Calculus one and several variables 10E Salas solutions manual ch19

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-19 JWDD027-Salas-v1 January 4, 2007 19:13

984 SECTION 19.1

CHAPTER 19

SECTION 19.1

1. y′ + xy = xy3 =⇒ y−3y′ + xy−2 = x. Let v = y−2, v′ = −2y−3y′.

−12v′ + xv = x

v′ − 2xv = −2x

e−x2v′ − 2xe−x2

v = −2xe−x2

e−x2v = e−x2

+ C

v = 1 + Cex2

y2 =1

1 + Cex2 .

2. y′ − y = −(x2 + x + 1)y2 =⇒ y−2y′ − y−1 = −(x2 + x + 1). Let v = y−1, v′ = −y−2y′.

−v′ − v = −(x2 + x + 1)

v′ + v = x2 + x + 1

exv =∫

ex(x2 + x + 1) dx = x2ex − xex + 2ex + C

v = x2 − x + 2 + Ce−x

y =1

x2 − x + 2 + Ce−x.

3. y′ − 4y = 2exy12 =⇒ y−

12 y′ − 4y

12 = 2ex. Let v = y

12 , v′ =

12y−

12 y′.

2v′ − 4v = 2ex

v′ − 2v = ex

e−2xv′ − 2e−2xv = e−x

e−2xv = −e−x + C

v = −ex + Ce2x

y = (Ce2x − ex)2.

4. y′ =1

2xy+

y

2x=⇒ yy′ − 1

2xy2 =

12x

. Let v = y2, v′ = 2yy′.

12v′ − 1

2xv =

12x

v′ − 1xv =

1x

1xv′ − 1

x2v =

1x2

1xv = − 1

x+ C

v = Cx− 1

y2 = Cx− 1.



SECTION 19.1 985

5. (x− 2)y′ + y = 5(x− 2)2y12 =⇒ y−

12 y′ +

1x− 2

y12 = 5(x− 2). Let v = y

12 , v′ =

12y−

12 y′.

2v′ +1

x− 2v = 5(x− 2)

v′ +1

2(x− 2)v =

52(x− 2)

√x− 2v′ +

12√x− 2

v =52(x− 2)

32

√x− 2v = (x− 2)

52 + C

v = (x− 2)2 +C√x− 2

y =[(x− 2)2 +

C√x− 2

]2

.

6. yy′ − xy2 + x = 0. Let v = y2, v′ = 2yy′.

12v′ − xv = −x

v′ − 2xv = −2x


v = −2xe−x2

e−x2v = e−x2

+ C

v = 1 + Cex2

y =√

1 + Cex2 .

7. y′ + xy = y3ex2

=⇒ y−3y′ + xy−2 = ex2. Let v = y−2, v′ = −2y−3y′.

−12v′ + xv = ex

2

v′ − 2xv = −2ex2


v = −2

e−x2v = −2x + C

v = −2xex2+ Cex

2

y−2 = Cex2 − 2xex

2.

C = 4 =⇒ y−2 = 4ex2 − 2xex

2.

8. y′ +1xy =

lnx

xy2 =⇒ y−2y′ +

1xy−1 =

lnx

x. Let v = y−1, v′ = −y−2y′.

−v′ +1xv =

lnx

x

v′ − 1xv = − lnx

x

1xv′ − 1

x2v = − lnx

x2



986 SECTION 19.1

1xv = −

∫lnx

x2dx =

1x

(lnx + 1) + C

v = lnx + 1 + Cx

y =1

lnx + 1 + Cx.

1 =1

ln 1 + 1 + C=⇒ C = 0 =⇒ y =

1lnx + 1

.

9. 2x3y′ − 3x2y = y3 =⇒ y−3y′ − 32x

y−2 =1

2x3. Let v = y−2, v′ = −2y−3y′.

−12v′ − 3

2xv =

12x3

v′ +3xv = − 1

x3

x3v′ + 3x2v = −1

x3v = −x + C

v =C − x

x3

y2 =x3

C − x

1 =1

C − x=⇒ C = 2 =⇒ y2 =

x3

2 − x.

10. y′ + tanx y = y2 sec3 x =⇒ y−2 y′ + tanxy−1 = sec3 x. Let v = y−1, v′ = −y−2y′.

−v′ + tanxv = sec3 x

v′ − tanx v = − sec3 x

cosxv′ − sinx v = − sec2 x

cosx v = − tanx + C

cosxy

= − tanx + C

cos 03

= − tan 0 + C =⇒ C =13

=⇒ cosxy

=13− tanx.

11. y′ − y

xln y = xy =⇒ y′

y− 1

xln y = x. Let u = ln y, u′ =

y′

y.

u′ − 1xu = x

1xu′ − 1

x2u = 1

1xu = x + C

u = x2 + Cx

ln y = x2 + Cx.



SECTION 19.1 987

12. (a) y′ + yf(x) ln y = g(x)yy′

y+ f(x) ln y = g(x)

u′ + f(x)u = g(x).

(b) cos y y′ + g(x) sin y = f(x). Let u = sin y, u′ = cos y y′.

Thus we have u′ + g(x)u = f(x).

13. f(x, y) =x2 + y2

2xy; f(tx, ty) =

(tx)2 + (ty)2

2(tx)(ty)=

t2(x2 + y2)t2(2xy)

=x2 + y2

2xy= f(x, y)

Set vx = y. Then, v + xv′ = y′ and

v + xv′ =x2 + v2x2

2vx2=

1 + v2

2v

v − 1 + v2

2v+ xv′ = 0

v2 − 1 + 2xvv′ = 0

1xdx +

2vv2 − 1

dv = 0

∫1xdx +

∫2v

v2 − 1dv = C

ln |x | + ln |v2 − 1 | = K or x(v2 − 1) = C

Replacing v by y/x, we get

x

(y2

x2− 1

)= C or y2 − x2 = Cx

14. f(tx, ty) =(ty)2

(tx)(ty) + (tx)2=

y2

xy + x2= f(x, y).


v + xv′ =v2

1 + v∫dx

x+

∫v + 1v

dv = C

ln |x| + v + ln |v| = C

v + ln |xv| = C

y

x+ ln |y| = C

15. f(x, y) =x− y

x + y; f(tx, ty) =

(tx) − (ty)tx + ty

=t(x− y)t(x + y)

=x− y

x + y= f(x, y)


v + xv′ =x− vx

x + vx=

1 − v

1 + v

v2 + 2v − 1 + x(1 + v)v′ = 0



988 SECTION 19.1

1xdx +

1 + v

v2 + 2v − 1dv = 0

∫1xdx +

∫1 + v

v2 + 2v − 1dv = C

ln |x | + 12 ln |v2 + 2v − 1 | = K or x

√v2 + 2v − 1 = C


x

√y2

x2+ 2

y

x− 1 = C or y2 + 2xy − x2 = C

16. f(tx, ty) =tx + ty

tx− ty=

x + y

x− y= f(x, y).


v + xv′ =x + vx

x− vx=

1 + v

1 − v∫dx

x+

∫v − 1v2 + 1

dv = C1

ln |x| + 12

ln |v2 + 1| − arctan v = C1

lnx2 + ln(v2 + 1) − 2 arctan v = C (= 2C1)

ln[x2(v2 + 1)] − 2 arctan v = C

ln[x2 + y2] − 2 arctan(y

x

)= C

17. f(x, y) =x2ey/x + y2

xy; f(tx, ty) =

(tx)2 − e(ty)/(tx) + (ty)2

(tx)(ty)=

t2(x2ey/x + y2

)t2(xy)

= f(x, y)


v + xv′ =x2ev + v2x2

vx2=

ev + v2

v

v2 + xvv′ = ev + v2

−ev + xvv′ = 0

1xdx = ve−v dv

∫1xdx =

∫ve−v dv

ln |x | = −ve−v − e−v + C

Replacing v by y/x, and simplifying, we get

y + x = xey/x(C − ln |x |)



SECTION 19.1 989

18. f(tx, ty) =(tx)2 + 3(ty)2

4(tx)(ty)=

x2 + 3y2

4xy= f(x, y).


v + xv′ =x2 + 3x2v2

4x2v∫dx

x+

∫4v

v2 − 1dv = C1

ln |x| + 2 ln |v2 − 1| = C1

x(v2 − 1)2 = C (= eC1)

(y2 − x2)2 = Cx3

19. f(x, y) =y

x+ sin(y/x); f(tx, ty) =

(ty)tx

+ sin[(ty/tx)] =y

x+ sin(y/x) = f(x, y)


v + xv′ =vx

x+ sin[(vx)/x] = v + sin v

xv′ = sin v

csc v dv =1xdx∫

csc v dv =∫

1xdx

ln | csc v − cot v | = ln |x | + K or csc v − cot v = Cx

Replacing v by y/x, and simplifying, we get

1 − cos(y/x) = Cx sin(y/x)

20. f(x, y) =y

x

(1 + ln

(y

x

)); f(tx, ty) =

ty

tx

(1 + ln

(ty

tx

))=

y

x

(1 + ln

(y

x

))= f(x, y)


v + xv′ =vx

x

(1 + ln

(vx

x

))= v(1 + ln v)

xv′ = v ln v

1v ln v

dv =1xdx∫

1v ln v

dv =∫

1xdx

ln |ln v| = ln |x| + K

ln(y

x

)= Cx

y

x= eCx or y = xeCx

21. The differential equation is homogeneous since

f(x, y) =y3 − x3

xy2; f(tx, ty) =

(ty)3 − (tx)3

(tx)(ty)2=

t3(y3 − x3)t3(xy2)

=y3 − x3

xy2= f(x, y)



990 SECTION 19.2


v + xv′ =(vx)3 − x3

v2x3=

v3 − 1v2

1 + xv2v′ = 0

1xdx + v2 dv = 0∫

1xdx +

∫v2 dv = 0

ln |x | + 13v3 = C


y3 + 3x3 ln |x | = Cx3

Applying the side condition y(1) = 2, we have

8 + 3 ln 1 = C =⇒ C = 8 and y3 + 3x3 ln |x | = 8x3

22.dy

dx=

1sin(y/x)

+y

x. Set y = vx. Then y′ = v + xv′ and

v + xv′ =1

sin v+ v∫

dx

x−

∫sin v dv = C

ln |x| + cos v = C

ln |x| + cos(yx

)= C

y(1) = 0 =⇒ 0 + cos 0 = C =⇒ C = 1 =⇒ ln |x| + cos(y

x

)= 1

SECTION 19.2

1.∂P

∂y= 2xy − 1 =

∂Q

∂x; the equation is exact on the whole plane.

∂f

∂x= xy2 − y =⇒ f(x, y) = 1

2 x2y2 − xy + ϕ(y)

∂f

∂y= x2y − x + ϕ′(y) = x2y − x =⇒ ϕ′(y) = 0 =⇒ ϕ(y) = 0 (omit the constant)∗

Therefore f(x, y) = 12 x

2y2 − xy, and a one-parameter family of solutions is:

12 x

2y2 − xy = C

∗ We will omit the constant at this step throughout this section.

2.∂

∂y(exsin y) = ex cos y =

∂

∂x(ex cos y); the equation is exact on the whole plane.

f(x, y) = ex sin y, and ex sin y = C is a one-parameter family of solutions.



SECTION 19.2 991

3.∂P

∂y= ey − ex =

∂Q


∂f

∂x= ey − yex =⇒ f(x, y) = xey − yex + ϕ(y)

∂f

∂y= xey − ex + ϕ′(y) = xey − ex =⇒ ϕ′(y) = 0 =⇒ ϕ(y) = 0

Therefore f(x, y) = xey − yex, and a one-parameter family of solutions is:

xey − yex = C

4.∂

∂y(sin y) = cos y =

∂

∂x(x cos y + 1); the equation is exact on the whole plane.

f(x, y) = x sin y + y, and x sin y + y = C is a one-parameter family of solutions.

5.∂P

∂y=

1y

+ 2x =∂Q

∂x; the equation is exact on the upper half plane.

∂f

∂x= ln y + 2xy =⇒ f(x, y) = x ln y + x2y + ϕ(y)

∂f

∂y=

x

y+ x2 + ϕ′(y) =

x

y+ x2 =⇒ ϕ′(y) = 0 =⇒ ϕ(y) = 0

Therefore f(x, y) = x ln y + x2y, and a one-parameter family of solutions is:

x ln y + x2y = C

6.∂

∂y(2x arctan y) =

2x1 + y2

=∂

∂x

(x2

1 + y2

); the equation is exact on the whole plane.

f(x, y) = x2 arctan y, and x2 arctan y = C is a one-parameter family of solutions.

7.∂P

∂y=

1x

=∂Q

∂x; the equation is exact on the right half plane.

∂f

∂x=

y

x+ 6x =⇒ f(x, y) = y lnx + 3x2 + ϕ(y)

∂f

∂y= lnx + ϕ′(y) = lnx− 2 =⇒ ϕ′(y) = −2 =⇒ ϕ(y) = −2y

Therefore f(x, y) = y lnx + 3x2 − 2y, and a one-parameter family of solutions is:

y lnx + 3x2 − 2y = C

8.∂

∂y(ex + ln y +

y

x) =

1y

+1x

=∂

∂x(x

y+ lnx + sin y);

the equation is exact in the first quadrant, not including the axes.

f(x, y) = ex + x ln y + y lnx− cos y and ex + x ln y + y lnx− cos y = C is a

one-parameter family of solutions.

9.∂P

∂y= 3y2 − 2y sinx =

∂Q


∂f

∂x= y3 − y2 sinx− x =⇒ f(x, y) = xy3 + y2 cosx− 1

2 x2 + ϕ(y)



992 SECTION 19.2

∂f

∂y= 3xy2 + 2y cosx + ϕ′(y) = 3xy2 + 2y cosx + e2y =⇒ ϕ′(y) = e2y =⇒ ϕ(y) = 1

2 e2y

Therefore f(x, y) = xy3 + y2 cosx− 12 x

2 + 12 e

2y, and a one-parameter family of solutions is:

xy3 + y2 cosx− 12 x

2 + 12 e

2y = C

10.∂

∂y(e2y − y cosxy) = 2e2y − cosxy + xy sinxy =

∂

∂x(2xe2y − x cosxy + 2y);

the equation is exact on the whole plane.

f(x, y) = xe2y − sinxy + y2 and xe2y − sinxy + y2 = C is a

one-parameter family of solutions.

11. (a) Yes:∂

∂y[p(x)] = 0 =

∂

∂x[q(y)].

(b) For all x, y such that p(y)q(x) �= 0,1

p(y)q(x)is an integrating factor.

Multiplying the differential equation by1

p(y)q(x), we get

1q(x)

+1

p(y)y′ = 0

which has the form of the differential equation in part (a).

12. Mimic the proof of the first part.

13.∂P

∂y= ey−x − 1 and

∂Q

∂x= ey−x − xey−x; the equation is not exact.

Since1Q

(∂P

∂y− ∂Q

∂x

)=

1xey−x − 1

(xey−x − 1

)= 1, μ(x) = e

∫dx = ex is

an integrating factor. Multiplying the given equation by ex, we get

(ey − yex) + (xey − ex) y′ = 0

This is the equation given in Exercise 3. A one-parameter family of solutions is:

xey − yex = C

14. w =1P

(∂P

∂y− ∂Q

∂x

)=

1x + ey

(ey + x) = 1 doesn’t depend on x, so e∫

−dy = e−y is an integrating

factor.

(xe−y + 1) − 12x2e−yy′ = 0 is exact; f(x, y) = 1

2x2e−y + x and a one-parameter family of solutions

is 12x

2e−y + x = C.

15.∂P

∂y= 6x2y + ey =

∂Q

∂x; the equation is exact.

∂f

∂x= 3x2y2 + x + ey =⇒ f(x, y) = x3y2 + 1

2 x2 + xey + ϕ(y)



SECTION 19.2 993

∂f

∂y= 2x3y + xey + ϕ′(y) = 2x3y + y + xey =⇒ ϕ′(y) = y =⇒ ϕ(y) = 1

2 y2

Therefore f(x, y) = x3y2 + 12 x

2 + xey + 12 y

2, and a one-parameter family of solutions is:

x3y2 + 12 x

2 + xey + 12 y

2 = C

16.∂

∂y(sin 2x cos y) = − sin 2x sin y = −2 sinx cosx sin y =

∂

∂x(− sin2 x sin y); exact.

f(x, y) = sin2 x cos y and sin2 x cos y = C is a one-parameter family of solutions.

17.∂P

∂y= 3y2 and

∂Q

∂x= 0; the equation is not exact.

Since1Q

(∂P

∂y− ∂Q

∂x

)=

13y2

(3y2) = 1, μ(x) = e∫

dx = ex is an

an integrating factor. Multiplying the given equation by ex, we get(y3ex + xex + ex

)+

(3y2ex

)y′ = 0

∂f

∂x= y3ex + xex + ex =⇒ f(x, y) = y3ex + xex + ϕ(y)

∂f

∂y= 3y2ex + ϕ′(y) = 3y2ex =⇒ ϕ′(y) = 0 =⇒ ϕ(y) = 0

Therefore f(x, y) = y3ex + xex, and a one-parameter family of solutions is:

y3ex + xex = C

18. v =1Q

(∂P

∂y− ∂Q

∂x

)=

1xe2x+y + 1

(−2xe2x+y − 2

)= −2, independent of y, so e

∫−2dx = e−2x

is an integrating factor. Thus (ey − 2ye−2x) + (xey + e−2x)y′ = 0 is exact.

f(x, y) = xey + ye−2x, and xey + ye−2x = C is a one-parameter family of solutions.

19.∂P

∂y= 1 =

∂Q


∂f

∂x= x2 + y =⇒ f(x, y) = 1

3 x3 + xy + ϕ(y)

∂f

∂y= x + ϕ′(y) = x + ey =⇒ ϕ′(y) = ey =⇒ ϕ(y) = ey

Therefore f(x, y) = 13 x

3 + xy + ey, and a one-parameter family of solutions is:

13 x

3 + xy + ey = C

Setting x = 1, y = 0, we get C = 43 and

13 x

3 + xy + ey = 43 or x3 + 3xy + 3ey = 4



994 SECTION 19.2

20.∂

∂y(3x2 − 2xy + y3) = −2x + 3y2 =

∂

∂x

(3xy2 − x2

); exact

f(x, y) = x3 − x2y + xy3 =⇒ x3 − x2y + xy3 = C.

Substituting x = 1, y = −1 we get 1 + 1 − 1 = C =⇒ x3 − x2y + xy3 = 1

21.∂P

∂y= 4y and

∂Q

∂x= 2y; the equation is not exact.

Since1Q

(∂P

∂y− ∂Q

∂x

)=

12xy

(2y) =1x, μ(x) = e

∫(1/x)dx = elnx = x is an

integrating factor. Multiplying the given equation by x, we get(2xy2 + x3 + 2x

)+

(2x2y

)y′ = 0

∂f

∂y= 2x2y =⇒ f(x, y) = x2y2 + ϕ(x)

∂f

∂x= 2xy2 + ϕ′(x) = 2xy2 + x3 + 2x =⇒ ϕ′(x) = x3 + 2x =⇒ ϕ = 1

4 x4 + x2

Therefore f(x, y) = x2y2 + 14 x

4 + x2, and a one-parameter family of solutions is:

x2y2 + 14 x

4 + x2 = C

Setting x = 1, y = 0, we get C = 54 and

x2y2 + 14 x

4 + x2 = 54 or 4x2y2 + x4 + 4x2 = 5

22. v =1Q

(∂P

∂y− ∂Q

∂x

)=

2 − 6xy2

3x2y2 − x= − 2

xdoesn’t depend on y, so e

∫− 2

x dx = x−2

is an integrating factor. Thus (1 + yx−2) + (3y2 − x−1)y′ = 0 is exact.

f(x, y) = x− y

x+ y3 =⇒ x− y

x+ y3 = C

Substituting x = 1, y = 1 we get 1 − 1 + 1 = C =⇒ x− y

x+ y = 1.

23.∂P

∂y= 3y2 and

∂Q

∂x= y2; the equation is not exact.

Since1P

(∂P

∂y− ∂Q

∂x

)=

1y3

(2y2) =2y, w(y) = e−

∫(2/y)dy = e−2 lny = y−2 is an

integrating factor. Multiplying the given equation by y−2, we get

y +(y−2 + x

)y′ = 0

∂f

∂x= y =⇒ f(x, y) = xy + ϕ(y)

∂f

∂y= x + ϕ′(y) = y−2 + x =⇒ ϕ′(y) = y−2 =⇒ ϕ(y) = − 1

y

Therefore f(x, y) = xy − 1y, and a one-parameter family of solutions is: xy − 1

y= C

Setting x = −2, y = −1, we get C = 3 and the solution xy − 1y

= 3.



SECTION 19.2 995

24.∂

∂y(x + y)2 = 2(x + y) =

∂

∂x(2xy + x2 − 1); exact.

f(x, y) =x3

3+ x2y + xy2 − y =⇒ x3

3+ x2y + xy2 − y = C

Setting x = 1, y = 1, we get C = 43 and the solution

x3

3+ x2y + xy2 − y =

43.

25.∂P

∂y= −2y sinh(x− y2) =

∂Q


∂f

∂x= cosh(x− 2y2) + e2x =⇒ f(x, y) = sinh(x− y2) + 1

2 e2x + ϕ(y)

∂f

∂y= −2y cosh(x− y2) + ϕ′(y) = y − 2y cosh(x− y2) =⇒ ϕ′(y) = y =⇒ ϕ(y) = 1

2 y2

Therefore f(x, y) = sinh(x− y2) + 12 e

2x + 12 y

2, and a one-parameter family of solutions is:

sinh(x− y2) + 12 e

2x + 12 y

2 = C

Setting x = 2, y =√

2, we get C = 12e

4 + 1 and the solution

sinh(x− y2) + 12 e

2x + 12 y

2 = 12e

4 + 1

26. Write the linear equation as p(x)y − q(x) + y′ = 0. Then P (x) = p(x)y − q(x), Q(x) = 1

and v =1Q

(∂P

∂y− ∂Q

∂x

)= p(x) depends only on x. Therefore v = e

∫p(x)dx is

an integrating factor.

27. (a)∂P

∂y= 2xy + kx2 and

∂Q

∂x= 2xy + 3x2 =⇒ k = 3.

(b)∂P

∂y= e2xy + 2xye2xy and

∂Q

∂x= ke2xy + 2kxye2xy =⇒ k = 1.

28. (a) We need g′(y) sinx = y2f ′(x). Take g(y) = 13y

3 and f(x) = − cosx

(b) We need g′(y)ey + g(y)ey = y, that isd

dy[g(y)ey] = y.

It follows that g(y)ey = 12y

2 + C, =⇒ g(y) = e−y( 12y

2 + C).

29. y′ = y2x3; the equation is separable.

y−2 dy = x3 dx =⇒ − 1y

= 14 x

4 + C =⇒ y =−4

x4 + C

30. y y′ = 4xe2x+y = 4xe2xey =⇒ the equation is separable.

ye−y dy = 4xe2x dx =⇒ −ye−y − e−y = 2xe2x − e2x + C

31. y′ +4xy = x4; the equation is linear.

H(x) =∫

(4/x) dx = 4 lnx = lnx4, integrating factor: elnx4= x4



996 SECTION 19.3

x4y′ + 4x3y = x8

ddx

[x4y

]= x8

x4y = 19 x

9 + C

y = 19 x

5 + Cx−4

32. y′ + 2xy = 2x3; the equation is linear with integrating factor e∫

2xdx = ex2

=⇒ d

dx(ex

2y) = 2x3ex

2=⇒ ex

2y = ex

2(x2 − 1) + C =⇒ y = x2 − 1 + Ce−x2

.

33.∂P

∂y= exy + xyexy =

∂Q


∂f

∂x= yexy − 2x =⇒ f(x, y) = exy − x2 + ϕ(y)

∂f

∂y= xexy + ϕ′(y) =

2y

+ xexy =⇒ ϕ′(y) =2y

=⇒ ϕ(y) = 2 ln | y |

Therefore f(x, y) = exy − x2 + 2 ln | y |, and a one-parameter family of solutions is:

exy − x2 + 2 ln | y | = C

34. w =1P

(∂P

∂y− ∂Q

∂x

)=

1y(1 − 2y) depends only on y, so an integrating factor is

e∫

−w(y)dy = e∫

[2−(1/y)]dy = e2y−lny =1ye2y.

Then e2y dx +(

2xe2y − 1y

)dy = 0 is exact.

f(x, y) = xe2y − ln y, and a one-parameter family of solutions is xe2y − ln y = C .

SECTION 19.3

1. y′ = y =⇒ y = Cex. Also, y(0) = 1 =⇒ C = 1

Thus y = ex and y(1) = 2.71828

(a) 2.48832, relative error= 8.46%.

(b) 2.71825, relative error= 0.001%.

2. y′ = x + y =⇒ y = Cex − x− 1, y(0) = 2 =⇒ C = 3

Thus y = 3ex − x− 1 and y(1) � 6.15485

(a) 5.46496, relative error = 11.2%.

(b) 6.15474, relative error = 0%.

3. (a) 2.59374, relative error= 4.58%.

(b) 2.71828, relative error= 0%.

4. (a) 5.78124, relative error= 6.07%.




SECTION 19.3 997

5. y′ = 2x =⇒ y = x2 + C. Also, y(2) = 5 =⇒ C = 1

Thus y = x2 + 1 and y(1) = 2.



6. y′ = 3x2 =⇒ y = x3 + C. Also, y(1) = 2 =⇒ C = 1

Thus y = x3 + 1 and y(0) = 1.



7. y′ =12y

Thus y =√x and y(2) =

√2 � 1.41421.

(a) 1.42052, relative error= −0.45%.


8. y′ =1

3y2

Thus y = x13 and y(2) � 1.25992.

(a) 1.26494, relative error= −0.4%.


9. (a) 2.65330, relative error= 2.39%.


10. (a) 5.95989, relative error= 3.17%.


PROJECT 19.3

1. (a) and (b)

−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3

y' = y

(c) y − y′ = 0 H(x) =∫−dx = −x; integrating factor: e−x

e−xy′ − e−xy = 0

d

dx(e−xy) = 0

e−xy = C

y = Cex

y(0) = 1 =⇒ C = 1. Thus y = ex.



998 SECTION 19.3

2. (a) and (b)

−1 −0.5 0 0.5 1 1.5 2

−1

0

1

2

3

4

5

6

7

8

z

y' = x+2y

(c) y′ − 2y = x H(x) =∫−2 dx = −2x; integrating factor: e−2x

e−2xy′ − 2e−2xy = xe−2x

d

dx(e−2xy) = xe−2x

e−2xy = −12xe−2x − 1

4e−2x + C

y = Ce2x − 12x− 1

4

y(0) = 1 =⇒ C =54. Thus y =

54e2x − 1

2x− 1

4.

3. (a) and (b)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3

−1

0

1

2

3

4

5

6

7

8

y' = 2xy



SECTION 19.4 999

(c) y′ − 2xy = 0 H(x) =∫−2x dx = −x2; integrating factor: e−x2

e−x2y′ − 2xex

2y = 0

d

dx(e−x2

y) = 0

e−x2y = C

y = Cex2

y(0) = 1 =⇒ C = 1. Thus y = ex2.

4. (a) and (b)

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−3

−2

−1

0

1

2

3

y' = 4x/y

(c) y′ = −4xy

;12y2 = −2x2 + C or x2 +

14y2 = C

y(1) = 1 =⇒ C =54. Thus 4x2 + y2 = 5.

SECTION 19.4

1. First consider the reduced equation. The characteristic equation is:

r2 + 5r + 6 = (r + 2)(r + 3) = 0

and u1(x) = e−2x, u2(x) = e−3x are fundamental solutions. A particular solution of the given

equation has the form

y = Ax + B.

The derivatives of y are: y′ = A, y′′ = 0.

Substituting y and its derivatives into the given equation gives

0 + 5A + 6(Ax + B) = 3x + 4.



1000 SECTION 19.4

Thus,

6A = 3

5A + 6B = 4

The solution of this pair of equations is: A = 12 , B = 1

4 , and y = 12 x + 1

4 .

2. The constant yp = − 12 is a solution.


r2 + 2r + 5 = 0

and u1(x) = e−x cos 2x, u2(x) = e−x sin 2x are fundamental solutions. A particular solution of

the given equation has the form

y = Ax2 + Bx + C.

The derivatives of y are: y′ = 2Ax + B, y′′ = 2A.


2A + 2(2Ax + B) + 5(Ax2 + Bx + C) = x2 − 1.

Thus,

5A = 1

4A + 5B = 0

2A + 2B + 5C = −1

The solution of this system of equations is: A = 15 , B = − 4

25 , C = − 27125 , and

y = 15 x

2 − 425 x− 27

125 .

4. We try y = Ax3 + Bx2 + Cx + D :

y′′ + y′ − 2y = (6Ax + 2B) + (3Ax2 + 2Bx + C) − 2(Ax3 + Bx2 + Cx + D) = x3 + x.

=⇒ −2A = 1, 3A− 2B = 0, 6A + 2B − 2C = 1, 2B + C − 2D = 0

=⇒ A = − 12 , B = − 3

4 , C = − 114 , D = − 17

8 ; yp = − 12x

3 − 34x

2 − 114 x− 17

8 .


r2 + 6r + 9 = (r + 3)2 = 0

and u1(x) = e−3x, u2(x) = xe−3x are fundamental solutions. A particular solution of the given


y = Ae3x.

The derivatives of y are: y′ = 3Ae3x, y′′ = 9Ae3x.



SECTION 19.4 1001


9Ae3x + 18Ae3x + 9Ae3x = e3x.

Thus, 36A = 1 =⇒ A =136

, and y = 136 e

3x.

6. Since −3 is a double root of the characteristic equation r2 + 6r + 9 = 0, we try

y = Ax2e−3x. Then y′ = A(−3x2 + 2x)e−3x, y′′ = A(9x2 − 12x + 2)e−3x, and

[A(9x2 − 12x + 2) + 6A(−3x2 + 2x) + 9Ax2]e−3x = e−3x, or 2Ae−3x = e−3x

Thus A = 12 and yp = 1

2x2e−3x.


r2 + 2r + 2 = 0

and u1(x) = e−x cosx, u2(x) = e−x sinx are fundamental solutions. A particular solution of the

given equation has the form

y = Aex.

The derivatives of y are: y′ = Aex, y′′ = Aex.


Aex + 2Aex + 2Aex = ex.

Thus, 5A = 1 =⇒ A = 15 and y = 1

5 ex.

8. Try y = (A + Bx)e−x. Substituting into y′′ + 4y′ + 4y = xe−x gives

A = −2, B = 1; yp = (x− 2)e−x


r2 − r − 12 = (r − 4)(r + 3) = 0

and u1(x) = e4x, u2(x) = e−3x are fundamental solutions. A particular solution of the given


y = A cosx + B sinx.

The derivatives of y are: y′ = −A sinx + B cosx, y′′ = −A cosx−B sinx.


−A cosx−B sinx− (−A sinx + B cosx) − 12(A cosx + B sinx) = cosx.

Thus,

−13A−B = 1

A− 13B = 0



1002 SECTION 19.4

The solution of this system of equations is: A = − 13170 , B = − 1

170 , and

y = − 13170 cosx− 1

170 sinx.

is a particular solution of the complete equation.

10. Try y = A cosx + B sinx. Substituting into y′′ − y′ − 12y = sinx gives

A = 1170 , B = −13

170 ; yp = 1170 cosx− 13

170 sinx.


r2 + 7r + 6 = (r + 6)(r + 1) = 0

and u1(x) = e−6x, u2(x) = e−x are fundamental solutions. A particular solution of the given


y = A cos 2x + B sin 2x.

The derivatives of y are: y′ = −2A sin 2x + 2B cos 2x, y′′ = −4A cos 2x− 4B sin 2x.


−4A cos 2x− 4B sin 2x + 7(−2A sin 2x + 2B cos 2x) + 6(A cos 2x + B sin 2x) = 3 cos 2x.

Thus,

2A + 14B = 3

−14A + 2B = 0

The solution of this system of equations is: A = 3100 , B = 21

100 and

y = 310 cos 2x + 21

100 sin 2x.

12. Try y = A cos 3x + B sin 3x. Substituting into y′′ + y′ + 3y = sin 3x gives

A = − 115 , B = − 2

15 ; yp = − 115 cos 3x− 2

15 sin 3x.


r2 − 2r + 5 = 0

and u1(x) = ex cos 2x, u2(x) = ex sin 2x are fundamental solutions. A particular solution of the

given equation has the form

y = Ae−x cos 2x + Be−x sin 2x

The derivatives of y are: y′ = −Ae−x cos 2x− 2Ae−x sin 2x−Be−x sin 2x + 2Be−x cos 2x,

y′′ = 4Ae−x sin 2x− 3Ae−x cos 2x− 4Be−x cos 2x− 3Be−x sin 2x.


4Ae−x sin 2x− 3Ae−x cos 2x− 4Be−x cos 2x− 3Be−x sin 2x−2 (−Ae−x cos 2x− 2Ae−x sin 2x−Be−x sin 2x + 2Be−x cos 2x) +

5 (Ae−x cos 2x + Be−x sin 2x) = e−x sin 2x.



SECTION 19.4 1003

Equating the coefficients of e−x cos 2x and e−x sin 2x we get,

8A + 4B = 1

4A− 8B = 0

The solution of this system of equations is: A = 110 , B = 1

20 and

y = 110 e

−x cos 2x + 120 e

−x sin 2x.

14. Try y = e2x(A cosx + B sinx). Substituting into y′′ + 4y′ + 5y = e2x cosx gives

A = 120 , B = 1

40 ; yp = e2x(

120 cosx + 1

40 sinx).


r2 + 6r + 8 = (r + 4)(r + 2) = 0

and u1(x) = e−4x, u2(x) = e−2x are fundamental solutions. A particular solution of the given


y = Axe−2x.

The derivatives of y are: y′ = Ae−2x − 2Axe−2x, y′′ = −4Ae−2x + 4Axe−2x.


−4Ae−2x + 4Axe−2x + 6(Ae−2x − 2Axe−2x

)+ 8Axe−2x = 3e−2x

Thus, 2A = 3 =⇒ A = 32 and y = 3

2 xe−2x.

16. Try y = ex (A cosx + B sinx) . Substituting into y′′ − 2y′ + 5y = ex sinx gives

A = 0, B = 13 ; yp = 1

3ex sinx.

17. First consider the reduced equation: y′′ + y = 0. The characteristic equation is:

r2 + 1 = 0

and u1(x) = cosx, u2(x) = sinx are fundamental solutions. A particular solution of the given


y = Aex.

The derivatives of y are: y′ = y′′ = Aex.

Substitute y and its derivatives into the given equation:

Aex + Aex = ex =⇒ A =12

and y = 12 e

x.

The general solution of the given equation is: y = C1 cosx + C2 sinx + 12 e

x.



1004 SECTION 19.4

18. r2 − 2r + 1 = 0 =⇒ r = 1 =⇒ y = C1ex + C2xe

x is the general solution of the reduced

equation. To find a particular solution, we try y = A cos 2x + B sin 2x. Substituting into

y′′ − 2y′ + y = −25 sin 2x gives A = −4, B = 3, so yp = 3 sin 2x− 4 cos 2x.

Therefore the general solution is: y = C1ex + C2xe

x + 3 sin 2x− 4 cos 2x.

19. First consider the reduced equation: y′′ − 3y′ − 10y = 0. The characteristic equation is:

r2 − 3r − 10 = (r − 5)(r + 2) = 0

and u1(x) = e5x, u2(x) = e−2x are fundamental solutions. A particular solution of the given


y = Ax + B.

The derivatives of y are: y′ = A, y′′ = 0.


−3A− 10(Ax + B) = −x− 1 =⇒ A = 110 , B = 7

100 and y = 110 x + 7

100

The general solution of the given equation is:

y = C1e5x + C2e

−2x + 110 x + 7

100

20. r2 + 4 = 0 =⇒ r = ±2i =⇒ y = C1 cos 2x + C2 sin 2x, general solution of reduced equation.

Particular solution: try y = x(A + Bx) cos 2x + x(C + Dx) sin 2x.

Substituting into y′′ + 4y = x cos 2x gives A = 116 , B = 0, C = 0, D = 1

8 ;

yp =116

x cos 2x +18x2 sin 2x. General solution: y = C1 cos 2x + C2 cos 2x + 1

16x cos 2x

+ 18x

2 sin 2x.

21. First consider the reduced equation: y′′ + 3y′ − 4y = 0. The characteristic equation is:

r2 + 3r − 4 = (r + 4)(r − 1) = 0

and u1(x) = ex, u2(x) = e−4x are fundamental solutions. A particular solution of the given equa-

tion has the form

y = Axe−4x.

The derivatives of y are: y′ = Ae−4x − 4Axe−4x, y′′ = −8Ae−4x + 16Axe−4x.


−8Ae−4x + 16Axe−4x + 3(Ae−4x − 4Axe−4x

)− 4Axe−4x = e−4x.

This implies −5A = 1, so A = − 15 and y = − 1

5 xe−4x.

The general solution of the given equation is: y = C1ex + C2e

−4x − 15 xe

−4x.



SECTION 19.4 1005

22. r2 + 2r = 0 =⇒ r = 0,−2 =⇒ y = C1 + C2e−2x, general solution of reduced equation.

Particular solution: try y = A cos 2x + B sin 2x.

Substituting into y′′ + 2y′ = 4 sin 2x gives A = B = − 12 ; yp = − 1

2 cos 2x− 12 sin 2x.

General solution: y = C1 + C2e−2x − 1

2 cos 2x− 12 sin 2x.

23. First consider the reduced equation: y′′ + y′ − 2y = 0. The characteristic equation is:

r2 + r − 2 = (r + 2)(r − 1) = 0

and u1(x) = e−2x, u2(x) = ex are fundamental solutions. A particular solution of the given


y = x(A + Bx)ex.

The derivatives of y are:

y′ = (A + (2B + A)x + Bx2)ex, y′′ = (2A + 2B + (4B + A)x + Bx2)ex.


(2A + 2B + (4B + A)x + Bx2 + A + (2B + A)x + Bx2 − 2Ax− 2Bx2)ex = 3xex.

This implies A = − 13 , B = 1

2 so y = x(− 13 + 1

2x)ex.

The general solution of the given equation is: y = C1e−2x + C2e

x − 13xe

x + 12x

2ex.

24. r2 + 4r + 4 = 0 =⇒ r = −2 =⇒ y = C1e−2x + C2xe

−2x, general solution of reduced equation.

Particular solution: try y = x2(A + Bx)e−2x.

Substituting into y′′ + 4y′ + 4y = xe−2x gives A = 0, B = 16 ; yp = 1

6x3e−2x.

General solution: y = C1e−2x + C2xe

−2x + 16x

3e−2x.

25. Let y1(x) be a solution of y′′ + ay′ + by = φ1(x), let y2(x) be a solution of y′′ + ay′ + by = φ2(x),

and let z = y1 + y2. Then

z′′ + az′ + bz = (y′′1 + y′′2) + a(y′1 + y′2) + b(y1 + y2)

= (y′′1 + ay′1 + by1) + (y′′2 + ay′2 + y2) = φ1 + φ2.

26. (a) y = − 115x− 2

225 is a particular solution of y′′ + 2y′ − 15y = x

y = − 17e

2x is a particular solution of y′′ + 2y′ − 15y = e2x

Therefore y = − 115x− 2

225 − 17e

2x is a particular solution of y′′ + 2y′ − 15y = x + e2x.

(b) y = − 14e

−x is a particular solution of y′′ − 7y′ − 12y = e−x.

y = 7226 cos 2x− 4

113 sin 2x is a particular solution of y′′ − 7y′ − 12y = sin 2x.

Therefore y = − 14e

−x + 7226 cos 2x− 4

113 sin 2x is a particular solution of

y′′ − 7y′ − 12y = e−x + sin 2x.



1006 SECTION 19.4

27. First consider the reduced equation: y′′ + 4y′ + 3y = 0. The characteristic equation is:

r2 + 4r + 3 = (r + 3)(r + 1) = 0

and u1(x) = e−3x, u2(x) = e−x are fundamental solutions. Since coshx = 12 (ex + e−x), a

particular solution of the given equation has the form

y = Aex + Bxe−x

The derivatives of y are: y′ = Aex + Be−x −Bxe−x y′′ = Aex − 2Be−x + Bxe−x.


Aex − 2Be−x + Bxe−x + 4 (Aex + Be−x −Bxe−x) + 3 (Aex + Bxe−x) = 12 (ex + e−x) .

Equating coefficients, we get A = 116 , B = 1

4 , and so y = 116 e

x + 14 xe

−x.

The general solution of the given equation is: y = C1e−3x + C2e

−x + 116 e

x + 14 xe

−x.

28. r2 + 1 = 0 =⇒ r = ±i. Fundamental solutions: u1 = cosx, u2 = sinx.

Wronskian: W = u1u′2 − u

′1u2 = 1; φ(x) = 3 sinx sin 2x

z1 = −∫

u2φ

Wdx = −

∫3 sin2 x sin 2x dx = −6

∫sin3 x cosx dx = −3

2sin4 x,

z2 =∫

u1φ

Wdx =

∫3 cosx sinx sin 2x dx =

32

∫sin2 2x dx =

316

(4x− sin 4x).

Therefore yp = z1u1 + z2u2 = − 32 sin4 x cosx + 3

16 (4x− sin 4x) sinx.

29. First consider the reduced equation y′′ − 2y′ + y = 0. The characteristic equation is:

r2 − 2r + 1 = (r − 1)2 = 0

and u1(x) = ex, u2(x) = xex are fundamental solutions. Their Wronskian is given by

W = u1u′2 − u2u

′1 = ex(ex + xex) − xex(ex) = e2x

Using variation of parameters, a particular solution of the given equation will have the form

y = u1z1 + u2z2,

where

z1 = −∫

xex(xex cosx)e2x

dx = −∫

x2 cosx dx = −x2 sinx− 2x cosx + 2 sinx,

z2 =∫

ex(xex cosx)e2x

dx =∫

x cosx dx = x sinx + cosx

Therefore,

y = ex(−x2 sinx− 2x cosx + 2 sinx

)+ xex (x sinx + cosx) = 2ex sinx− xex cosx.



SECTION 19.4 1007

30. r2 + 1 = 0. Fundamental solutions: u1 = cosx, u2 = sinx.

Wronskian: W = u1u′2 − u′

1u2 = 1; φ(x) = cscx.

z1 = −∫

u2φ

Wdx = −

∫sinx cscx dx = −x,

z2 =∫

u1φ

Wdx =

∫cosx cscx dx =

∫cotx dx = ln(sinx) [sinx > 0 since 0 < x < π].

Therefore yp = z1u1 + z2u2 = −x cosx + ln(sinx) sinx.

31. First consider the reduced equation y′′ − 4y′ + 4y = 0. The characteristic equation is:

r2 − 4r + 4 = (r − 2)2 = 0

and u1(x) = e2x, u2(x) = xe2x are fundamental solutions. Their Wronskian is given by

W = u1u′2 − u2u

′1 = e2x

(e2x + 2xe2x

)− xe2x(2e2x) = e4x.


y = u1z1 + u2z2,

where

z1 = −∫

xe2x(

13 x

−1e2x)

e4xdx = −1

3

∫dx = − 1

3 x,

z2 =∫

e2x(

13 x

−1e2x)

e4xdx =

13

∫1xdx = 1

3 ln |x|.Therefore,

y = e2x(− 1

3 x)

+ xe2x(

13 ln |x|

)= − 1

3 xe2x + 1

3 x ln |x| e2x.

Note: Since u = − 13 xe

2x is a solution of the reduced equation,

y = 13 x ln |x| e2x

is also a particular solution of the given equation.

32. r2 + 4 = 0 =⇒ r = ±2i. Fundamental solutions: u1 = cos 2x, u2 = sin 2x.


1u2 = 2 cos2 2x + 2 sin2 2x = 2; φ(x) = sec2 2x

z1 = −∫

u2φ

Wdx = −

∫sin 2x

2 cos2 2xdx = − 1

4sec 2x,

z2 =∫

u1φ

Wdx =

∫cos 2x

2 cos2 2xdx =

12

∫sec 2x dx =

14

ln | sec 2x + tan 2x|.

Therefore

yp = − 14

sec 2x cos 2x +14

ln | sec 2x + tan 2x| sin 2x = − 14

(1 − ln | sec 2x + tan 2x| sin 2x) .

33. First consider the reduced equation y′′ + 4y′ + 4y = 0. The characteristic equation is:

r2 + 4r + 4 = (r + 2)2 = 0



1008 SECTION 19.4

and u1(x) = e−2x, u2(x) = xe−2x are fundamental solutions. Their Wronskian is given by

W = u1u′2 − u2u

′1 = e−2x

(e−2x − 2xe2x

)− xe−2x(−2e−2x) = e−4x.


y = u1z1 + u2z2,

where

z1 = −∫

xe−2x(x−2e−2x

)e−4x

dx = −∫

1xdx = − ln |x|

z2 =∫

e−2x(x−2e−2x

)e−4x

dx =∫

1x2

dx = − 1x

Therefore,

y = e−2x (− ln |x|) + xe−2x

(− 1

x

)= − e−2x ln |x| − e−2x.

Note: Since u = − e−2x is a solution of the reduced equation, we can take

y = − ln |x |e2x.

34. r2 + 2r + 1 = 0 =⇒ r = −1. Fundamental solutions: u1 = e−x, u2 = xe−x.


1u2 = (1 − x)e−2x + xe−2x = e−2x; φ(x) = e−x lnx.

z1 = −∫

u2φ

Wdx = −

∫xe−xe−x lnx

e−2xdx = −

∫x lnx dx = −1

2x2 lnx +

x2

4,

z2 =∫

u1φ

Wdx =

∫e−xe−x lnx

e−2xdx =

∫lnx dx = x lnx− x.

Therefore

yp = z1u1 + z2u2 = e−x

(x2

4− 1

2x2 lnx

)+ xe−x(x lnx− x)

=14x2e−x (2 lnx− 3)

35. First consider the reduced equation y′′ − 2y′ + 2y = 0. The characteristic equation is:

r2 − 2r + 2 = 0

and u1(x) = ex cosx, u2(x) = ex sinx are fundamental solutions. Their Wronskian is given by

W = ex cosx [ex sinx + ex cosx] − ex sinx [ex cosx− ex sinx] = e2x


y = u1z1 + u2z2,

where

z1 = −∫

ex sinx · ex secxe2x

dx = −∫

tanx dx = − ln | secx| = ln | cosx|

z2 =∫

ex cosx · ex secxe2x

dx =∫

dx = x



SECTION 19.4 1009

Therefore,

y = ex cosx (ln | cosx|) + ex sinx(x) = ex cosx ln | cosx| + xex sinx.

36. v′′ + (2k + a)v′ + (k2 + ak + b)v = cnxn + cn−1x

n−1 + · · · + c1x + c0.

37. Assume that the forcing function F (t) = F0 (constant). Then the differential equation has a particular

solution of the form i = A. The derivatives of i are: i′ = i′′ = 0. Substituting i and its derivatives

into the equation, we get1CA = F0 =⇒ A = CF0 =⇒ i = CF0.

The characteristic equation for the reduced equation is:

Lr2 + Rr +1C

= 0 =⇒ r1, r2 =−R±

√R2 − 4L/C2L

=−R

√C ±

√CR2 − 4L

2L√C

(a) If CR2 = 4L, then the characteristic equation has only one root: r = −R/2L,

and u1 = e−(R/2L)t, u2 = t e−(R/2L)t are fundamental solutions.

The general solution of the given equation is:

i(t) = C1e−(R/2L)t + C2t e

−(R/2L)t + CF0

and its derivative is:

i′(t) = −C1(R/2L)e−(R/2L)t + C2e−(R/2L)t − C2(R/2L)t e−(R/2L)t.

Applying the side conditions i(0) = 0, i′(0) = F0/L, we get

C1 + CF0 = 0

(−R/2L)C1 + C2 = F0/L

The solution is C1 = −CF0, C2 =F0

2L(2 −RC).

The current in this case is:

i(t) = −CF0e−(R/2L)t +

F0

2L(2 −RC) t e−(R/2L)t + CF0.

(b) If CR2 − 4L < 0 then the characteristic equation has complex roots:

r1 = −R/2L± iβ, where β =

√4L− CR2

4CL2(here i2 = −1)

and fundamental solutions are: u1 = e−(R/2L)t cosβt, u2 = e−(R/2L)t sinβt.

The general solution of the given differential equation is:

i(t) = e−(R/2L)t (C1 cosβt + C2 sinβt) + CF0

and its derivative is:

i′(t) = (−R/2L)e−(R/2L)t (C1 cosβt + C2 sinβt) + βe−(R/2L)t (−C1 sinβt + C2 cosβt) .



1010 SECTION 19.4

Applying the side conditions i(0) = 0, i′(0) = F0/L, we get

C1 + CF0 = 0

(−R/2L)C1 + βC2 = F0/L

The solution is C1 = −CF0, C2 =F0

2Lβ(2 −RC).

The current in this case is:

i(t) = e−(R/2L)t

(F0

2Lβ(2 −RC) sinβt− CF0 cosβt

)+ CF0.

38. (a) x2y1′′ − xy1

′ + y1 = x2 · 0 − x · 1 + x = 0 : y1 is a solution.

x2y2′′ − xy2

′ + y2 = x2(

1x

)− x(lnx + 1) + x lnx = 0 : y2 is a solution.

W = y1y2′ − y1

′y2 = x(lnx + 1) − 1(x lnx) = x is nonzero on (0,∞).

(b) To use the method of variation of parameters as described in the text, we first re-write

the equation in the form

y′′ − 1xy′ +

1x2

y =4x

lnx.

Then, a particular solution of the equation will have the form yp = y1z1 + y2z2, where

z1 = −∫

x lnx · [(4/x) lnx]x

dx = −4∫

1x

(lnx)2 dx = −43(lnx)3

and

z2 =∫

x · [(4/x) lnx]x

dx = 4∫

lnx

xdx = 2(lnx)2

Thus, yp = − 43x (lnx)3 + x lnx · 2(lnx)2 which simplifies to: yp = 2

3x (lnx)3.

39. (a) Let y1(x) = sin(ln x2

). Then

y′1 =(

2x

)cos

(ln x2

)and y′′1 = −

(4x2

)sin

(ln x2

)−

(2x2

)cos

(ln x2

)Substituting y1 and its derivatives into the differential equation, we have

x2

[−

(4x2

)sin

(ln x2

)−

(2x2

)cos

(ln x2

)]+ x

[(2x

)cos

(ln x2

)]+ 4 sin

(ln x2

)= 0

The verification that y2 is a solution is done in exactly the same way.

The Wronskian of y1 and y2 is:

W (x) = y1y′2 − y2y

′1

= sin(ln x2

) [−

(2x

)sin

(ln x2

)]− cos

(ln x2

) [(2x

)cos

(ln x2

)]= − 2

x �= 0 on (0,∞)

(b) To use the method of variation of parameters as described in the text, we first re-write the equation

in the form

y′′ + x−1 y′ + 4x−2 y = x−2 sin(lnx).



SECTION 19.5 1011

Then, a particular solution of the equation will have the form y = y1z1 + y2z2, where

z1 = −∫

cos(lnx2)x−2 sin(lnx)−2/x

dx

= 12

∫cos(2 lnx)x−1 sin(lnx) dx

= 12

∫cos 2u sinu du (u = lnx)

= 12

∫(2 cos2 u− 1) sinu du

= − 13 cos3 u + 1

2 sinu

and

z2 =∫

sin(lnx2)x−2 sin(lnx)−2/x

dx

= − 12

∫sin(2 lnx)x−1 sin(lnx) dx

= − 12

∫sin 2u sinu du (u = lnx)

= −∫

sin2 u cosu du

= − 13 sin3 u

Thus, y = sin 2u(− 1

3 cos3 u + 12 sinu

)− cos 2u

(13 sin3 u

)which simplifies to:

y = 13 sinu = 1

3 sin(lnx).

SECTION 19.5

1. The equation of motion is of the form

x (t) = A sin (ωt + φ0) .

The period is T = 2π/ω = π/4. Therefore ω = 8. Thus

x (t) = A sin (8t + φ0) and v (t) = 8A cos (8t + φ0) .

Since x (0) = 1 and v (0) = 0, we have

1 = A sinφ0 and 0 = 8A cosφ0.

These equations are satisfied by taking A = 1 and φ0 = π/2.

Therefore the equation of motion reads

x (t) = sin(8t + 1

2π).

The amplitude is 1 and the frequency is 8/2π = 4/π.

2. x(t) = A sin(ωt + φ0). ω = 2πf = 2π(

1π

)= 2

0 = x(0) = A sinφ0, −2 = x′(0) = ωA cosφ0 =⇒ A = 1, φ0 = π.

Amplitude 1, period T = 1f = π.



1012 SECTION 19.5

3. We can write the equation of motion as

x (t) = A sin(

2πT

t

).

Differentiation gives

v (t) =2πAT

cos(

2πT

t

).

The object passes through the origin whenever sin [(2π/T )] = 0.

Then cos [(2π/T ) t] = ±1 and v = ±2πA/T .

4. x(t) = A sin(

2πT t + φ0

), v = x′(t) = 2π

T A cos(

2πT t + φ0

).

Note that x2 +(

T2πv

)2 = A2.

At x = x0, v = ±v0, so A =√x0

2 +(

T2πv0

)2 = (1/2π)√

4π2x20 + T 2v2

0 .

5. In this case φ0 = 0 and, measuring t in seconds, T = 6.

Therefore ω = 2π/6 = π/3 and we have

x (t) = A sin(π

3t), v (t) =

πA

3cos

(π

3t).

Since v (0) = 5, we have πA/3 = 5 and therefore A = 15/π.

The equation of motion can be written

x (t) = (15/π) sin(

13πt

)6. (a) A sin(ωt + φ0) = A cos(ωt + φ0 − π

2 ); take φ1 = φ0 − 12π.

(b) A sin(ωt + φ0) = A cosφ0 sinωt + A sinφ0 cosωt = B sinωt + C cosωt.

7. x (t) = x0 sin(√

k/m t + 12π

)

8. (a) maximum speed at x = 0.

(b) zero speed at x = ±x0.

(c) maximum acceleration (in absolute value) at x = ±x0.

(d) zero acceleration at x = 0 (when total force is zero).

9. The equation of motion for the bob reads

x (t) = x0 sin(t√k/m + 1

2π). (Exercise 7)

Since v (t) =√k/m x0 cos

(√k/m t + 1

2π)

, the maximum speed is√

k/m x0.

The bob takes on half of that speed where∣∣ cos

(√k/m t + 1

2π) ∣∣ = 1

2 . Therefore

∣∣ sin(√

k/m t + 12π

) ∣∣ =√

1 − 14 = 1

2

√3 and x (t) = ± 1

2

√3 x0.



SECTION 19.5 1013

10. v = −√

k

mx0 sin

(√k

mt

)has maximum value

√k

mx0, so the maximum kinetic energy is

12mv2 =

12m

k

mx0

2 =12kx0

2.

11. KE = 12m[v(t)]2 = 1

2m(k/m)x02 cos2

(√k/m t + 1

2π)

= 14kx0

2[1 + cos

(2√k/m t + π

)].

Average KE =1

2π√m/k

∫ 2π√

m/k

0

14kx0

2[1 + cos

(2√k/m t + π

)]dt

= 14kx0

2.

12. v(t) = −√

k

mx0 sin

(√k

mt

)= ±

√k

m

√x0

2 − [x(t)]2.

13. Setting y (t) = x (t) − 2, we can write x′′ (t) = 8 − 4x (t) as y′′ (t) + 4y (t) = 0.

This is simple harmonic motion about the point y = 0; that is, about the point x = 2. The equation

of motion is of the form

y (t) = A sin (2t + φ0) .

The condition x(0) = 0 implies y(0) = −2 and thus

(∗) A sinφ0 = −2

Since y′(t) = x′(t) and y′(t) = 2A cos(2t + φ0), the condition x′(0) = 0 gives y′(0) = 0, and thus

(∗∗) 2A cosφ0 = 0.

Equations (*) and (**) are satisfied by A = 2, φ0 = 32π. The equation of motion can therefore be

written

y(t) = 2 sin(

2t +32π

).

The amplitude is 2 and the period is π.

14. (a) Since limθ→0

sin θ

θ= 1, sin θ ∼= θ for small θ.

(b) The general solution is θ(t) = A sin(√

g/L t + φ0

)(i) Here A = θ0 and φ0 = π

2 , so θ(t) = θ0 sin(√

g/L t + π2

)= θ0 cos

(√g/L t

).

(ii) 0 = θ(0) = A sinφ0, −√

g

Lθ0 = θ′(0) = A

√g

Lcosφ0 =⇒ A = θ0, φ0 = π.

Therefore, the equation of motion becomes θ(t) = −θ0 sin(√

g/L t)

(c) ω =√

g

L, T =

2πω

= 2π

√L

g= 2 =⇒ L =

g

π2∼= 3.24 feet or 0.993 meters.



1014 SECTION 19.5

15. (a) Take the downward direction as positive. We

begin by analyzing the forces on the buoy at

a general position x cm beyond equilibrium.

First there is the weight of the buoy: F1 = mg.

This is a downward force. Next there is the

buoyancy force equal to the weight of the fluid

displaced; this force is in the opposite direc-

tion: F2 = −πr2 (L + x) ρ. We are neglecting

friction so the total force is

F = F1 + F2 = mg − πr2 (L + x) ρ =(mg − πr2Lρ

)− πr2xρ.

We are assuming at the equilibrium point that the forces (weight of buoy and buoyant force of

fluid) are in balance:

mg − πr2Lρ = 0.

Thus,

F = −πr2xρ.

By Newton’s

F = ma (force = mass × acceleration)

we have

ma = −πr2xρ and thus a +πr2ρ

mx = 0.

Thus, at each time t,

x′′ (t) +πr2ρ

mx (t) = 0.

(b) The usual procedure shows that

x(t) = x0 sin(r√πρ/m t + 1

2π).

The amplitude A is x0 and the period T is (2/r)√

mπ/ρ.

16. Uniform circular motion consists of simple harmonic motion in both x and y, the two being out of

phase by π2 .

17. From (19.5.4), we have

x(t) = Ae(−c/2m)t sin(ωt + φ0) =A

e(c/2m)tsin(ωt + φ0) where ω =

√4km− ω2

2m

If c increases, then both the amplitude,∣∣∣∣ A

e(c/2m)t

∣∣∣∣ and the frequencyω

2πdecrease.



SECTION 19.5 1015

18. Assume that r1 > r2. If C1 = 0 or C2 = 0, then x = C1er1t + C2e

r2t can never be zero.

If both C1 and C2 are nonzero, then C1er1t + C2e

r2t = 0 implies e(r1−r2)t = −C2

C1. Since

e(r1−r2)t is an increasing function (r1 > r2), it can take the value −C2

C1at most once.

By the same reasoning, x′(t) = C1r1er1t + C2r2e

r2t can be zero at most once. Therefore the

motion can change direction at most once.

19. Set x(t) = 0 in (19.5.6). The result is:

C1e(−c/2m)t + C2te

(−c/2m)t = 0 =⇒ C1 + C2t = 0 =⇒ t = −C1/C2

Thus, there is at most one value of t at which x(t) = 0.

The motion changes directions when x′(t) = 0:

x′(t) = −C1(c/2m)e(−c/2m)t + C2e(−c/2m)t − C2(c/2m)te(−c/2m)t.

Now,

x′(t) = 0 =⇒ −C1(c/2m) + C2 − C2t(c/2m) = 0 =⇒ t =C2 − C1(c/2m)

C2(c/2m)

and again we conclude that there is at most one value of t at which x′(t) = 0.

20. If γ �= ω, we try xp = A cos γt + B sin γt as a particular solution of x′′ + ω2x =F0

mcos γt.

Substituting xp into the equation, we get −γ2xp + ω2xp = F0m cos γt,

giving xp =F0/m

ω2 − γ2cos γt.

21. x(t) = A sin(ωt + φ0) +F0/m

ω2 − γ2cos(γt)

If ω/γ = m/n is rational, then 2πm/ω = 2πn/γ is a period.

22. If γ = ω, we try xp = At cosωt + Bt sinωt as a particular solution of x′′ + ω2x =F0

mcosωt.

Substituting xp into the equation, we have

(2Bω −Aω2t) cosωt− (2Aω + Bω2t) sinωt + ω2(At cosωt + Bt sinωt) =F0

mcosωt,

which gives A = 0, B =F0

2ωm, as required.

23. The characteristic equation is

r2 + 2αr + ω2 = 0; the roots are r1, r2 = −α±√α2 − ω2

Since 0 < α < ω, α2 < ω2 and the roots are complex. Thus, u1(t) = e−αt cosβt, u2(t) = e−αt sinβt,

where β =√ω2 − α2 are fundamental solutions, and the general solution is:

x(t) = e−αt(C1 cosβt + C2 sinβt); β =√α

2 − ω2

24. Straightforward computation.



1016 REVIEW EXERCISES

25. Set ω = γ in the particular solution xp given in Exercise 24. Then we have

xp =F0

2αγmsin γt

As c = 2αm → 0+, the amplitude∣∣∣ F02αγm

∣∣∣ → ∞

26.F0/m

(ω2 − γ2)2 + 4α2γ2

[(ω2 − γ2) cos γt + 2αγ sin γt

]

=F0/m√

(ω2 − γ2)2 + 4α2γ2

[ω2 − γ2√

(ω2 − γ2)2 + 4α2γ2cos γt +

2αγ√(ω2 − γ2)2 + 4α2γ2

sin γt

]

Setting φ = tan−1

(ω2 − γ2

2αγ

), this expression becomes.

F0/m√(ω2 − γ2)2 + 4α2γ2

(sinφ cos γt + cosφ sin γt) =F0/m√

(ω2 − γ2)2 + 4α2γ2sin(γt + φ)

27.(ω2 − γ2

)2 + 4α2γ2 = ω4 + γ4 + 2γ2(2α2 − ω2) increases as γ increases.

28. (a) To maximize the amplitude, we need to minimize

(ω2 − γ2)2 + 4α2γ2 = ω4 − 2(ω2 − 2α2)γ2 + γ4.

This is a parabola in γ2, and the minimum occurs when γ2 = ω2 − 2α2.

Therefore the maximum amplitude occurs when γ =√ω2 − 2α2

(b) f =2πγ

=2π√

ω2 − 2α2

(c) When γ2 = ω2 − 2α2, the amplitude is:F0/m√

(ω2 − γ2)2 + 4α2γ2=

F0/m

2α√ω2 − α2

(d) Since 2α = c/m, the resonant amplitude in (c) can be rewrittenF0

c√ω2 − c2/4m2

.

This gets large as c gets small.

REVIEW EXERCISES

1. The equation is linear: H(x) =∫

1dx = x =⇒ eH(x) = ex

d

dx(exy) = 2e−x =⇒ exy = −2e−x + C; the solution is: y = −2e−2x + Ce−x

2. Since∂(3x2y2)

∂y=

∂(2x3y + 4y3)∂x

, the equation is exact.

f(x, y) =∫

3x2y2 dx = x3y2 + φ(y).

∂f

∂y= 2x3y + φ′(y) = 2x3y + 4y3 =⇒ φ′(y) = 4y3 and φ(y) = y4.

The solution is: x3y2 + y4 = C



REVIEW EXERCISES 1017

3. The equation is separable:y

y2 + 1dy =

1cos2 x

dx = sec2 x dx

12

ln(y2 + 1) = tan x + C

The solution is: ln(1 + y2) = 2 tanx + C

4. The equation is separable:

y ln y dy = xex dx∫y ln y dy =

∫xex dx

12y2 ln y − 1

4y2 = xex − ex + C

The solution is:12y2 ln y − 1

4y2 = xex − ex + C

5. The equation can be written y′ − 2xy =

1x2

y2 a Bernoulli equation.

y−2y′ − 2xy−1 =

1x2

Let v = y−1. Then v′ = −y−2y′, and we get the linear equation

v′ +2xv = − 1

x2.

Integrating factor: H(x) =∫

(2/x) dx = lnx2 and eH(x) = x2.

x2v′ + 2xv = −1

x2v = −x + C

v = − 1x

+C

x2=

C − x

x2

The solution for the original equation is y =x2

C − x

6. Rewrite the equation as y′ +2xy =

cosxx2

, a linear equation.

H(x) =∫

(2/x) dx = lnx2; eH(x) = x2.

x2y′ + 2xy = cos x

x2y = sin x + C

y =sin x

x2+

C

x2

7. Since∂(y sinx + xy cosx)

∂y= sin x + x cos x =

∂(x sinx + y2)∂x


f(x, y) =∫

(y sinx + xy cosx) dx = xy sinx + φ(y).




∂f∂y = x sinx + φ′(y) = y2 + x sinx =⇒ φ′(y) = y2 =⇒ φ(y) = 1

3y3

The solution is x sinx + 13y

3 = C

8.1Q

(Py −Qx) =1xy

(2y − y) =1x

.

Therefore, e∫

(1/x)dx = x is an integrating factor;

(x3 + xy2 + x2) dx + (x2y dy) = 0

is exact.

f(x, y) =∫

x2y dy =12x2y2 + ψ(x).

∂f

∂x= xy2 + ψ′(x) = x3 + xy2 + x2 =⇒ ψ′(x) = x3 + x2 and ψ(x) =

14x4 +

13x3.

The solution is: 14x

4 + 13x

3 + 12x

2y2 = C or 3x4 + 4x3 + 6x2y2 = C


1 + y

ydy = (x2 − 1) dx

ln |y| + y =13x3 − x + C

10. The equation is a Bernoulli equation. We write it as

y−1/3y′ − 3xy2/3 = x4

Let v = y2/3. Then v′ = 23y

−1/3y′, and get the linear equation

v′ − 2xv =

23x4

H(x) = −∫

(2/x) dx = lnx−2; eH(x) = x−2.

x−2v′ − 2x−3v =23x2

x−2v =29x3 + C

v =29x5 + Cx2

Therefore, y2/3 = 29x

5 + Cx2 or y =(

29x

5 + Cx2)3/2.

11. The equation can be written as y′ +2xy = x2, a linear equation.

H(x) = lnx2; eH(x) = x2.

x2y′ + 2xy = x4

x2y =15x5 + C

y =15x3 + Cx−2




12. Rewrite the equation asdy

dx=

3y2 + 2xy2xy + x2

; a homogeneous equation.

Set v = y/x. Then y = vx anddy

dx= v + x

dv

dx.

v + xdv

dx=

3v2x2 + 2x2v

2x2v + x2=

3v2 + 2v2v + 1

xdv

dx=

3v2 + 2v2v + 1

− v =v2 + v

2v + 1

2v + 1v2 + v

dv =1xdx

ln |v2 + v| = ln |x| + C

v2 + v = Cx

Replacing v by y/x, we get y2 + xy = Cx3.

13. The differential equation is homogeneous.

Set v = y/x. Then y = vx anddy

dx= v + x

dv

dx.

v + xdv

dx=

x2 + x2v2

2x2v=

1 + v2

2v

xdv

dx=

1 + v2

2v− v =

1 − v2

2v

2v1 − v2

dv =1xdx

− ln |1 − v2| = ln |x| + C

1 − v2 =C

x

Replacing v by y/x, we get x2 − y2 = Cx.

Applying the initial condition y(1) = 2 gives C = −3. The solution of the initial-value problem is:

x2 + 3x− y2 = 0.

14. Write the equation as y′ +2yx

= x; the equation is linear.

H(x) =∫

(2/x) dx = lnx2 and eH(x) = x2.

x2y′ + 2xy = x3

x2y =14x4 + C

y =14x2 + Cx−2

Applying the initial condition y(1) = 0 gives C = −1/4 and y =x2

4− 1

4x2




15. Since∂(x + y)2

∂y= 2x + 2y =

∂(2xy + x2 − 1)∂x


f(x, y) =∫

(x + y)2 dx =∫

(x2 + 2xy + y2) dx =13x3 + x2y + xy2 + φ(y).

∂f

∂y= x2 + 2xy + φ′(y) = 2xy + x2 − 1 =⇒ φ′(y) = −1 =⇒ φ(y) = −y

The general solution is: 13x

3 + x2y + xy2 − y = C

Applying the initial condition y(1) = 1 gives C = 4/3. The solution of the initial-value problem is:

13x

3 + x2y + xy2 − y = 4/3

16. The equation is separable:y√

y2 + 1dy = 4x dx

√y2 + 1 = 2x2 + C

y2 = (2x2 + C)2 − 1

Applying the initial condition y(0) = 1 gives C =√

2. The solution of the initial-value problem is:

y2 = (2x2 +√

2)2 − 1.

17. The equation is a Bernoulli equation; rewrite it as: y−2y′ + xy−1 = x.

Set v = y−1. Then v′ = −y−2y′, and we have

v′ − xv = −x,

a linear equation. H(x) =∫

(−x) dx = − 12x

2 and eH(x) = e−x2/2

e−x2/2v′ − xe−x2/2v = −xe−x2/2

e−x2/2v = e−x2/2 + C

v = 1 + Cex2/2

y =1

1 + Cex2/2

Applying the initial condition y(0) = 2 gives C = −1/2. The solution of the initial-value problem

is: y =2

2 − ex2/2.


dy

y − y2= dx

∫ (1y

+1

1 − y

)dy = dx

ln |y| − ln |1 − y| = ln∣∣∣∣ y

1 − y

∣∣∣∣ = x + C




Applying the initial condition y(0) = 2 gives C = ln 2. The solution of the initial-value problem is:

ln∣∣∣∣ y

1 − y

∣∣∣∣ = x + ln 2 or y =2

2 + e−x.

19. The characteristic equation is: r2 − 2r + 2 = 0. The roots are: r1, r2 = 1 ± i.

The general solution is:

ex(C1 cosx + C2 sinx).

20. The characteristic equation is: r2 + r + 14 = 0. The roots are: r1 = r2 = −1/2.


y = C1e−x/2 + C2xe

−x/2.

21. The characteristic equation for the reduced equation is: r2 − r − 2 = 0. The roots are: r = 2,−1.

Use undetermined coefficients to find a particular solution of the nonhomogeneous equation:

z = A cos 2x + B sin 2x

z′ = −2A sin 2x + 2B cos 2x

z′′ = −4A cos 2x− 4B sin 2x

Substituting z, z′, z′′ into the differential equation yields the pair of equations:

−6A− 2B = 0, 2A− 6B = 1 =⇒ A =120

, B = − 320

.

The general solution is: y = C1e2x + C2e

−x + 120 cos 2x− 3

20 sin 2x

22. The characteristic equation is: r2 − 4r = 0. The roots are: r1 = 4, r2 = 0.


y = C1e4x + C2.

23. The characteristic equation for the reduced equation is: r2 − 6r + 9 = 0. The roots are: r1 = r2 = 3.

Use undetermined coefficients to find a particular solution of the nonhomogeneous equation.

Since z = e3x and z = xe3x are solutions of the reduced equation, set z = Ax2e3x.

z = Ax2e3x

z′ = 2Axe3x + 3Ax2e3x

z′′ = 2Ae3x + 12Axe3x + 9Ax2e3x

Substituting z, z′, z′′ into the differential equation gives:

2A = 3 =⇒ A =32.

The general solution is: y = C1e3x + C2xe

3x + 32x

2e3x.




24. The characteristic equation for the reduced equation is: r2 + 1 = 0. The roots are: r1 = i, r2 = −i.

Use variation of parameters to find a particular solution of the nonhomogeneous equation.

Set u1 = cos x and u2 = sin x. Then their Wronskian is W (x) = 1.

z1 = −∫

sinx sec3 x dx = −12

sec2 x, z2 =∫

cosx sec3 x dx = tanx

yp = − 12 secx + tanx sinx.

The general solution of the equation is: y = C1 cosx + C2 sinx− 12 secx + tanx sinx

25. The characteristic equation for the reduced equation is: r2 − 2r + 1 = 0. The roots are: r1 = r2 = 1.


Set u1 = ex and u2 = xex. Then their Wronskian is W (x) = e2x.

z1 = −∫

xex(1/x)ex

e2xdx = −

∫dx = −x z2 =

∫ex(1/x)ex

e2xdx =

∫(1/x) dx = lnx

yp = −xex + xex lnx

The general solution of the equation is: y = C1ex + C2xe

x + xex lnx, x > 0

26. The characteristic equation for the reduced equation is: r2 − 5r + 6 = 0. The roots are: r1 = 2, r2 = 3.


z = A cos x + B sin x + C

z′ = −A sin x + B cos x

z′′ = −A cos x−B sin x

Substituting z, z′, z′′ into the differential equation yields the equations:

5A− 5B = 0, 5A + 5B = 2, 6C = 4 =⇒ A =15, B =

15, C =

23.

The general solution is: y = C1e2x + C2e

3x + 15 cosx + 1

5 sinx + 23 .

27. The characteristic equation for the reduced equation is: r2 + 4r + 4 = 0. The roots are: r1 = r2 = −2.


z = Ax2e−2x + Be2x

z′ = 2Axe−2x − 2Ax2e−2x + 2Be2x

z′′ = 2Ae−2x − 8Axe−2x + 4Ax2e−2x + 4Be2x


2A = 4, 16B = 2 =⇒ A = 2, B =18.

The general solution is: y = C1e−2x + C2xe

−2x + 18e

2x + 2x2e−2x




28. The characteristic equation for the reduced equation is: r2 + 4 = 0. The roots are: r1 = 2i, r2 = −2i.


Set u1 = cos 2x and u2 = sin 2x. Then their Wronskian is W (x) = 2.

z1 = −∫

sin 2x tan 2x2

dx = −12

∫1 − cos2 2x

cos 2xdx = −1

4ln | sec 2x + tan 2x| + 1

4sin 2x

z2 =∫

cos 2x tan 2x2

dx =12

∫sin 2x dx = −1

4cos 2x

yp = − 14 cos 2x ln | sec 2x + tan 2x|

The general solution of the equation is: y = C1 cos 2x + C2 sin 2x− 14 cos 2x ln | sec 2x + tan 2x|

29. First find the general solution of the differential equation.

The characteristic equation for the reduced equation is: r2 + r = 0. The roots are: r1 = −1, r2 = 0.


Set z = Ax2 + Bx

z = Ax2 + Bx

z′ = 2Ax + B

z′′ = 2A


2A = 1, A + B = 0 =⇒ A = 1/2, B = −1.

The general solution of the differential equation is: y = C1e−x + C2 + 1

2x2 − x

Applying the initial conditions y(0) = 1, y′(0) = 0, we get the pair of equations

C1 + C2 = 1, −C1 − 1 = 0, =⇒ C1 = −1, C2 = 2.

The solution of the initial-value problem is: y = 2 − e−x + 12x

2 − x


The characteristic equation for the reduced equation is: r2 + 1 = 0. The roots are: r1 = i, r2 = −i.


Set z = A cos 2x + B sin 2x + Cx cos x + Dx sin x

z = A cos 2x + B sin 2x + Cx cos x + Dx sin x

z′ = −2A sin 2x + 2B cos 2x + C cos x− Cx sin x + D sin x + Dx cos x

z′′ = −4A cos 2x− 4B sin 2x− 2C sin x− Cx cos x + 2D cos x−Dx sin x


−3A = 4, −3B = 0, −2C = −4, 2D = 0 ⇒ A = −4/3, B = 0, C = 2, D = 0.




The general solution of the differential equation is:

y = C1 cos x + C2 sin x− 43

cos 2x + 2x cosx

Applying the initial conditions y(π/2) = −1, y′(π/2) = 0, we get C1 = −π − 16/13, C2 = −25/13.

The solution of the initial-value problem is:

y = −π cos x− 73

sin x− 43

cos 2x + 2x cos x


The characteristic equation for the reduced equation is: r2 − 5r + 6 = 0. The roots are: r1 = 2, r2 = 3.


Set z = Axe2x

z = Axe2x

z′ = Ae2x + 2Axe2x

z′′ = 4Ae2x + 4Axe2x

Substituting z, z′, z′′ into the differential equation gives: −A = 10, A = −10.

The general solution of the differential equation is: y = C1e2x + C2e

3x − 10xe2x

Applying the initial conditions y(0) = 1, y′(0) = 1, we get C1 = −8, C2 = 9.

The solution of the initial-value problem is: y = 9e3x − 8e2x− 10xe2x


The characteristic equation is: r2 + 4r + 4 = 0. The roots are: r1 = r2 = −2.

The general solution of the differential equation is: y = C1e−2x + C2xe

−2x

Applying the initial conditions y(−1) = 2, y′(−1) = 1, yields the equations

C1e2 − C2e

2 = 2, −2C1e2 + 3C2e

2 = 1, =⇒ C1 = 7e−2, C2 = 5e−2.

The solution of the initial-value problem is: y = 7e−2(x+1) + 5xe−2(x+1)

33. Assume x(t) = A sin(wt + φ0).

From T = 2π/ω = π/2, ω = 4 and x(t) = A sin(4t + φ0)

x(0) = 2 =⇒ A sin(φ)0) = 2; x′(0) = 0 =⇒ 4A cos(φ0) = 0 =⇒ φ0 =π

2and A = 2.

Therefore,

x(t) = 2 sin(4t + π/2); amplitude A = 2; frequency 2/π.

34. Assume x(t) = A sin(wt + φ0).

The period T = 8. Therefore 2π/ω = 8 which implies ω = π/4 and x(t) = A sin(

14π t + φ0

).

The condition x(0) = 0 implies that φ0 = 0. Therefore, x(t) = A sin(

14π t

).




The condition x′(0) = 8 = π4A cos 0 implies A = 32/π. Hence

x(t) =32π

sin(

14πt

).

35. Assume that the downward direction is positive. Then

4x′′(t) = −64x(t) + 8 sin 4t, x(0) = −12, x′(0) = 0

This equation can be written as

x′′ + 16x = 2 sin 4t

The characteristic equation for the reduced equation is: r2 + 16 = 0 and the roots are r = ±4i.


Set z = At cos 4t + Bt sin 4t

z = At cos 4t + Bt sin 4t

z′ = A cos 4t− 4At sin 4t + B sin 4t + 4Bt cos 4t

z′′ = −8A sin 4t− 16At cos 4t + 8B cos 4t− 16Bt sin 4t

Substituting z, z′, z′′ into the differential equation yields the equations, we get A = − 14 , B = 0.


x(t) = C1 cos 4t + C2 sin 4t− 14t cos 4t.

Applying the initial conditions x(0) = −1/2, x′(0) = 0, we get C1 = −1/2, C2 = 1/16.

The equation of motion is:

x(t) = −12

cos 4t− 14t cos 4t +

116

sin 4t

36. Assume that the downward direction is positive. Then

10x′′(t) = −60x(t) − 50x′(t) + 4 sin t, x(0) = 0, x′(0) = −1.

The differential equation can be written as

x′′ + 5x′ + 6x =25

sin t

The characteristic equation for the reduced equation is: r2 + 5r + 6 = 0 and the roots are:

r1 = −2, r2 = −3.


Set z = A cos t + B sin t

z = A cos t + B sin t

z′ = −A sin t + B cos t

z′′ = −A cos t−B sin t




Substituting z, z′, z′′ into the differential equation gives A = −1/25, B = 1/25.


x(t) = C1e−2t + C2e

−3t − 125 cos t + 1

25 sin t.

Applying the initial conditions x(0) = 0, x′(0) = −1, we get C1 = −23/25, C2 = 24/25.

The equation of motion is:

x(t) = −2325

e−2t +2425

e−3t − 125

cos t +125

sin t.

Calculus one and several variables 10E Salas solutions manual ch19

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Transcript of Calculus one and several variables 10E Salas solutions manual ch19