial-Value Problems 13

7 the initial conditions we obtain the system

c 1 =− 1 c 2 = 8

The solution of the initial-value problem isx=−cost+ 8 sint.

8 the initial conditions we obtain the system

c 2 = 0−c 1 = 1

The solution of the initial-value problem isx=−cost.

9 the initial conditions we obtain √ 3 2 c 1 +

1

2 c 2 =

1 2 −

1

2 c 2 +

√

3 2 = 0

Solving, we findc 1 =

√

3 /4 andc 2 = 1/4. The solution of the initial-value problem is

x= (

√

3 /4) cost+ (1/4) sint.

10 the initial conditions we obtain √ 2 2 c 1 +

√

2

2 c 2 =

√

2

[6pt]−

√

2

c 1 +

√

2

c 2 = 2

√

2.

Solving, we findc 1 = −1 andc 2 = 3. The solution of the initial-value problem is x= −cost+ 3 sint. In Problems11–14, we usey=c 1 ex+c 2 e−xandy′=c 1 ex−c 2 e−xto obtain a system of two equations in the two unknownsc 1 andc 2. 11 the initial conditions we obtain

c 1 +c 2 = 1

c 1 −c 2 = 2.

Solving, we findc 1 = 32 andc 2 = − 12. The solution of the initial-value problem isy = 3 2 e

x− 1 2 e

−x.

12 the initial conditions we obtain

ec 1 +e− 1 c 2 = 0

ec 1 −e− 1 c 2 =e.

Solving, we findc 1 = 12 andc 2 =− 12 e 2. The solution of the initial-value problem is

y= 1 2

ex− 1 2

e 2 e−x= 1 2

ex− 1 2

e 2 −x.

Full file at TestbankDirect/

14 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

13 the initial conditions we obtain

e− 1 c 1 +ec 2 = 5

e− 1 c 1 −ec 2 =− 5.

Solving, we findc 1 = 0 andc 2 = 5e− 1. The solution of the initial-value problem isy= 5 e− 1 e−x= 5e− 1 −x.

14 the initial conditions we obtain

c 1 +c 2 = 0

c 1 −c 2 = 0.

Solving, we findc 1 =c 2 = 0. The solution of the initial-value problem isy= 0.

15 solutions arey= 0 andy=x 3.

16 solutions arey= 0 andy=x 2. (Also, any constant multiple ofx 2 is a solution.)

17(x, y) =y 2 / 3 we have∂f ∂y

= 2

3

y− 1 / 3. Thus, the differential equation will have a unique solution in any rectangular region of the plane wherey 6 = 0.

18(x, y) =√xywe have∂f /∂y= 12

√

x/y. Thus, the differential equation will have a unique solution in any region wherex >0 andy >0 or wherex <0 andy <0.

19(x, y) =

y xwe have

∂f ∂y=

1

x. Thus, the differential equation will have a unique solution in any region wherex 6 = 0.

20(x, y) =x+y we have

∂f ∂y = 1. Thus, the differential equation will have a unique solution in the entire plane.

21(x, y) =x 2 /(4−y 2 ) we have∂f /∂y= 2x 2 y/(4−y 2 ) 2. Thus the differential equation will have a unique solution in any region wherey <−2,− 2 < y <2, ory >2.

22(x, y) =

x 2 1 +y 3 we have

∂f ∂y =

− 3 x 2 y 2 (1 +y 3 ) 2

. Thus, the differential equation will have a

unique solution in any region wherey 6 =−1.

23(x, y) =

y 2 x 2 +y 2 we have

∂f ∂y=

2 x 2 y (x 2 +y 2 ) 2

. Thus, the differential equation will have a

unique solution in any region not containing (0,0).

Full file at TestbankDirect/

16 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

(a)Applyingy(1) = 1 toy=− 1 /(x+c) gives

1 =− 1 1 +c

or 1 +c=− 1

Thusc=−2 and

y=−x− 12 = 2 − 1 x.

(b)Applyingy(3) =−1 toy=− 1 /(x+c) gives

−1 =−

1

3 +c or 3 +c= 1. Thusc=−2 and

y=− 1 x− 2

= 1

2 −x

.

(c) No, they are not the same solution. The intervalIof definition for the solution in part (a) is (−∞,2); whereas the intervalIof definition for the solution in part (b) is (2,∞). See the figure.

(a)Differentiating 3x 2 −y 2 =cwe get 6x− 2 yy′= 0 oryy′= 3x.

(b)Solving 3x 2 −y 2 = 3 forywe get

y=φ 1 (x) =

√

3(x 2 −1), 1 < x <∞,

y=φ 2 (x) =−

√

3(x 2 −1), 1 < x <∞,

y=φ 3 (x) =

√

3(x 2 −1), −∞< x <− 1 ,

y=φ 4 (x) =−

√

3(x 2 −1), −∞< x <− 1.

4 –2 2 4 x
4

(a)Setting x = 2 and y = −4 in 3x 2 −y 2 = cwe get 12 −16 =−4 =c, so the explicit solution is

y=−

√

3 x 2 + 4, −∞< x <∞.

4 –2 2 4 x
4

Full file at TestbankDirect/

1 Initial-Value Problems 17

(b)Settingc= 0 we havey=

√

3 xandy=−

√

3 x, both defined on (−∞,∞).

In Problems35–38, we consider the points on the graphs withx-coordinates x 0 =− 1 ,x 0 = 0, andx 0 = 1. The slopes of the tangent lines at these points are compared with the slopes given by y′(x 0 )in(a)through(f).

35 graph satisfies the conditions in (b) and (f).

36 graph satisfies the conditions in (e).

37 graph satisfies the conditions in (c) and (d).

38 graph satisfies the conditions in (a). In Problems 39–44y=c 1 cos 2x+c 2 sin 2xis a two parameter family of solutions of the second- order differential equation y′′+ 4y = 0. In some of the problems we will use the fact that y′=− 2 c 1 sin 2x+ 2c 2 cos 2x.

39 the boundary conditionsy(0) = 0 andy

(π 4

)

= 3 we obtain

y(0) =c 1 = 0

(π 4

)

=c 1 cos

(π 2

)

+c 2 sin

(π 2

)

=c 2 = 3.

Thus,c 1 = 0,c 2 = 3, and the solution of the boundary-value problem isy= 3 sin 2x.

40 the boundary conditionsy(0) = 0 andy(π) = 0 we obtain

y(0) =c 1 = 0

y(π) =c 1 = 0.

Thus,c 1 = 0, c 2 is unrestricted, and the solution of the boundary-value problem isy = c 2 sin 2x, wherec 2 is any real number.

41 the boundary conditionsy′(0) = 0 andy′

(π 6

)

= 0 we obtain

y′(0) = 2c 2 = 0

y′

(π 6

)

=− 2 c 1 sin

(π 3

)

=−

√

3 c 1 = 0.

Thus,c 2 = 0,c 1 = 0, and the solution of the boundary-value problem isy= 0.

42 the boundary conditionsy(0) = 1 andy′(π) = 5 we obtain

y(0) =c 1 = 1

y′(π) = 2c 2 = 5.

Thus,c 1 = 1,c 2 = 5 2

, and the solution of the boundary-value problem isy= cos 2x+ 5 2

sin 2x.

Full file at TestbankDirect/

1 Initial-Value Problems 19

48 note that the initial conditiony(0) = 0,

0 =

ˆy

√ 1

t 3 + 1

is satisfied only wheny= 0. For anyy >0, necessarily ˆy

√ 1

t 3 + 1

dt > 0

because the integrand is positive on the interval of integration. Then from (12) of Section 1. and the Chain Rule we have:

d dxx=

d dx

ˆy

√ 1

t 3 + 1

1 =√ 1

y 3 + 1

dy dx

and

dy dx=

√

y 3 + 1

y′(0) =dy dx

∣∣

∣x=0=

√

(y(0)) 3 + 1 =

√

0 + 1 = 1.

Computing the second derivative, we see that:

d 2 y dx 2 =

d dx

√

y 3 + 1 =

3 y 2 2

√

y 3 + 1

dy dx=

3 y 2 2

√

y 3 + 1

· √

y 3 + 1 =

3

2 y

d 2 y dx 2

= 3

2

y 2.

This is equivalent to 2

d 2 y dx 2 − 3 y

2 = 0.

49 the solution is tangent to thex-axis at (x 0 ,0), then y′ = 0 whenx= x 0 andy= 0. Substituting these values intoy′+ 2y= 3x−6 we get 0 + 0 = 3x 0 −6 orx 0 = 2.

50 theorem guarantees a unique (meaning single) solution through any point. Thus, there cannot be two distinct solutions through any point.

51= 161 x 4 ,y′= 14 x 3 =x( 14 x 2 ) =xy 1 / 2 , andy(2) = 161 (16) = 1. When







0 , x < 0 1 16 x

4 , x≥ 0

we have

y′=







0 , x < 0 1 4 x 3 , x≥ 0







0 , x < 0 1 4 x 2 , x≥ 0

=xy 1 / 2 ,

andy(2) = 161 (16) = 1. The two different solutions are the same on the interval (0,∞), which is all that is required by Theorem 1. 2 .1.

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20 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

1 Differential Equations as Mathematical Models

1.

dP dt =kP+r;

dP dt =kP−r 2 the rate of births anddthe rate of deaths. Thenb=k 1 P andd=k 2 P. Since dP/dt=b−d, the differential equation isdP/dt=k 1 P−k 2 P.

3 the rate of births anddthe rate of deaths. Thenb=k 1 P andd=k 2 P 2. Since dP/dt=b−d, the differential equation isdP/dt=k 1 P−k 2 P 2.

dPdt =k 1 P−k 2 P 2 −h, h > 0

5 the graph in the text we estimateT 0 = 180◦ andTm= 75◦. We observe that when T= 85,dT /dt≈−1. From the differential equation we then have

k=TdT /dt−T m

= 85 −− 175 =− 0. 1.

6 inspecting the graph in the text we takeTmto beTm(t) = 80−30 cos (πt/12). Then the temperature of the body at timetis determined by the differential equation

dT dt

[

T−

(

80 −30 cos

(π 12

))]

, t > 0.

7 number of students with the flu isxand the number not infected is 1000−x, sodx/dt= kx(1000−x).

8 analogy, with the differential equation modeling the spread of a disease, we assume that the rate at which the technological innovation is adopted isproportional to the number of people who have adopted the innovation and also to the numberof people,y(t), who have not yet adopted it. If one person who has adopted the innovation is introduced into the population, thenx+y=n+ 1 and

dx dt

=kx(n+ 1−x), x(0) = 1.

9 rate at which salt is leaving the tank is

Rout(3 gal/min)·

(

A

300 lb/gal

)

=

A

100 lb/min.

ThusdA/dt=A/100. The initial amount isA(0) = 50.

10 rate at which salt is entering the tank is

Rin= (3 gal/min)·(2 lb/gal) = 6 lb/min.

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Solution Manual for A First Course in Differential Equations with Modeling Applications 11th Edition by Zill

Course: 미분적분학 (BNMU0233)

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Chapter 1

Introduction to Differential Equations

1.1 Definitions and Terminology

1. Second order; linear

2. Third order; nonlinear because of (dy/dx)4

3. Fourth order; linear

4. Second order; nonlinear because of cos(r+u)

5. Second order; nonlinear because of (dy/dx)2or p1 + (dy/dx)2

6. Second order; nonlinear because of R2

7. Third order; linear

8. Second order; nonlinear because of ˙x2

9. Writing the differential equation in the form x(dy/dx) + y2= 1, we see that it is nonlinear

in ybecause of y2. However, writing it in the form (y2−1)(dx/dy) + x= 0, we see that it is

linear in x.

10. Writing the differential equation in the form u(dv/du) +(1+u)v=ueuwe see that it is linear

in v. However, writing it in the form (v+uv −ueu)(du/dv)+u= 0, we see that it is nonlinear

in u.

11. From y=e−x/2we obtain y′=−1

2e−x/2. Then 2y′+y=−e−x/2+e−x/2= 0.

12. From y=6

5−6

5e−20twe obtain dy/dt = 24e−20t, so that

dt + 20y= 24e−20t+ 20 6

5−6

5e−20t= 24.

13. From y=e3xcos 2xwe obtain y′= 3e3xcos 2x−2e3xsin 2xand y′′ = 5e3xcos 2x−12e3xsin 2x,

so that y′′ −6y′+ 13y= 0.

Solution Manual for A First Course in Differential Equations with Modeling Applications 11th Edition by Zill

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