Zill Calculo 4e Manual de Solucionario c16

Chapter 16

Higher-Order DifferentialEquations

16.1 Exact First-Order Equations

1. Since Py = 0 = Qx, the equation is exact.

fx = 2x+ 4, f = x2 + 4x+ g(y), fy = g�(y) = 3y − 1, g(y) =3

2y2 − y

The solution is x2 + 4x+3

2y2 − y = C.

2. Since Py = 1 and Qx = −1, the equation is not exact.

3. Since Py = 4 = Qx, the equation is exact.

fx = 5x+ 4y, f =5

2x2 + 4xy + g(y), fy = 4x+ g�(y) = 4x− 8y3, g(y) = −2y4

The solution is5

2x2 + 4xy − 2y4 = C.

4. Since Py = cos y − sinx = Qx, the equation is exact.fx = sin y−y sinx, f = x sin y+y cosx+g(y), fy = x cos y+cosx+g�(y) = cosx+x cos y−y,

g(y) = −1

2y2

The solution is x sin y + y cosx− 1

2y2 = C.

5. Since Py = 4xy = Qx, the equation is exact.fx = 2y2x− 3, f = y2x2 − 3x+ g(y), fy = 2yx2 + g�(y) = 2yx2 + 4, g(y) = 4yThe solution is y2x2 − 3x+ 4y = C.

6.� y

x2− 4x3 + 3y sin 3x

�dx +

�2y − 1

x+ cos 3x

�dy = 0. Since Py =

1

x2+ 3 sin 3x and Qx =

1

x2− 3 sin 3x, the equation is not exact.

7. (x2− y2)dx+(x2− 2xy)dy = 0. Since Py = −2y and Qx = 2x− 2y, the equation is not exact.

1027

1028 CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

8.�1 + lnx+

y

x

�dx + (lnx − 1)dy = 0. Since Py =

1

x= Qx, the equation is exact. fy =

lnx−1, f = y lnx−y+g(x), fx =y

x+g�(x) = 1+lnx+

y

x, g�(x) = 1+lnx, g(x) = x lnx

The solution is y lnx− y + x lnx = C.

9. (y3 − y2 sinx− x)dx+ (3xy2 + 2y cosx)dy = 0. Since Py = 3y2 − 2y sinx = Qx, the equationis exact.fx = y3 − y2 sinx − x, f = xy3 + y2 cosx − 1

2x2 + g(y), fy = 3xy2 + 2y cosx + g�(y) =

3xy2 + 2y cos, g(y) = 0The solution is xy3 + y2 cosx− 1

2x2 = C

10. Since Py = 3y2 = Qx, the equation is exact. fx = x3 + y3, f =1

4x4 + xy3 + g(y), fy =

3xy2 + g�(y) = 3xy2, g(y) = 0

The solution is1

4x4 + xy3 = C.

11. Since Py = 1 + ln y + xe−xy and Qx = ln y, the equation is not exact.

12. Since Py = 3x2+ey = Qx, the equation is exact. fx = 3x2y+ey, f = x3y+xey+g(y), fy =x3 + xey + g�(y) = x3 + xey − 2y, g(y) = −y2

The solution is x3y + xey − y2 = C.

13. (2xex − y + 6x2)dx− xdy = 0. Since Py = −1 = Qx, the equation is exact. fx = 2xex − y +6x2, f = 2xex − 2ex − yx+ 2x3 + g(y), fy = −x+ g�(y) = −x, g(y) = 0The solution is 2xex − 2ex − yx+ 2x3 = C.

14.

�1− 3

x+ y

�dx+

�1− 3

y+ x

�dy = 0. Since Py = 1 = Qx, the equation is exact.

fx = 1− 3

x+ y, f = x− 3 ln |x|+ xy + g(y), fy = x+ g�(y) = 1− 3

y+ x, g�(y) = 1− 3

y,

g(y) = y − 3 ln |y|The solution is x− 3 ln |xy|+ xy + y = C.

15. Since Py = 3x2y2 = Qx, the equation is exact.

fy = x3y2, f =1

3x3y3 + g(x), fx = x2y3 + g�(x) = x2y3 − 1

1 = 9x2, g�(x) = − 1

1 + 9x2=

−1

9

1

1/9 + x2, g(x) = −1

9

1

1/3tan−1 x

1/3= −1

3tan1 3x

The solution is1

3x3y3 − 1

3tan−1 3x = C or x3y3 = tan−13x = C1.

16. 2ydx− (5y − 2x)dy = 0. Since Py = 2 = Qx, the equation is exact.

fx = 2y, f = 2xy + g(y), fy = 2x+ g�(y) = −5y + 2x, g(y)= − 5

2y2

The solution is 2xy − 5

2y2 = C.

17. Since Py = sinx cos y = Qx, the equation is exact.fy = cosx cos y, f = cosx sin y+g(x), fx = − sinx sin y+g�(x) = tanx−sinx sin y, g�(x) =tanx, g(x) = ln | secx|The solution is cosx sin y + ln | secx| = C or cosx sin y − ln | cosx|+ C.

16.1. EXACT FIRST-ORDER EQUATIONS 1029

18. (2y sinx cosx − y + 2y2exy2)dx + (sin2 x + 4xyexy

2 − x)dy = 0. Since Py = 2 sinx cosx −1 + 4xy3exy

2+ 4yexy

2= Qz, the equation is exact. fx = 2y sinx cosx − y + 2y2exy

2=

y sin 2x− y + 2y2exy2, f = −1

2y cos 2x− xy + 2exy

2+ g(y),

fy = −1

2cos 2x− x+ 4xyexy

2

+ g�(y) = −1

2(1− 2 sin2 x)− x+ 4xyexy

2

+ g�(y)

= −1

2+ sin2 x− x+ 4xyexy

2

+ g�(y) = sin2 x+ 4xyexy2 − x

g�(y) =1

2, g(y) =

1

2y

The solution is −1

2y cos 2x− xy + 2exy

2+

1

2y = C.

19. Since Py = 4t3 − 1 = Qt, the equation is exact.ft = 4t3y − 15t2 − y, f = t4y − 5t3 − yt+ g(y),fy = t4 − t+ g�(y) = t4 + 3y2 − t, g�(y)− 3y2, g(y) = y3.The solution is t4y − 5t3 − yt+ y3 = C.

20. Since Py = − (t2 + y2) + y(2y)

(t2 + y2)2=

y2 − t2

(t2 + y2)2

and Qy =−2t

(t2 + y2)2, the equation is not exact.

21. Since Py = 2(x+ y) = Qx, the equation is exact.

fx = (x+y)2 = x2+2xy+y2, f =1

3x3+x2y+xy2+g(y), fy = x2+2xy+g�(y) = 2xy+x2−1

g�(y) = −1, g(y) = −y A family of solutions is1

3x3 + x2y+ xy2 − y = C. Substituting x = 1

and y = 1 we obtain1

3+ 1 + 1 − 1 =

4

3= C. The solution subject to the given condition is

1

3x3 + x2y + xy2 − y =

4

3.

22. Since Py = 1 = Qx, the equation is exact.fx = ex + y, f = ex + xy + g(y), fy = x + g�(y) = 2 + x + yey, g�(y) = 2 + yey Usingintegration by parts, g(y) = 2y+yey −y. A family of solutions is ex+xy+2y+yey − ey = C.Substituting x = 0 and y = 1 we obtain 1 + 2 + e − e = 3 = C. The solution subject to thegiven condition is ex + xy + 2y + yey − ey = 3.

23. Since Py = 4 = Qt, the equation is exact.ft = 4y+2t−5, f = 4ty+t2−5t+g(y), fy = 4t+g�(y) = 6y+4t−1, g�(y) = 6y−1, g(y) =3y2 − y A family of solutions is 4ty + t2 − 5t + 3y2 − y = C. Substituting t = −1 and y = 2we obtain −8 + 1 + 5 + 12 − 2 = 8 = C. The solution subject to the given condition is4ty + t2 − 5t+ 3y2 − y = 8.

24. Since Py = 2y cosx− 3x2 = Qx, the equation is exact.fx = y2 cosx − 3x2y − 2x, f = y2 sinx − x3y − x2 + g(y), fy = 2y sinx − x3 + g�(y) =2y sinx− x3 + ln y,g�(y) = ln y, g(y) = y ln y − y A family of solutions is y2 sinx − x3y − x2 + y ln y − y = C.Substituting x = 0 and y = e we obtain e − e = 0 = C. The solution subject to the givencondition is y2 sinx− x3y − x2 + y ln y − y = 0.


25. We want Py = Qx or 3y2 + 4kxy3 = 3y2 + 40xy3. Thus, 4k = 40 and k = 10.

26. We want Py = Qx or 18xy2 − sin y = 4kxy2 − sin y. Thus 4k = 18 and k = 92 .

27. We need Py = Qx, so we must have∂M

∂y= exy + xyexy + 2y − 1

x2 . This gives M(x, y) =

1xe

xy +(yx− 1)exy

x+ y2 − y

x2+ g(x) for some function g.

28. We need Py = Qx, so we must have∂N

∂x= 1

2x−1/2y−1/2 − x

(x2 + y2)2. This gives N(x, y) =

x1/2y−1/2 +1

2(x2 + y)+ g(y) for some function g.

29. Let µ(x, y = y3. Then∂

∂y[µ(x, y)M(x, y)] =

∂

∂y

�xy4

�= 4xy3

∂

∂z[µ(x, y)N(x, y)] =

∂

∂x

�2x2y3 + 3y5 − 20y3

�= 4xy3

Therefore, µ(x, y)M(x, y)dx+µ(x, y)N(x, y) = 0 is exact, and µ(x, y) is an integrating factor.Now, if y3

�xydx+ (2x2 + 3y2 − 20)dy

�= 0, then xydx + (2x2 + 3y2 − 20)dy = 0, provided

y �= 0. Therefore, to solve the original DE, we solve xy4dx+�2x2y3 + 3y5 − 20y3

�dy = 0.

fx = xy4, f = 12x

2y4 + g(y), fy = 2x2y+g�(y) = 2x2y3 + 3y5 − 20y3,g�(y) = 3y5 − 20y, g(y) = 1

2y6 − 5y4, f = 1

2x2y4 + 1

2y6 − 5y4.

The solution is therefore 12x

2y4 + 12y

6 − 5y4 = C.

30. True; a separable equation can be written as1

h(y)dy − g(x)dx = 0. Since g is a function of x

only and h is a function of y only, we have Py = Qx = 0.

16.2 Homogeneous Linear Equations

1. 3m2 −m = 0 =⇒ m(3m− 1) = 0 =⇒ m = 0, 1/3; y + C1 + C2ex/3

2. 2m2 + 5m = 0 =⇒ m(2m+ 5) = 0 =⇒ m = 0, −5/2; y = C1 + C2e−5x/2

3. m2 − 16 = 0 =⇒ m2 = 16 =⇒ m = −4, 4; y = C1e−4x + C2e4x

4. m2 − 8 = 0 =⇒ m2 = 8 =⇒ m = −2√2, 2

√2; y = C1e−2

√2x + C2e2

√2x

5. m2 + 9 = 0 =⇒ m2 = −9 =⇒ m = −3i, 3i; y = C1 cos 3x+ C2 sin 3x

6. 4m2 + 1 = 0 =⇒ m2 = −1/4 =⇒ m = −i/2, i/2; y = C1 cos1

2x+ C2 sin

1

2x

7. m2 − 3m+ 2 = 0 =⇒ (m− 1)(m− 2) = 0 =⇒ m = 1, 2; y = C1ex + C2e2x

8. m2 −m− 6 = 0 =⇒ (m+ 2)(m− 3) = 0 =⇒ m = −2, 3; y = C1e−2x + C233x

9. m2 + 8m+ 16 = 0 =⇒ (m+ 4)2 = 0 =⇒ m = −4, −4; y = C1e−4x + C2xe−4x

10. m2 − 10m+ 25 = 0 =⇒ (m− 5)2 = 0 =⇒ m = 5, 5; y = C1e5x + C2xe5x

16.2. HOMOGENEOUS LINEAR EQUATIONS 1031

11. m2 + 3m− 5 = 0 =⇒ m = −3/2±√29/2; y = C1e(−3/2−

√29/2)x + C2e(−3/2+

√29/2)x

12. m2 + 4m− 1 = 0 =⇒ m = −2±√5; y = C1e(−2−

√5)x + C2e(−2+

√5)x

13. 12m2 − 5m− 2 = 0 =⇒ (3m− 2)(4m+ 1) = 0 =⇒ m = −1/4, 2/3; y = C1e−x/4 + C2e2x/3

14. 8m2 + 2m− 1 = 0 =⇒ (4m− 1)(2m+ 1) = 0 =⇒ m = −1/2, 1/4; y = C1e−x/2 + C2ex/4

15. m2 − 4m+ 5 = 0 =⇒ m = 2± i; y = e2x(C1 cosx+ C2 sinx)

16. 2m2 − 3m+ 4 = 0 =⇒ m = 3/4± (√23/4)i; y = e3x/4

�C1 cos

√23

4x+ C2 sin

√23

4x

�

17. 3m2 + 2m+ 1 = 0 =⇒ m = −1/3± (√2/3)i; y = e−x/3

�C1 cos

√2

3x+ C2 sin

√2

3x

�

18. 2m2 + 2m+ 1 = 0 =⇒ m = −1/2± (1/2)i; y = e−x/2

�C1 cos

1

2x+ C2 sin

1

2x

�

19. 9m2 + 6m+ 1 = 0 =⇒ (3m+ 1)2 = 0 =⇒ m = −1/3, −1/3; y = C1e−x/3 + C2xe−x/3

20. 15m2 − 16m− 7 = 0 =⇒ (3m+1)(5m− 7) = 0 =⇒ m = −1/3, 7/5; y = C1e−x/3 +C2e7x/5

21. m2 + 16 = 0 =⇒ m2 = −16 =⇒ m = ±4i; y = C1 cos 4x + C2 sin 4x; y� = −4C1 sin 4x +C2 cos 4xUsing y(0) = 2 we obtain 2 = C1. Using y�(0) = −2 we obtain −2 = 4C2 or C2 = −1/2. The

solution is y = 2 cos 4x− 1

2sin 4x.

22. m2 − 1 = 0 =⇒ m2 = 1 =⇒ m = ±1; y = C1ex + C2e−x; y� = C1ex − C2e−x. Usingy(0) = Y �(0) = 1 we obtain the system C1+C2 = 1, C1−C2 = 1. Thus, C1 = 1 and C2 = 0.The solution is y = ex.

23. m2 + 6m+ 5 = 0 =⇒ (m+ 1)(m+ 5) = 0 =⇒ m = −5, −1; y = C1e−5x + C2e−x;y� = −5C1e−5x − C2e−x. Using y(0) = 0 and y�(0) = 3 we obtain the system C1 + C2 = 0,

− 5C1 − C2 = 3. Thus, C1 = −3/4 and c2 = 3/4. The solution is y = −3

4e−5x +

3

4e−x.

24. m2 − 8m+ 17 = 0 =⇒ m = 4± i; y = e4x(C1 cosx+ C2 sinx);y� = e4x [(4C1 + C2) cosx+ (−C1 + 4C2) sinx] .Using y(0) = 4 and y�(0) = −1 we obtain the system C1 = 4, 4C1 + C2 = −1. Thus, C1 = 4and C2 = −17. The solution is y = e4x(4 cosx− 17 sinx).

25. 2m2−2m+1 = 0 =⇒ m = 1/2± (1/2)i; y = ex/2(C1 cos1

2x+C2 sin

1

2x); y� = ex/2[

1

2(C1+

C2) cos1

2x−1

2(C1−C2) sin

1

2x]. Using y(0) = −1 and y�(0) = 0 we obtain the system C1 = −1,

1

2C1 +

1

2C2 = 0. Thus, C1 = −1 and C2 = 1. The solution is y = ex/2

�sin

1

2x− cos

1

2x

�.


26. m2 − 2m+ 1 = 0 =⇒ (m− 1)2 = 0 =⇒ m = 1, 1; y = C1ex + C2xex; y� = (C1 + C2)ex +C2xex.Using y(0) = 5 and y�(0) = 10 we obtain the system C1 = 5, C1 + C2 = 10. Thus,C1 = C2 = 5. The solution is y = 5ex + 5xex.

27. m2 +m+ 2 = 0 =⇒ m = −1/2± (√7/2)i; y = e−x/2(C1 cos

√7

2x+ C2 sin

√7

2x);

y� = e−x/2

�(−1

2C1 +

√7

2C2) cos

√7

2x+ (−

√7

2C1 −

1

2C2) sin

√7

2x

�.

Using y(0) = y�(0) = 0 we obtain the system C1 = 0, −1

2C1+

√7

2C2 = 0. Thus, C1 = C2 = 0.

The solution is y = 0.

28. 4m2 − 4− 3 = 0 =⇒ (2m− 3)(2m+ 1) = 0 =⇒ m = −1/2, 3/2; y = C1e−x/2 + C2e3x/2;

y� = −1

2C1e−x/2+

3

2C2e3x/2. Using y(0) = 1 and y�(0) = 5 we obtain the system C1+C2 = 1,

− 1

2C1 +

3

2C2 = 5. Thus, c1 = −7/4 and C2 = 11/4. The solution is y = −7

4e−x/2 +

11

4e3x/2.

29. m2−3m+2 = 0 =⇒ (m−1)(m−2) = 0 =⇒ m = 1, 2; y = C1ex+C2e2x; y� = C1ex+2C2e2x.Using y(1) = 0 and y�(1) = 1 we obtain the system eC1 + e2C2 = 0, eC1 + 2e2C2 = 1. Thus,C1 = −e−1 and C2 = e−2. The solution is y = −ex−1 + e2x−2.

30. m2+1 = 0 =⇒ m2 = −1 =⇒ m = ±i; y = C1 cosx+C2 sinx; y� = C1 sinx+C2 cosx. Using

y(π/3) = 0 and y�(π/3) = 2 we obtain the system1

2C1 +

√3

2C2 = 0, −

√3

2C1 +

1

2C2 = 2.

Thus, C1 = −√3 and C2 = 1. The solution is y = −

√3 cosx+ sinx.

31. The auxiliary equation is (m − 4)(m + 5) = m2 + m − 20 = 0. The differential equation isy�� + y� − 20y = 0.

32. The auxiliary equation is [(m− 3)− i] [(m− 3) + i] = (m−3)2− i2 = m2−6m+10 = 0. Thedifferential equation is y�� = 6y� + 10y = 0.

33. The auxiliary equation ism2+1 = 0, som = ±i. The general solution is y = C1 cosx+C2 sinx.The boundary conditions yield y(0) = C1 = 0, y(π) = −C1 = 0, so y = C2 sinx.

34. The general solution is y = C1 cosx + C2 sinx. The boundary conditions yield y(0) = C1 =0, y(π) = −C1 = 1, which is a contradiction. No solution.

35. The general solution is y = C1 cosx + C2 sinx. The boundary conditions yield y�(0) = C2 =0, y�

�12

�= −C1 = 2, so y = −2 cosx.

36. The auxiliary equation is m2 − 1 = 0, so m = ±1. The general solution is y = C1ex +C2e−x.The boundary conditions yield y(0) = C1 + C2 = 1, y(1) = C1e + C2e−1 = −1, or C1 =−1− e−1

e− e−1and C2 =

e+ 1

e− e−1, so y =

�−1− e−1

e− e−1

�ex +

�e+ 1

e− e−1

�e−x.

37. The auxiliary equation is m2 − 2m + 2 = 0, so m = 1 ± i. The general solution is y =ex (C1 cosx+ C2 sinx) . The boundary conditions yield y(0) = C1 = 1 and y(π) = −eπC1 =−1, which is a contradiction. No solution.

16.2. HOMOGENEOUS LINEAR EQUATIONS 1033

38. The general solution is y = ex (C1 cosx+ C2 sinx) . The boundary conditions yield y(0) =C1 = 1 and y (π/2) = C2eπ/2 = 1, so y = ex

�cosx+ e−π/2 sinx

�

39. The auxiliary equation is m2 − 4m+4 = 0, so m = 2 is a repeated root. The general solutionis y = C1e2x + C2xe2x. The boundary conditions yield y(0) = C1 = 0 and y(1) = C2e2 = 1,so y = xe−2e2x = xe2(x−1).

40. The general solution is y = C1e2x+C2xe2x. The boundary conditions yield y�(0) = 2C1+C2 =1 and y(1) = (C1 + C2) e2 = 2, or C1 = 1− 2e−2 and C2 = −1+4e−2, so y = (1− 2e−2)e2x+(−1 + 4e−2)xe2x.

41. Assuming a solution of the form y = emx we obtain the auxiliary equation m3−9m2+25m−17 = 0. Since y1 = ex is a solution we know that m1 = 1 is a root of the auxiliary equation.The equation can then be written as (m−1)(m2−8m+17) = 0. The roots of this equation are 1and 4±i. The general solution of the differential equation is y = C1ex+e4x(C2 cosx+C3 sinx).

42. Assuming a solution of the form y = emx we obtain the auxiliary equationm3+6m2+m−34 =0. Since y1 = e−4x cosx is a solution, we know that m1 = −4 + i is a root of the auxiliaryequation. Using the fact that complex roots of real polynomial equations occur in conjugatepairs we have thatm2 = −4−i is also a root. Thus [m−(−4+i)][m−(−4−i)] = m2+8m+17 isa factor of the auxiliary equation and we can write it asm3+6m2+m−34 = (m2+8m+17)(m−2) = 0. The general solution of the differential equation is y = C1e2x+e−4x(C1 cosx+C2 sinx).

43. y� = memx, y�� = m2emx, y�� = m3emx; m3emx − 4m2emx − 5memx = 0 =⇒ (m3 − 4m2 −5m)emx = 0 =⇒ m3 − 4m2 − 5m = 0 =⇒ m(m − 5)(m + 1) = 0 =⇒ m = 0, −1, 5; y =C1 + C2e−x + C3e5x

44. y� = memx, y�� = m2emx, y�� = m3emx; m3emx + 3m2emx − 4memx − 12emx = 0 =⇒(m3 + 3m2 − 4m − 12)emx = 0 =⇒ m3 + 3m2 − 4m − 12 = 0 =⇒ m2(m + 3) − 4(m + 3) =0 =⇒ (m2 − 4)(m+ 3) = 0 =⇒ m = −3, −2, 2; y = C1e−3x + C2e−2x + C3e2x

45. Case 1: λ = −α2 < 0Auxiliary equation is m2 − α2 = 0, so m = ±α and general solution is y = C1eαx + C2e−αx.Boundary conditions yield y(0) = C1+C2 = 0 and y(1) = C1eα+C2e−α = 0, or C1 = C2 = 0.So Case 1 yields no nonzero solutions.

Case 2: λ = 0Auxiliary equation ism2 = 0, som = 0 is a repeated root and general solution is y = C1+C2x.Boundary conditions yield y(0) = C1 = 0 and y(1) = C2 = 0. So Case 2 yields no nonzerosolutions.

Case 3: λ = α2 > 0Auxiliary equation is m2 + α2 = 0, so m = ±αi and the general solution is y = C1 cosαx +C2 sinx. Boundary conditions yield y(0) = C1 = 0 and y(1) = C2 sinα = 0. Hence, nonzerosolutions exist only when sinα = 0, which implies α = ±nπ so that λ = n2π2 for n = 1, 2, 3, . . .(n = 0 is excluded since that would give λ = 0).


46. (a) If the earth has density ρ then M = ρ4

3πR3 and Mr = ρ

4

3πr3, so that M/Mr = R3/r3

and Mr = r3M/R3. Then

F = −kMrm

r2= −k

r3Mm/R3

r2= −k

mM

R3r.

(b) Since a = d2r/dt2,F = ma = md2r

dt2= −k

mM

R3r =⇒ d2r

dt2+

kM

R3r = 0 =⇒ d2r

dt2+ ω2r =

0whereω2 = kM/R3. Since kmM/R2 = mg we have ω2 = kM/R3 = g/R.

(c) The general solution of the differential equation in part (b) is r(t) = c1 cosωt+ c2 sinωt.The initial conditions r(0) = R and r�(0) = 0 imply c1 = R and c2 = 0. Then r(t) =R cosωt. The mass oscillates back and forth from one side of the earth to the other witha period of T = 2π/ω. If we use R=3960 mi and g=32 ft/s2, then T ≈ 5079 s or 1.41 h.

47.

48.

16.3 Nonhomogeneous Linear Equations

1. m2 − 9 = 0 =⇒ m = −3, 3; yc = C1e−3x + C2e3x; yp = A, y�p = y��p = 0; −9A = 54 =⇒A = −6; yp = −6; y = C1e−3x + C2e3x − 6

2. 2m2 − 7m + 5 = 0 =⇒ (2m − 5)(m − 1) = 0 =⇒ m = 1, 5/2; yc = C1ex + C2e5x/2; yp =

A, y�p = y��p = 0; 5A = −29 =⇒ A = −29/5; y = C1ex + C2e5x/2 −29

5

3. m2 + 4m + 4 = 0 =⇒ (m + 2)2 = 0 =⇒ m = −2, −2; yc = C1e−2x + C2xe−2x; yp =Ax+B, y�p = A, y��p = 0; 4A+ 4(Ax+B) = 2x+ 6 =⇒ 4Ax+ 4(A+B) = 2x+ 6Solving 4A = 2, 4A+4B = 6, we obtain A = 1/2 and B = 1. Thus, y = C1e−2x+C2xe−2x+1

2x+ 1.

4. m2 − 2m+ 1 = 0 =⇒ (m− 1)2 = 0 =⇒ m = 1, 1; yc = C1ex + C2xex; yp = Ax3 + Bx2 +Cx+ d, y�p = 3Ax2 + 2Bx+ C, y��p = 6Ax+ 2B(6Ax+ 2B)− 2(3Ax2 + 2Bx+ c) + (Ax3 +Bx2 + Cx+D) = x3 + 4x=⇒ Ax3 + (−6A+B)x2 + (6A− 4B + C)x+ (2B − 2C +D) = x3 + 4xSolving A = 1, −6A+B = 0, 6A− 4B + c = 4, 2B − 2C +D = 0, we obtain A = 1, B =6, C = 22, and D = 32. Thus, y = C1ex + C2xex + x3 + 6x2 + 22x+ 32.

5. m2 + 25 = 0 =⇒ m = ±5i; yc = C1 cos 5x+ C2 sin 5x; yp = A sinx+ B cosx, y�p = A cosx−B sinx, y��p = −A sinx − B cosx; −A sinx − B cosx + 25(A sinx + B cosx) = 6 sinx =⇒24A sinx+ 24B cosx = 6 sinx; A = 1/4, B = 0; y = C1 cos 5x+ C2 sin 5x+

1

4sinx

6. m2 − 4 = 0 =⇒ m = −2, 2; yc = C1e−2x + C2e2x; yp = Ae4x, y�p = 4Ae4x, y��p =

16Ae4x − 4Ae4x = 7e4x =⇒ 12Ae4x = 7e4x =⇒ A = 7/12; y = C1e−2x + C2e2x +7

12e4x

16.3. NONHOMOGENEOUS LINEAR EQUATIONS 1035

7. m2 − 2m− 3 = 0 =⇒ (m− 3)(m+ 1) = 0 =⇒ m = −1, 3; yc = C1e−x + C2e3x

yp = Ae2x +Bx3 +Cx2 +Dx+E, y�p = 2Ae2x +3Bx2 +2Cx+D, y��p = 4Ae2x +6Bx+2C(4Ae2x + 6Bx + 2C) − 2(2Ae2x + 3Bx2 + 2Cx + D) − 3(Ae2x + Bx3 + Cx2 + Dx + E) =4e2x+2x3 =⇒ −3Ae2x3Bx3+(−6B−3C)x2+(6B−4C−3D)x+(2C−2D−3E) = 4e2x+2x3

Solving −3A = 4, −3B = 2, −6B − 3C = 0, 6B − 4C − 3D = 0, 2C − 2D − 3E = 0, weobtain A = −4/3, B = −2/3, C = 4/3, D = −28/9, and E = 80/27. Thus,

y = C1e−x + C2e

3x − 4

3e2x − 2

3x3 +

4

3x2 − 28

9x+

80

27.

8. m2 +m+ 1 = 0 =⇒ m = −1

2±

√3

2i; yc = e−x/2(C1 cos

√3x/2 + C2 sin

√3x/2)

yp = Ax2ex +Bxex + Cex +D, y�p = Ax2ex + (2A+B)xex + (B + C)ex

y��p = Ax2ex + (4A+B)xex + (2A+ 2B + C)ex�Ax2ex + (4A+B)xex + (2A+ 2B + C)ex

�+

�Ax2ex + (2A+B)xex + (B + C)ex

�

+�Ax2ex +Bxex + Cex +D

�= x2ex+3 =⇒ 3Ax2ex+(6A+3B)xex(2A+3B+3C)ex+D =

x2ex + 3Solving 3A = 1, 6A + 3B = 0, 2A + 3B + 3C = 0, D = 3, we obtain A = 1/3, B =−2/3, C = 4/9, and D = 3. Thus,

y = e−x/2(C1 cos√3x/2 + C2 sin

√3x/2) +

1

3x2ex − 2

3xex +

4

9ex + 3.

9. m2 − 8m+ 25 = 0 =⇒ m = 4± 3i; yc = e4x(C1 cos 3x+ C2 sin 3x); yp = Ae3x + B sin 2x+C cos 2x, y�p = 3Ae3x + 2B cos 2x− 2C sin 2x, y��p = 9Ae3x − 4B sin 2x− 4C cos 2x(9Ae3x − 4B sin 2x − 4C cos 2x) − 8(3Ae3x + 2B cos 2x − 2C sin 2x) + 25(Ae3x + B sin 2x +C cos 2x) = e3x − 6 cos 2x =⇒ 10Ae3x + (21B + 16C) sin 2x + (−16B + 21C) cos 2x = e3x −6 cos 2xSolving 10A = 1, 21B + 16C = 0, −16B + 21C = −6, we obtain A = 1/10, B = 96/697,and C = −126/697. Thus,

y = e4x(C1 cos 3x+ C2 sin 3x) +1

10e3x +

96

697sin 2x− 126

697cos 2x.

10. m2 − 5m+ 4 = 0 =⇒ (m− 1)(m− 4) = 0 =⇒ m = 1, 4; yc = C1ex + C2e4x

yp = A sinh 3x+B cosh 3x, y�p = 3A cosh 3x+ 3B sinh 3x, y��p = 9A sinh 3x+ 9B cosh 3x(9A sinh 3x+9B cosh 3x)−5(3A cosh 3x+3B sinh 3x)+4(A sinh 3x+B cosh 3x) = 2 sinh 3x =⇒(13A− 15B) sinh 3x+ (−15A+ 13B) cosh 3x = 2 sinh 3xSolving 13A− 15B = 2, −15A+ 13B = 0, we obtain A = −13/28 and B = −15/28. Thus,

y + C1ex + C2e

4x − 13

28sinh 3x− 15

28cosh 3x.

11. m2 − 64 = 0 =⇒ m = −8, 8; yc = C1e−8x + C2e8x; yp = A, y�p = y��p = 0

− 64A = 16 =⇒ A = −1/4; y = C1e−8x + C2e8x − 1

4, y� = −8C1e−8x + 8C2e8x.

Using y(0) = 1 and y�(0) = 0 we obtain C1+C2−1

4= 1, −8C1+8C2 = 0, or C1 = C2 = 5/8.

Thus, y =5

8e−8x +

5

8e8x − 1

4.


12. m2 + 5m − 6 = 0 =⇒ (m + 6)(m − 1) = 0 =⇒ m = −6, 1; yc = C1e−6x + C2ex; yp =Ae2x, y�p = 2Ae2x, y��p = 4Ae2x; Ae2x+5(2Ae2x)−6(Ae2x) = 10e2x =⇒ 8Ae2x = 10e2x =⇒A = 5/4; y = C1e−6x + C2ex +

5

4e2x, y� = −6C1e−6x + C2ex +

5

2e2x.

Using y(0) = 1 and y�(0) = 0 we obtain C1 + C2 +5

4= 1, −6C1 + C2 +

5

2= 0, or C1 =

9

28

and C2 = −4

7. Thus, y =

9

28e−6x − 4

7ex +

5

4e2x.

13. m2 + 1 = 0 =⇒ m = −i; i; yc = C1 cosx+ C2 sinx; W =

��cosx sinx− sinx cosx

�� = 1

u�1 = − sinx secx = − tanx, u1 = ln | cosx|; u�

2 = cosx secx = 1, u2 = xyp = cosx ln | cosx|+ x sinx; y = C1 cosx+ C2 sinx+ cosx ln | cosx|+ x sinx

14. m2 + 1 = 0 =⇒ m = −i, i; yc = C1 cosx+ C2 sinx; W =


�� = 1

u�1 = − sinx tanx = − sin2 x

cosx= −1− cos2 x

cosx= − secx+cosx, u1 = − ln | secx+tanx|+sinx

u�2 = cosx tanx = sinx; u2 = − cosx

yp = − cosx ln | secx+ tanx|+ sinx cosx− sinx cosx = − cosx ln | secx+ tanx|y = C1 cosx+ C2 sinx− cosx ln | secx+ tanx|



�� = 1

u�1 = − sin2 x, u1 = −1

2x+

1

2sinx cosx; u�

2 = sinx cosx, u2 =1

2sin2 x

yp = −1

2x cosx+

1

2sinx cos2 x+

1

2sin3 x

y = C1 cosx+ C2 sinx− 1

2x cosx+

1

2sinx cos2 x+

1

2sin3 x

= C1 cosx+ C2 sinx− 1

2x cosx+

1

2sinx(cos2 x+ sin2 x) = C1 cosx+ C3 sinx− 1

2x cosx



�� = 1

u�1 = − sinx secx tanx = − tan2 x = 1−sec2 x, u1 = x−tanx; u�

2 = cosx secx tanx = tanxu2 = − ln | cosx|; yp = cosx(tanx− x) = x cosx− sinx− sinx ln | cosx|y = C1 cosx+ C2 sinx+ x cosx− sinx− sinx ln | cosx| = C1 cosx+ C3 sinx+ x cosx− sinx ln | cosx|



�� = 1

u�1 = − sinx cos2 x u1 = 1

3 cos3 x; u�

2 = cosx cos2 x = cosx−cosx sin2 x, u2 = sinx− 13 sin

3 xyp = 1

3 cos4 x+ sin2 x− 1

3 sin4 x = sin2 x+ 1

3 (cos2 x− sin2 x) = sin2 x+ 1

3 cos 2x

y = C1 cosx+ C2 sinx+ sin2 x+1

3cos 2x = C1 cosx+ C2 sinx+

1

2− 1

2cos 2x+

1

3cos 2x

= C1 cosx+ C2 sinx+1

2− 1

6cos 2x



�� = 1


u�1 = − sinx sec2 x = − tanx secx, u1 = − secx; u�

2 = cosx sec2 x = secx;u2 = ln | secx+ tanx|yp = − cosx secx+ sinx ln | secx+ tanx| = −1 + sinx ln | secx+ tanx|y = C1 cosx+ C2 sinx− 1 + sinx ln | secx+ tanx|

19. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1e−x + C2ex; W =

��e−x ex

−e−x ex

�� = 2

u�1 =

1

2ex coshx = −1

4(e2x + 1), u1 = −1

8e2x − 1

4x;

u�2 =

1

2e−x coshx =

1

4(1 + e−2x) u2 =

1

4x− 1

8e−2x,

yp = e−x(−1

8e2x − 1

4x) + ex(

1

4x− 1

8e−2x = −1

8ex − 1

4xe−x +

1

4xex − 1

8d−x

= −1

8ex − 1

8e−x +

1

2x sinhx

y = C1e−x + C2ex − 1

8ex − 1

8e−x +

1

2x sinhx = C3e−x + C4ex +

1

2x sinhx

20. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1e−x + C2ex; W =

��e−x ex

−e−x ex

�� = 2

u�1 = −1

2ex sinh 2x = −1

4(e3x − e−x), u1 = − 1

12e3x − 1

4e−x

u�2 =

1

2d−x sinh 2x =

1

4(ex − e−3x), u2 =

1

4ex +

1

12e−3x

yp = e−x(− 1

12e3x − 1

4e−x) + ex(

1

4ex +

1

12e−3x) = − 1

12e2x − 1

4e−2x +

1

4e2x +

1

12e−2x

=1

6e2x − 1

6e−2x =

1

3sinh 2x

y = C1e−x + C2ex +1

3sinh 2x

21. m2 − 4 = 0 =⇒ m = −2, 2; yc = C1e−2x + C2e2x; W =

��e−2x e2x

−2e−2x 2e2x

�� = 4

u�1 = −1

4e2x(

e2x

x) = −1

4

e4x

x, u1 = −1

4

� x

x0

e4t

tdt; u�

2 =1

4e−2x

�e2x

x=

1

4x

�, u2 =

1

4ln |x|

yp = −1

4e−2x

� x

x0

e4t

tdt+

1

4e2x ln |x|; y = C1e

−2x + C2e2x − 1

4e2x

� x

x0

e4t

tdt+

1

4e2x ln |x|

22. m2 − 9 = 0 =⇒ m = −3, 3; yc = C1e−3x + C2e3x; W =

��e−3x e3x

−3e−3x 3e3x

�� = 6

u�1 = −1

6e3x(

9x

e3x= −3

2x, u1 = −3

4x2; u�

2 =1

6e−3x(

9x

e3x) =

3

2xe−6x

u2 =

�3

2xe−6xdx Integration by parts

= −1

4xe−6x − 1

24e−6x

yp = −3

4x2e−3x − 1

4xe−3x − 1

24e−3x; y = C3e−3x + C2e3x − 3

4x2e−3x − 1

4xe−3x


23. m2 + 3m+ 2 = 0 =⇒ (m+ 2)(m+ 1) = 0 =⇒ m = −2, −1; yc = C1e−2x + C2e−x

W =

��e−2x e−x

−2e−2x −e−x

�� = e−3x; u�1 = − 1

e−3x

e−x

1 + ex= − e2x

1 + ex

u1 = −�

e2x

1 + exdx v = 1 + ex, dv = exdx, ex = v − 1

= −�

v − 1

vdv = −v + ln |v| = −1− ex + ln(1 + ex)

yp = e−2x[−1− ex + ln(1 + ex)] + e−x ln(1 + ex) = −e−2x − e−x + e−2x ln(1 + ex)

+ e−x ln(1 + ex)

y = C1e−2x + C2e

−x − e−2x − e−x + e−2x ln(1 + ex) + e−x ln(1 + ex)

= C3e−2x + C4e

−x + e−2x ln(1 + ex) + e−x ln(1 + ex)

24. m2 − 3m+ 2 = 0 =⇒ (m− 1)(m− 2) = 0 =⇒ m = 1, 2; yc = C1ex + C2e2x;

W =

��ex e2x

ex 2e2x

�� = e3x; u�1 = − 1

e3xe2x

e3x

1 + ex= − e2x

1 + ex

u1 = −�

e2x

1 + exdx v = 1 + ex, dv = exdx, ex = v − 1

= −�

v − 1

vdv = −v + ln |v| = −1− ex + ln(1 + ex)

u�2 =

1

e3xex

e3x

1 + ex=

ex

1 + ex, u2 = ln(1 + ex)

yp = ex[−1− ex + ln(1 + ex)] + e2x ln(1 + ex) = −ex − e2x + ex ln(1 + ex) + e2x ln(1 + ex)y = C1e

x + C2e2x − ex − e2x + ex ln(1 + ex) + e2x ln(1 + ex)

= C3ex + C4e

2x + ex ln(1 + ex) + e2x ln(1 + ex)

25. m2 + 3m+ 2 = 0 =⇒ (m+ 2)(m+ 1) = 0 =⇒ m = −2, −1; yc = C1e−2x + C2e−x

W =

��e−2x e−x

−2e−2x −e−x

�� = e−3x; u�1 = − 1

e−3xe−x sin ex = −e2x sin ex

u1 = −�

e2x sin exdx Integration by parts

= ex cos ex − sin ex

u�2 =

1

e−3xe−2x sin ex = ex sin ex, u2 = − cos ex

yp = e−2x(ex cos ex−sin ex)+e−x(− cos ex) = −e−2x sin ex; y = C1e−2x+C2e−x−e−2x sin ex

26. m2 − 2m+ 1 = 0 =⇒ (m− 1)2 = 0 =⇒ m = 1, 1; yc = C1ex + C2xex;

W =

��ex xex

ex xex + ex

�� = e2x; u�1 = − 1

e2xxexex tan−1 x = −x tan−1 x


u1 = −�

x tan−1 xdx Integration by parts

= −1

2x2 tan−1 x− 1

2tan−1 x+

1

2x

u�2 =

1

e2xexex tan−1 x = tan−1 x, u2 = x tan−1 x− 1

2ln(1 + x2)

yp = ex�−1

2x2 tan−1 x− 1

2tan−1 x+

1

2x

�+ xex

�x tan−1 x− 1

2ln(1 + x2)

�

=1

2x2ex tan−1 x− 1

2ex tan−1 x+

1

2xex − 1

2xex ln(1 + x2)

y = C1ex + C3xe

x +1

2x2ex tan−1 x− 1

2ex tan−1 x− 1

2xex ln(1 + x2)

27. m2 − 2m+ 1 = 0 =⇒ (m− 1)2 = 0 =⇒ m = 1, 1; yc = C1ex + C2xex;

W =

��ex xex

ex xex + ex

�� = e2x; u�1 = − 1

e2xxex

ex

1 + x2= − x

1 + x2, u1 = −1

2ln(1 + x2);

u�2 =

1

e2xex

ex

1 + x2=

1

1 + x2, u2 = tan−1 x

yp = −1

2ex ln(1 + x2) + xex tan−1 x; y = C1ex + C2xex − 1

2ex ln(1 + x2) + xex tan−1 x

28. m2 − 2m+ 2 = 0 =⇒ m = 1± i; yc = ex(C1 cosx+ C2 sinx);

W =

��ex cosx ex sinx

−ex sinx+ ex cosx ex cosx+ ex sinx

�� = e2x

u�1 = − 1

e2xex sinx ex secx = − tanx, u1 = ln | cosx|; u�

2 =1

e2xex secx = 1, u2 = x

yp = ex cosx ln | cosx|+ xex sinx; y = ex(C1 cosx+ C2 sinx) + ex cosx ln | cosx|+ xex sinx

29. m2 + 2m+ 1 = 0 =⇒ (m+ 1)2 = 0 =⇒ m = −1, −1; yc = C1e−x + C2xe−x

W =

��e−x xe−x

−e−x −xe−x + e−x

�� = e−2x; u�1 = − 1

e−2xxe−xe−x lnx = −x lnx

u1 = −�

x lnxdx Integration by parts

=1

4x2 − 1

2x2 lnx

u�2 =

1

e−2xe−xe−x lnx = lnx, u2 = x lnx− x

yp = e−x

�1

4x2 − 1

2x2 lnx

�+ xe−x(x lnx− x) =

1

2x2e−x lnx− 3

4x2e−x

y = C1e−x + C2xe−x +1

2x2e−x lnx− 3

4x2e−x

30. m2 + 10m+ 25 = 0 =⇒ (m+ 5)2 = 0 =⇒ m = −5, −5; yc = C1e−5x + C2xe−5x

W =

��e−5x xe−5x

−5e−5x −5xe−5x + e−5x

�� = e−10x; u�1 =

1

e−10xxe−5x e

−10x

x2= −e−5x

x

u1 = −�

e−5x

xdx = −

� x

x0

e−5t

tdt; u�

2 =1

e−10xe−5x e

−10x

x2=

e−5x

x2,

u2 =

�e−5x

x2dx =

� x

x0

e−5t

t2dt; yp = −e−5x

� x

x0

e−5x

tdt+ xe−5x

� x

x0

e−5t

t2dt


y = C1e−5x + C2xe

−5x − e−5x

� x

x0

e−5t

tdt+ xe−5x

� x

x0

e−5t

t2dt

31. 4m2 − 4m+ 1 = 0 =⇒ (2m− 1)2 = 0 =⇒ m = 1/2, 1/2; yc = C1ex/2 + C2xex/2

W =

��ex/2 xex/2

1

2ex/2

1

2xex/2 + ex/2

�� = ex; u�1 = − 1

exxex/2(2e−x +

1

4x) = −2xe−3x/2 − 1

4x2e−x/2

u1 = −2

�xe−3x/2dx− 1

4

�x2e−x/2dx Integration by parts

=4

3xe−3x/2 +

8

9e−3x/2 +

1

2x2e−x/2 + 2xe−x/2 + 4e−x/2

u�2 =

1

exex/2(2e−x +

1

4x) = 2e−3x/2 − 1

4xe−x/2

u2 = 2

�e−3x/2dx+

1

4

�xe−x/2dx Integration by parts

= −4

3e−3x/2 − 1

2xe−x/2 − e−x/2

yp = ex/2(4

3xe−3x/2 +

8

9e−3x/2 +

1

2x2e−x/2 + 2xe−x/2 + 4e−x/2)

+ xex/2(−4

3e−3x/2 − 1

2xe−x/2 − e−x/2) =

8

9e−x + x+ 4

y = C1ex/2 + C2xex/2 +8

9e−x + x+ 4

32. 4m2 − 4m+ 1 = 0 =⇒ (2m− 1)2 = 0 =⇒ m = 1/2, 1/2; yc = C1ex/2 + C2xex/2

W =

��ex/2 xex/2

1

2ex/2

1

2xex/2 + ex/2

�� = ex; u�1 = − 1

exxex/2

ex/2

4

√1− x2 = −x

√1− x2

4u�1 =

1

12(1− x2)3/2; u�

2 =1

exex/2

ex/2

4

√1− x2 =

√1− x2

4

u2 =1

4

� �1− x2dx Trig substitution

=1

8sin−1 x+

1

8x�1− x2

yp =1

12ex/2(1− x2)3/2 +

1

8xex/2 sin−1 x+

1

8x2ex/2

√1− x2

y = C1ex/2 + C2xex/2 +1

12ex/2(1− x2)3/2 +

1

8xex/2 sin−1 x+

1

8x2ex/2

√1− x2

33. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1e−x + C2ex; W =

��e−x ex

−e−x ex

�� = 2

u�1 =

1

2exxex = −1

2xe2x

u1 = −1

2

�xe2xdx Integration by parts

=1

8e2x − 1

4xe2x

u�2 =

1

2e−xxex =

1

2x, u2 =

1

4x2; yp = e−x(

1

8e2x − 1

4xe2x) + ex(

1

4x2) =

1

8ex − 1

4xex +

1

4x2ex


y = C1e−x + C3ex − 1

4xex +

1

4x2ex; y� = −C1e−x + C3ex − 1

4ex +

1

4xex +

1

4x2ex

Using y(0) = 1 and y�(0) = 0 we have C1 + C3 = 1, −C1 + C3 −1

4= 0, or C1 = 3/8 and

C3 = 5/8. Thus, y =3

8e−x +

5

8ex − 1

4xex +

1

4x2ex.

34. 2m2 +m− 1 = 0 =⇒ (2m− 1)(m+ 1) = 0 =⇒ m = −1, 1/2; yc = C1e−x + C2ex/2

W =

��e−x ex/2

−e−x 1

2ex/2

�� =3

2e−x/2; u�

1 = − 2

3e−x/2ex/2

(x+ 1)

2= −1

3(xex + ex), u1 = −1

3xex

u�2 =

2

3e−x/2e−x (x+ 1)

2=

1

3e−x/2(x+ 1)

u2 =1

3

�e−x/2(x+ 1)dx Integration by parts

= −2

3xe−x/2 − 2e−x/2

yp = e−x(−1

3xex) + ex/2(−2

3xe−x/2 − 2e−x/2) = −x− 2

y = C1e−x + C2ex/2 − x− 2; y� = −C1e−x +1

2C2ex/2 − 1

Using y(0) = 1 and y�(0) = 0 we obtain C1 + C2 − 2 = 1, −C1 +1

2C2 − 1 = 0, or C1 = 1/3

and C2 = 8/3. Thus, y =1

3e−x +

8

3ex/2 − x− 2.

35. y�� − 1

xy� +

1

x2y =

4

xlnx; yc = C1x+ C2x lnx; W =

��x x lnx1 1 + lnx

�� = x

u�1 = − 1

x(x lnx)(

4

xlnx) = − 4

x(lnx)2, u1 = −4

3(lnx)3; u�

2 =1

x(x)(

4

xlnx) =

4

xlnx,

u2 = 2(lnx)2; yp = −4

3x(lnx)3 + 2x(lnx)3 =

2

3x(lnx)3; y + C1x+ C2x lnx+

2

3x(lnx)3

36. y�� − 4

xy� +

6

x2y =

1

x3; yc = C1x2 + C2x3; W =

��x2 x3

2x 3x2

�� = x4;

u�1 = − 1

x4(x3)

�1

x3

�= − 1

x4; u1 =

1

3x3; u�

2 =1

x4(x2)

�1

x3

�=

1

x5, u2 = − 1

4x4;

yp = x2

�1

3x3

�+ x3

�− 1

4x4

�=

1

12x

y = C1x2 + C2x3 +1

12x

37. Writing the differential equation in the form d2C/dx2 − (1/λ2)C = −C(∞)/λ2 we see thatthe auxiliary equation is m2 − 1/λ2 = 0. Thus, Cc = c1ex/λ + c2e−x/λ. Using undeterminedcoefficients with Cp = A we find that A = C(∞). Then C(x) = c1ex/λ + c2e−x/λ + C(∞).Since C(0) = c1+c2+C(∞) = 0 and lim

x→∞C(x) = C(∞) we see that c1 = 0 and c2 = −C(∞).

Thus, C(x) = C(∞)(1− e−x/λ).

38. If yc is the complementary function and yp is a particular solution, we have

ay��c + by�c + cyc = 0 and ay��p + by�p + cyp = g(x).


Therefore, letting y = yc + yp, we have

ay�� = by� + cy� = a(yc + yp)�� + b(yc + yp)

� + c(yc + yp)

= ay��c + ay��p + by�c + by�p + cyc + cyp

= [ay��c + by�c + cyc] + [ay��p + by�p + cyp]

= ay��p + by�p + cyp = g(x)

39. (a) Substituting Aex in for y in the DE, we have Aex + 2Aex − 3Aex = 10ex or 0 = 10ex,which is a contradiction for any value of A.

(b) Substituting Axex for y, we have

A(x+ 2)ex +A(2x+ 2)ex − 3Axex = 10ex.

Equating coefficients of xex and coefficients of ex, we get

A+ 2A− 3A = 0 and 2A+ 2A = 10

which gives A =5

2. Therefore, yp =

5

2xex.

(c) The auxiliary equation is m2 + 2m − 3 = 0, so m = −3 or m = 1. This gives yc =C1e−3x + C2ex. Therefore, the general solution is

y = yc + yp = C1e−3x + C2e

x +5

2xex

40. The auxiliary equation is m2−1 = 0, so m = ±1. This gives yc = C1e−x+C2ex. We look for aparticular solution of the form yp = Axex+B(x−2)e−x−Axex−Bxex = e−x−ex. Equatingcoefficients of xex, ex, xe−x, and e−x, we get A−A = 0, 2A = −1, B−B = 0, −2B = 1,

which gives A = −1

2, B = −1

2. Therefore, yp = −1

2xex − 1

2xe−x and the general solution is

y = yc + yp = C1e−x + C2e

x − 1

2xex − 1

2xe−x

16.4 Mathematical Models

1. A weight of 4 pounds is pushed up 3 feet above the equilibrium position. At t = 0 it is givenan initial speed upward of 2 feet per second.

2. A mass of 2 pounds is pulled down 0.7 feet below the equilibrium position and held. At t = 0it is released from rest.

3. Using m = W/g = 8/32 = 1/4, the initial value problem is1

4x�� + x = 0; x(0) =

1

2, x�(0) =

3

2. The auxiliary equation is

1

4m2 + 1 = 0, so m = ±2i and x = C1 cos 2t+ C2 sin 2t,

x� = −2C1 sin 2t + 2C2 cos 2t. Using the initial condition, we obtain C1 = 1/2 and C2 =3

4.

The equation of motion is x(t) =1

2cos 2t+

3

4sin 2t.

16.4. MATHEMATICAL MODELS 1043

4. From Hooke’s law we have 24 = k(1/3), so k = 72. Using m = W/g = 24/32 = 3/4, the initial

value problem is3

4x��+72x = 0; x(0) = −3, x�(0) = 0. The auxiliary equation is

3

4m2+72 =

0, som = ±4√6i and x = C1 cos 4

√6t+C2 sin 4

√6t, x� = −4

√6C1 sin 4

√6+4

√6C2 cos 4

√6t.

Using the initial conditions, we obtain C1 = −1/4 and C2 = 0. Thus, x(t) = −1

4cos 4

√6t.

5. From Hooke’s law we have 400 = k(2), so k = 200. The initial value problem is 50x��+200x =0; x(0) = 0, x�(0) = −1 = . The auxiliary equation is 50m2 + 200 = 0, so m = ±2i andx = C1 cos 2x + C2 sin 2x, x� = −2C1 sin 2x + 2C2 cos 2x. Using the initial conditions, weobtain C1 = 0 and C2 = −5. Thus, x(1) = −5 sin 2x.


16x�� + 4x = 0; x(0) =

2

3, x�(0) = −4

3. The auxiliary equation is m2/16 + 4 = 0, so m = ±8i and x = C1 cos 8x +

C2 sin 8x, x� = −8C1 sin 8x + 8C2 cos 8x. Using the initial conditions, we obtain C1 = 2/3

and C2 = −1/6. Thus, x(t) =2

3cos 8t− 1

6sin 8t.

7. A 2 pound weight is released from the equilibrium position with an upward speed of 1.5 ft/s.A damping force numerically equal to twice the instantaneous velocity acts on the system.

8. A 16 pound weight is released from 2 feet above the equilibrium position with a downwardspeed of 1 ft/s. A damping force numerically equal to the instantaneous velocity acts on thesystem.


8x�� + x� + 2x = 0; x(0) =

−1, x�(0) = 8. The auxiliary equation is m2/8+m+2 = 0 or (m+4)2 = 0, so m = −4, −4and x = C1e−4t +C2te−4t, x� = (C2 − 4C1)e−4t − 4C2te−4t. Using the initial conditions, weobtain C1 = −1 and C2 = 4. Thus, x(t) = −e−4t+4te−4t. Solving x(t) = −e4t+4te−4t = 0, wesee that the weight passes through the equilibrium position at t = 1/4s. To find the maximumdisplacement we solve x�(t) = 8e−4t − 16te−4t = 0. This gives t = 1/2. Since x(1/2) = e−2 ≈0.14, the maximum displacement is approximately 0.14 feet below the equilibrium position att = 1/2s.

10. From Hooke’s law we have 40(980) = k(10), so k = 3920. The initial value problem is 40x�� +560x� + 3920x = 0; x(0) = 0, x�(0)− 2. The auxiliary equation is 40m2 + 560m+ 3920 = 0or m2 + 14m+ 98 = 0, so m = −7± 7i and x = e−7t(C1 cos 7t+ C2 sin 7t),x� = −7(C1 + C2)e−7t sin 7t − 7(C1 − C2)e−7t cos 7t. Using the initial conditions, we obtain

C1 = 0 and C2 = 2/7. Thus, x(t) =2

7e−7t sin 7t.

11. From Hooke’s law we have 10 = k(7 − 5), so k = 5. Using m = W/g = 8/32 = 1/4, the

initial value problem is1

4x�� + x� + 5x = 0; x(0) =

1

2; x�(0) = 1. The auxiliary equation is

m2/4+m+5 = 0 orm2+4m+20 = 0, som = −2±4i. Thus, x = e−2t(C1 cos 4t+C2 sin 4t) andx� = −2(C1 − 2C2)e−2t cos 4t− 2(2C1 +C2)e−2t sin 4t. Using the initial conditions, we obtain1

2= C1 and 1 = −2(

1

2− 2C2), so C1 = 1/2 and C2 = 1/2. Therefore x(t) =

1

2e−2t(cos 4t +

sin 4t).


12. From Hooke’s law we have 24 = k(4), so k = 6. Using m = W/g = 24/32 = 3/4, the initial

value problem is3

4x�� + βx� + 6x = 0; x(0) = 0, x�(0) = −2. The auxiliary equation

is3

4m2 + βm + 6 = 0. Using the quadratic formula, m = (−β ±

�β2 − 18/(3/2). When

β >√18 = 3

√2, we have m1 = −2

3β +

2

3

�β2 − 18 and m2 = −2

3β − 2

3

�β2 − 18. Thus,

x(t) = C1e−2βt/3+2t

√β2−18/3 + C2e

−2βt/3−2t√

β2−18/3

= e−2βt/3

�C3 cosh

�2

3

�β2 − 18

�t+ C4 sinh

�2

3

�β2 − 18

�t

�

see Example 5 in Section 16.2.

From x(0) = 0 we obtain C3 = 0 so that x(t) = C4e−2βt/3 sinh(2

3

�β2 − 18t). The velocity is

x�(t) =2

3

�β2 − 18C4e

−2βt/3 cosh(2

3

�β2 − 18t)− 2β

3C4e

−2βt/3 sinh(2

3

�β2 − 18t).

From x�(0) = −2 we obtain −2 =2

3

�β2 − 18C4 or C4 = −3/

�β2 − 18. Therefore,

x(t) =−3�

β2 − 18e−2βt/3 sinh(

2

3

�β2 − 18t).

13. From Hooke’s law we have 10 = k(2), so k = 5. Using m = W/g = 10/32 = 5/16, the

differential equation is5

16x�� + βx� + 5 = 0. The auxiliary is

5

16m2 + βm + 5 = 0 Using the

quadratic formula, m = (−β ±�β2 − 25/4)/(5/8). For β > 0 the motion is

(a) overdamped when β2 − 25/4 > 0 or β > 5/2

(b) critically damped when β2 − 25/4 = 0 or β = 5/2

(c) underdamped when β2 − 25/4 < 0 or β < 5/2.

14. Since W = mg = 1(32) = 32, we have from Hooke’s law 32 = k(2), so k = 16. The initialvalue problem is x�� + 8x� + 16x = 8 sin 4t; x(0) = x�(0) = 0. The auxiliary equation ism2 + 8m + 16 = (m + 4)2 = 0 so m = −4, −4, and xc = C1e−4t + C2te−4t. Usingxp = A sin 4t+B cos 4t we find A = 0 and B = −1/4. Thus,

x(t) = C1e−4t + C2te

−4t − 1

4cos 4t and x�(t) = −4C1e

−4t − 4C2te−4t + C2e

−4t + sin 4t.

Using the initial conditions, we obtain 0 = C1 −1

4and 0 = −4C1 + C2. Thus, C1 = 1/4 and

C2 = 4C1 = 1. Therefore x(t) =1

4e−4t + te−4t − 1

4cos 4t.

16.4. MATHEMATICAL MODELS 1045

15. The initial value problem is x�� + 8x� + 16x = e−t sin 4t; x(0) = x�(0) = 0. Using xp =Ae−t sin 4t+Be−t cos 4t we find A = −7/625 and B = −24/625. Thus,

x(t) = C1e−4t + C2te

−4t − 7

625e−t sin 4t− 24

625e−t cos 4t,

x�(t) = −4C1e−4t − 4C2te

−4t + C2e−4t − 28

625e−t cos 4t+

7

625e−t sin 4t+

96

625e−t sin 4t

+24

625e−t cos 4t.

Using the initial conditions, we obtain C1 = 24/625 and C2 = 100/625. Thus,

x(t) =24

625e−4t +

100

625te−4t − 7

625e−t sin 4t− 24

626e−t cos 4t.

As t −→ ∞, e−t −→ 0 and x(t) −→ 0.

16. A 32 pound weight is pulled 2 feet below the equilibrium position and held. At time t =0 an external force equal to 5 sin 3t is applied to the system. The auxiliary equation ism2 + 9 = 0, so m = ±3i and xc = C1 cos 3t + C2 sin 3t. Using variation of parameters

xp = −5

6t cos 3t+

5

18sin 3t, so

x(t) = C1 cos 3t+ C3 sin 3t−5

6t cos 3t

x�(t) = −3C1 sin 3t+ 3C2 cos 3t+5

2t sin 3t− 5

6cos 3t.

Using the initial conditions, we obtain C1 = 2 and C2 = 5/18. Thus, x(t) = 2 cos 3t +5

18sin 3t− 5

6t cos 3t. The spring-mass system is in pure resonance.

17. The DE describing charge is .05q�� + 2q� + 100q = 0. The auxiliary equation is 0.5m2 + 2m+100 = 0, so m = −20 ± 40i. The general solution is q = e−20t(C1 cos 40t + C2 sin 40t). Theinitial conditions yield q(0) = C1 = 5 and i(0) = q�(0) = −20C1 + 40C2 = 0, which gives

C1 = 5 and C2 = 52 . Therefore q(t) = e−20t

�5 cos 40t+

5

2sin 40t

�, and q(0.01) = 4.568C.

q(t) = 0 when 5 cos 40t+ 52 sin 40t = 0 which first occurs at t = 0.0509 s.

18. The DE describing charge is 14q

��+20q�+300q = 0. The auxiliary equation is 14m

2+20m+300 =0, so m = −20 or m = −60. The general solution is q(t) = C1e−20t + C2e−60t. The initialconditions yield q(0) = C1+C2 = 4 and i(0) = q�(0) = −20C1− 60C2 = 0, which give C1 = 6and C2 = −2. Therefore, q(t) = 6e−20t − 2e−60t. The charge is never equal to zero.

19. The DE is5

3q�� + 10q� + 30q = 300. The auxiliary equation is 5

3m2 + 10m + 30 = 0. so

m = −3± 3i. This gives qc = e−3t(C1 cos 3t+ C2 sin 3t). Assume a particular solution of theform qp = A. Substituting into the DE, we have 30A = 300 so that A = 10 and thereforeqp = 10. Thus, the general solution is q = qc+ qp = e−3t(C1 cos 3t+C2 sin 3t)+10. The initialconditions yield q(0) = C1+10 = 0 and i(0) = q�(0) = −3(C1−C2) = 0, which gives C1 = −10and C2 = −10. Therefore, q(t) = e−3t(−10 cos 2t−10 sin 3t)+10−10−10e−3t(cos 3t+sin 3t),i(t) = q�(t) = 60e−3t sin 3t. The charge q(t) attains a maximum of 10.432 C at t = π

3 .


20. The DE is q��+100q�+2500q = 30. The auxiliary equation is m2+100m+250000, so m = −50is a repeated root. This gives qc = C1e−50t + C2te−50t. Assume a particular solution of theform qp = A. Substituting into the DE, we have 2500A = 30 so that A = 3

250 and therforeqp = 3

250 . The general solution is q = qc+ qp = C1e−50t+C2te−50t+ 3250 . The general solution

is q = qc + qp = C1e−50t + C2te−50t + 3250 . The initial conditions yield q(0) = C1 = 0 and

i(0) = q�(0) = −50C!+C2 = 2 which give C1 = 0 and C2 = 2. Therefore, q(t) = 2te−50t+ 3250

and i(t) = q�(t) = (2− 100t)e−50t. The charge q(t) attains a maximum of 0.0267 C at t = 150

s.

21.

16.5 Power Series Solutions

1.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+∞�

n=0

cnxn =

∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=0

ckxk

=∞�

k=0

[(k + 2)(k + 1)ck+2 + ck]xk = 0

(k + 2)(k + 1)ck+2 + ck = 0; ck+2 = − ck(k + 2)(k + 1)

, k = 0, 1, 2, . . .

c2 = −c02

= −c02!, c3 = − c1

3 · 2 = −c13!, c4 = − c2

4 cos 3=

c04 · 3 · 2! =

c04!,

c5 = − c35 · 4 =

c15 · 4 · 3! =

c15!, c6 = − c4

6 · 5 = − c06 · 5 · 4! = −c0

6!,

c7 = − c57 · 6 = − c1

7 · 6 · 5! = −c17!

y = c0

�1− 1

2!x2 +

1

4!x4 − 1

6!x6 + · · ·

�+ c1

�x− 1

3!x3 +

1

5!x5 − 1

7!x7 + · · ·

�

= c0

∞�

n=0

(−1)n1

(2n)!x2n + c1

∞�

n=0

(−1)n1

(2n+ 1)!x2n+1

2.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−∞�

n=0

cnxn =

∞�

k=0

(k + 2)(k + 1)ck+2xk −

∞�

k=0

ckxk

=∞�

k=0

[(k + 2)(k + 1)ck+2 − ck]xk = 0

(k + 2)(k + 1)ck+2 − ck = 0; ck+2 =ck

(k + 2)(k + 1), k = 0, 1, 2, . . . ; c2 =

c02!,

c3 =c13 · 2 =

c13!, c4 =

c24 · 3 =

c04!, c5 =

c35 · 4 =

c15!, c6 =

c46 · 5 =

c06!, c7 =

c57 · 6 =

c17!

y = c0

�1 +

1

2!x2 +

1

4!x4 +

1

6!x6 + · · ·

�+ c1

�x+

1

3!x3 +

1

5!x5 +

1

7!x7 + · · ·

�

= c0

∞�

n=0

1

(2n)!x2n + c1

∞�

n=0

1

(2n+ 1)!x2n+1

16.5. POWER SERIES SOLUTIONS 1047

3.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−∞�

n=1

ncnxn−1

� �� k=n−1

=∞�

k=0

(k + 2)(k + 1)ck+2xk −

∞�

k=0

(k + 1)ck+1xk

=∞�

k=0

[(k + 2)(k + 1)ck+2 − (k + 1)ck+1]xk = 0

(k + 2)(k + 1)ck+2 − (k + 1)ck+1 = 0; ck+2 =ck+1

(k + 2), k = 0, 1, 2, . . . ; c2 =

c12

=c12!,

c3 =c23

=c13!, c4 =

c34

=c14!,

y = c0 + c1

�x+

1

2!x2 +

1

3!x3 + · · ·

�= c0 + c1

�∞n=1

1

n!xn

4.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−∞�

n=1

ncnxn−1

� �� k=n−1

= 2∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=0

(k + 1)ck+1xk

=∞�

k=0

[2(k + 2)(k + 1)ck+2 − (k + 1)ck+1]xk = 0

2(k + 2)(k + 1)ck+2 + (k + 1)ck+1 = 0; ck+2 = − ck+1

2(k + 2), k = 0, 1, 2, . . . ;

c2 = − c12 · 2 = − c1

2 · 2! , c3 = − c22 · 3 =

c122 · 3! , c4 = − c3

2 · 4 = − c123 · 4! ,

y = c0 + c1

�x− 1

2 · 2!x2 +

1

22 · 3!x3 − 1

23 · 4!x4 + · · ·

�

5.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−x∞�

n=0

cnxn

� �� k=n+1

=∞�

k=1

(k + 2)(k + 1)ck+2xk −

∞�

k=1

ck−1xk

= 2c2 +∞�

k=0

[(k + 2)(k + 1)ck+2 − ck−1]xk = 0

c2 = 0; (k + 2)(k + 1)ck+2 − ck−1 = 0; ck+2 =ck−1

(k + 2)(k + 1), k = 1, 2, 3, . . . ;

c3 =c03 · 2 , c5 =

c25 · 4 = 0, c6 =

c36 · 5 =

c06 · 5 · 3 · 2 , c7 =

c47 · 6 =

c17 · 6 · 4 · 3

c9 =c58 · 7 = 0, c9 =

c69 · 8 = c9 =

c09 · 8 · 6 · 5 · 3 · 2 , c10 =

c710 · 9 =

c110 · 9 · 7 · 6 · 4 · 3

y = c0

�1 +

1

3 · 2x3 +

1

6 · 5 · 3 · 2x6 +

1

9 · 8 · 6 · 5 · 3 · 2x9 + · · ·

�

+ c1

�x+

1

4 · 3x4 +

1

7 · 6 · 4 · 3x7 +

1

10 · 9 · 7 · 6 · 4 · 3x10 + · · ·

�


6.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+x2∞�

n=0

cnxn

� �� k=n+2

=∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=2

ck−2xk

= 2c2 ++c3x+∞�

k=2

[(k + 2)(k + 1)ck+2 + ck−2]xk = 0

c2 = c3 = 0; (k + 2)(k + 1)ck+2 + ck−2 = 0; ck+2 = − ck−2

(k + 2)(k + 1), k = 2, 3, 4, . . . ;

c4 = − c04 · 3 , c5 = − c1

5 · 4 , c6 = − c27 · 6 = 0, c7 = − c3

7 · 6 , c8 = − c48 · 7 =

c08 · 7 · 4 · 3

c9 = − c59 · 8 =

c19 · 8 · 5 · 4 , c10 = − c6

10 · 9 = 0, c11 −c7

11 · 10 = 0,

c12 = − c812 · 11 = − c0

12 · 11 · 8 · 7 · 4 · 3 , c13 = − c913 · 12 = − c1

13 · 12 · 9 · 8 · 5 · 4 ,

y = c0

�1− 1

4 · 3x4 +

1

8 · 7 · 4 · 3x8 − · · ·

�c1

�x− 1

5 · 4x5 +

1

9 · 8 · 5 · 4x9 − · · ·

�

7.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−2x∞�

n=1

cnxn−1 +

∞�

n=0

cnxn

=∞�

k=0

(k + 2)(k + 1)ck+2xk −

∞�

k=1

kckxk +

∞�

k=0

ckxk

= c0 + 2c2 +∞�

k=1

[(k + 2)(k + 1)ck+2 − (2k − 1)ck]xk = 0

c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 − (2k − 1)ck = 0; c2 = −c02

ck+2 =(2k − 1)ck

(k + 2)(k + 1), k = 1, 2, 3, . . . ; c3 =

c13 · 2 =

c13!, c4 =

3c24 · 3 = −3c0

4!,

c5 =5c35 · 4 =

5c15!

, c6 =7c46 · 5 =

7 · 3c06!

, c7 =9c57 · 6 =

9 · 5c17!

y = c0

�1− 1

2!x2 − 3

4!x4 − 7 · 3

6!x6 − · · ·

�+ c1

�x+

1

3!x3 +

5

5!x5 +

9 · 57!

x7 + · · ·�

8.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−x∞�

n=1

ncnxn−1 + 2

∞�

n=0

cnxn

=∞�

k=0

(k + 2)(k + 1)ck+2xk −

∞�

k=1

kckxk + 2

∞�

k=0

ckxk

= 2c0 + 2c2 +∞�

k=1

[(k + 2)(k + 1)ck+2 − (k − 2)ck]xk = 0

2c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 − (k − 2)ck = 0; c2 = −c0

ck+2 =(k − 2)ck

(k + 2)(k + 1), k = 1, 2, 3, . . . ; c3 = − c1

3 · 2 = −c13!, c4 = 0, c5 =

c35 · 4 = −c1

5!,


c6 = c8 = c10 = 0, c7 =3c57 · 6 = −3c1

7!, c9 =

5c79 · 8 = −5 · 3c1

9!

y = c0(1− x2) + c1

�1− 1

3!x3 − 1

5!x5 − 3

7!x7 − 5 · 3

9!x9 + · · ·

�

9.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+x2∞�

n=1

ncnxn−1

� �� k=n+1

+x∞�

n=0

cnxn

� �� k=n+1

=∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=2

(k − 1)ck−1xk +

∞�

k=1

ck−1xk

= 2c2 + (6c3 + c0)x+∞�

k=2

[(k + 2)(k + 1)ck+2 + kck−1]xk = 0

2c2 = 0, 6c3 + c0 = 0 (k + 2)(k + 1)ck+2 + kck−1 = 0; c2 = 0, c3 = − c03 · 2

ck+2 = − kck−1

(k + 2)(k + 1), k = 2, 3, 4, . . . ; c4 = − 2c1

4 · 3 , c5 = 0, c6 = − 4c36 · 5 =

4c06 · 5 · 3 · 2

c7 = − 5c47 · 6 =

5 · 2c17 · 6 · 4 · 3 , c8 = c11 = c14 = · · · = 0, c9 = − 7c6

9 · 8 = − 7 · 4c09 · 8 · 6 · 5 · 3 · 2

c10 = − 8c710 · 9 = − 8 · 5 · 2c1

10 · 9 · 7 · 6 · 4 · 3y = c0

�1− 1

3!x3 +

42

6!x6 − 72 · 42

9!x9 · · ·

�+ c1

�x− 22

4x4 +

52 · 227!

x7 − 82 · 52 · 2210!

x10 + · · ·�

10.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+2x∞�

n=1

ncnxn−1 + 2

∞�

n=0

cnxn

=∞�

k=0

(k + 2)(k + 1)ck+2xk + 2

∞�

k=1

kckxk + 2

∞�

k=0

ckxk

= 2c2 + 2c0 +∞�

k=1

[(k + 2)(k + 1)ck+2 + 2(k + 1)ck]xk = 0

2c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 + 2(k + 1)ck = 0; c2 = −c0

ck+2 = − ckk + 2

, k = 1, 2, 3, . . . ; c3 = −2c13

, c4 = −2c24

=2c04

, c5 = −2c35

=22c15 · 3 ,

c6 = −2c46

= −22c06 · 4 , c7 = −2c5

7= − 23c1

7 · 5 · 3 , c8 = −2c68

=23c0

8 · 6 · 4y = c0

�1− x2 +

2

4x4 − 22

6 · 4x6 +

23

8 · 6 · 4x8 + · · ·

�+c1

�x− 2

3x3 +

22

5 · 3x5 − 23

7 · 5 · 3x7 + · · ·

�


11. (x− 1)∞�

n=2

n(n− 1)cnxn−2 +

∞�

n=1

ncnxn−1

=∞�

n=2

n(n− 1)cnxn−1

� �� k=n−1

−∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+∞�

n=2

ncnxn−1

� �� k=n−1

=∞�

k=1

(k + 1)kck+1xk −

∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=0

(k + 1)ck+1xk

= c1 − 2c2 +∞�

k=1

[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1]xk = 0

c1 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 = 0; c2 =c12

ck+2 =(k + 1)ck+1

k + 2, k = 1, 2, 3, . . . ; c3 =

2c23

=c13, c4 =

3c34

=c14

y = c0 + c1

�x+

1

2x2 +

1

3x3 +

1

4x4 + · · ·

�= c0 + c1

∞�

n=1

1

nxn

12. (x+ 2)∞�

n=2

n(n− 1)cnxn−2 + x

∞�

n=1

ncnxn−1 −

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn−1

� �� k=n−1

+2∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+∞�

n=1

ncnxn −

∞�

n=0

cnxn

=∞�

k=1

(k + 1)kck+1xk + 2

∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=1

kckxk −

∞�

k=0

ckxk

= 4c2 − c0 +∞�

k=1

[(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck]xk = 0

4c2 − c0 = 0; (k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck = 0; c2 =c04

ck+2 = − (k + 1)kck+1 + (k − 1)ck2(k + 2)(k + 1)

= − kck+1

2(k + 2)− (k − 1)ck

2(k + 2)(k + 1), k = 1, 2, 3, . . .

c3 = − c22 · 3 = − c0

2 · 3 · 4 , c4 = − 2c32 · 4 − c2

2 · 4 · 3 =c0

2 · 3 · 42 − c02 · 3 · 42 = 0

c5 = 0− 2c32 · 5 · 4 =

c95 · 42 · 3 · 2 , c6 = − 4c5

2 · 6 − 0 = − c06 · 5 · 4 · 3 · 22

y = c0

�1 +

1

4x2 − 1

4 · 3 · 2x3 +

1

5 · 42 · 3 · 2x5 − · · ·

�+ c1x


13. (x2 − 1)∞�

n=2

n(n− 1)cnxn−2 + 4x

∞�

n=1

ncnxn−1 + 2

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn −

∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+4∞�

n=1

ncnxn + 2

∞�

n=0

cnxn

=∞�

k=2

k(k − 1)ckxk −

∞�

k=0

(k + 2)(k + 1)ck+2xk + 4

∞�

k=1

kckxk + 2

∞�

k=0

ckxk

= (2c0 − 2c2) + (2c1 + 4c1 − 6c3)x+∞�

k=2

[k(k − 1)ck − (k + 2)(k + 1)ck+2 + 4kck + 2ck]xk

= 02c0 − 2c2 = 0; 6c1 − 6c3 = 0; (k + 2)(k + 1)ck − (k + 2)(k + 1)ck+2 = 0; c2 = c0, c3 = c1;ck+2 = ck, k = 2, 3, 4, . . . ; c4 = c2 = c0, c5 = c3 = c1, c6 = c4 = c0, c7 = c5 = c1y = c0

�1 + x2 + x4 + · · ·

�+ c1[x+ x3 + x5 + · · · ] = c0

�∞n=0 x

2n + c1�∞

n=0 x2n+1

14. (x2 + 1)∞�

n=2

n(n− 1)cnxn−2 − 6

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn +

∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−6∞�

n=0

ncnxn

=∞�

k=2

k(k − 1)ckxk +

∞�

k=0

(k + 2)(k + 1)ck+2xk − 6

∞�

k=0

ckxk

= (2c2 − 6c0) + (6c3 − 6c1)x+∞�

k=2

[k(k − 1)ck + (k + 2)(k + 1)ck+2 − 6ck]xk = 0

2c2 − 6c0 = 0; 6c3 − 6c1 = 0; (k− 3)(k+ 2)ck + (k+ 2)(k+ 1)ck+2 = 0; c2 = 3c0, c3 = c1;

ck+2 = − (k − 3)ckk + 1

, k = 2, 3, 4, . . . ; c4 = −−c23

= c0, c5 = 0, c6 = −c45

= −c05

c7 = c9 = c11 = · · · = 0, c8 = −3c67

=3c07 · 5 , c10 = −5c8

9= − 5 · 3c0

9 · 7 · 5y = c0

�1 + 3x2 + x4 − 3

5 · 3x6 + · · ·

�+ c1(x+ x3)


15. (x2 + 2)∞�

n=2

n(n− 1)cnxn−2 + 3x

∞�

n=1

ncnxn−1 −

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn + 2

∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+3∞�

n=1

ncnxn −

∞�

n=0

cnxn

=∞�

k=2

k(k − 1)ckxk + 2

∞�

k=0

(k + 2)(k + 1)ck+2xk + 3

∞�

k=1

kckxk −

∞�

k=0

ckxk

= (4c2 − c0) + (12c3 + 3c1 − c1)x+∞�

k=2

[k(k − 1)ck + 2(k + 2)(k + 1)ck+2 + 3kck − ck]xk

= 0

4c2−c0 = 0; 12c3+2c1 = 0; 2(k+2)(k+1)ck+2+(k2+2k−1)ck = 0; c2 =c04, c3 = −c1

6;

ck+2 = − (k2 + 2k − 1)ck2(k + 2)(k + 1)

, k = 2, 3, 4, . . . ; c4 = − 7c22 · 4 · 3 = − 7

4 · 4!c0, c5 = − 14c32 · 5 · 4 =

14

2 · 5!c1

c6 = − 23c42 · 6 · 5 =

23 · 723 · 6!c0, c7 = − 34c5

2 · 7 · 6 = −34 · 144 · 7! c1y = c0

�1 +

1

4x2 − 7

4 · 4!x4 +

23 · 78 · 6! x

6 − · · ·�+

c1

�x− 1

6x3 +

14

2 · 5!x5 − 34 · 14

4 · 7! x7 + · · ·�

16. (x2 − 1)∞�

n=2

n(n− 1)cnxn−2 + x

∞�

n=0

ncnxn−1 −

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn −

∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+∞�

n=1

ncnxn −

∞�

n=0

cnxn

=∞�

k=2

k(k − 1)ckxk −

∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=1

kckxk −

∞�

k=0

ckxk

= −(2c2 + c0) + (c1 − 6c3 − c1)x+∞�

k=2

[k(k − 1)ck − (k + 2)(k + 1)ck+2 + kck − ck]xk

= 0

2c2 + c0 = 0; −6c3 = 0; (k + 1)(k − 1)ck − (k + 2)(k + 1)ck+2 = 0; c2 = −c02, c3 = 0;

ck+2 =(k − 1)ckk + 2

, k = 2, 3, 4, . . . ; c4 =c24

= − c04 · 2 , c5 = c7 = c9 = · · · = 0, c6 =

3c46

=

− c04 · 22

y = c0

�1− 1

2x2 − 1

8x4 − 1

16x6 − · · ·

�+ c1x


17.∞�

n=2

n(n− 1)cnxn−2 − (x+ 1)

∞�

n=1

ncnxn−1 −

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−∞�

n=1

ncnxn −

∞�

n=1

ncnxn−1

� �� k=n−1

−∞�

n=0

cnxn

=∞�

k=0

(k + 2)(k + 1)ck+2xk −

∞�

k=1

kckxk −

∞�

k=0

(k + 1)ck+1xk −

∞�

k=0

ckxk

= 2c2 − c1 − c0 +∞�

k=1

[(k + 2)(k + 1)ck+2 − kck − (k + 1)ck+1 − ck]xk = 0

2c2 − c1 − c0 = 0; (k + 2)(k + 1)ck+2 − (k + 1)ck+1 − (k + 1)ck = 0; c2 =c0 + c1

2

ck+2 =ck + ck+1

k + 2, k = 1, 2, 3, . . . ; c3 =

c1 + c23

=c1 + c0/2 + c1/2

3=

c0 + 3c16

c4 =c2 + c3

4=

c0/2 + c1/2 + c0/6 + c1/2

4=

2c0 + 3c112

c5 =c3 + c4

5=

c0/6 + c1/2 + c0/6 + c1/4

5=

4c0 + 9c160

y = c0

�1 +

1

2x2 +

1

6x3 +

1

6x4 + · · ·

�+ c1

�x+

1

2x2 +

1

2x3 +

1

4x4 + · · ·

�

18.∞�

n=2

n(n− 1)cnxn−2 − x

∞�

n=1

ncnxn−1 − (x+ 2)

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−∞�

n=1

ncnxn −

∞�

n=0

cnxn+1

� �� k=n+1

−2∞�

n=0

cnxn

=∞�

k=0

(k + 2)(k + 1)ck+2xk −

∞�

k=1

kckxk −

∞�

k=1

ck−1xk − 2

∞�

k=0

ckxk

= 2c2 − 2c0 +∞�

k=1

[(k + 2)(k + 1)ck+2 − kck − ck−1 − 2ck]xk = 0

2c2 − 2c0 = 0; (k + 2)(k + 1)ck+2 − (k + 2)ck − ck−1 = 0; c2 = 0

ck+2 =ck

k + 1+ +

ck−1

(k + 2)(k + 1), k = 1, 2, 3, . . . ; c3 =

c12

+c03 · 2 =

c03!

+c12

c4 =c23

+c14 · 3 =

2c03!

+2c14!

, c5 =c34

+c25 · 4 =

c14 · 2 +

c04!

+c05 · 4 =

11c05!

+3c14!

c6 =c45

+c36 · 5 =

2c05 · 3! +

2c15!

+c1

6 · 5 · 2 +c0

6 · 5 · 3! =52c06!

+4c15!

y = c0

�1 + x2 +

1

3!x3 +

2

3!x4 +

11

5!x5 + · · ·

�+ c1

�x+

1

2x3 +

2

4!x4 +

3

4!x5 + · · ·

�


19. (x− 1)∞�

n=2

n(n− 1)cnxn−2 − x

∞�

n=1

ncnxn−1 +

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn−1

� �� k=n−1

−∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−∞�

n=1

ncnxn +

∞�

n=0

cnxn

=∞�

k=1

(k + 1)kck+1xk −

∞�

k=0

(k + 2)(k + 1)ck+2xk −

∞�

k=1

kckxk +

∞�

k=0

ckxk

= c0 − 2c2 +∞�

k=1

[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 − kck + ck]xk = 0

c0 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 − (k − 1)ck = 0; c2 =1

2c0

ck+2 =(k + 1)kck+1 − (k − 1)ck

(k + 2)(k + 1), k = 1, 2, 3, . . . ; c3 =

2c23 · 2 =

c03 · 2

c4 =3 · 2c3 − c2

4 · 3 =c0 − c0/2

4 · 3 =c0

4 · 3 · 2y = c0

�1 +

1

2x2 +

1

3 · 2x3 +

1

4 · 3 · 2x4 + · · ·

�+ c1x; y� = c0

�x+

1

2x2 + · · ·

�+ c1

Using the initial conditions, we obtain −2 = y(0) = c0 and 6 = y�(0) = c1. The solution is

y = −2

�1 +

1

2x2 +

1

3 · 2x3 +

1

4 · 3 · 2x4 + · · ·

�+ 6x = −2 + 6x− x2 − 1

3x3 − 1

4 · 3x4 + · · ·

20. (x− 1)∞�

n=2

n(n− 1)cnxn−2 − 2x

∞�

n=1

ncnxn−1 + 8

∞�

n=0

cnxn

=∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

−2∞�

n=1

ncnxn + 8

∞�

n=0

cnxn

=∞�

k=0

(k + 2)(k + 1)ck+2xk − 2

∞�

k=1

kckxk + 8

∞�

k=0

ckxk

= 2c2 + 8c0 +∞�

k=1

[(k + 2)(k + 1)ck+2 − 2kck + 8ck]xk = 0

2c2 + 8c0 = 0; (k + 2)(k + 1)ck+2 − 2(k − 4)ck = 0; c2 = −4c0

ck+2 =2(k − 4)ck

(k + 2)(k + 1), k = 1, 2, 3, . . . ; c3 =

−2 · 3c13 · 2 = −c1, c4 =

−2 · 2c24 · 3 =

4

3c0

c5 =−2 · 1c35 · 4 =

1

5 · 2c1, c6 =2 · 0c46 · 5 = 0, c8 = c10 = c12 = · · · = 0

y = c0

�1− 4x2 +

4

3x4

�+ c1

�x− x3 +

1

10x5 + · · ·

�

y� = c0

�−8x+

16

3x3

�+ c1

�1− 3x2 +

1

2x4 + · · ·

�

CHAPTER 16 IN REVIEW 1055

Using the initial conditions, we obtain 3 = y(0) = c0 and 0 = y�(0) = c1. The solution is

y = 3

�1− 4x2 +

4

3x4

�= 3− 12x2 + 4x4

Chapter 16 in Review

A. True/False

1. True

2. True. We know a general solution is y = Aex +Be−x. Now

C1 coshx+ C2 sinhx = C1

�ex + e−x

2

�+ C2

�ex − e−x

2

�

=

�C1

2+

C2

2

�ex +

�C1

2− C2

2

�e−x.

By varying C1 and C2, we see that the two equations are different forms of the same generalsolution.

3. False. y2 is a constant multiple of y1. Specifically, y2 = 0 · y1.

4. False. Plugging yp = A into the DE gives 0 = 10, a contradiction.

5. True. Any constant function solves the DE.

6. False. Py = 2x while Qx = −2x.

7. True

8. True

B. Fill in the Blanks

1. By inspection, the constant function y = 0 solves the DE.

2. The auxiliary equation is m2 − m = 0, so m = 0 or m = 1. The general solution is y =C1 + C2ex. Boundary conditions yield y(0) = C1 + C2 = 1 and y(1) = C1 + C2e = 0, which

give C1 =e

e− 1and C2 =

−1

e− 1. Therefore, y =

e

e− 1−�

1

e− 1

�ex.

3. 10 = k(2.5) =⇒ k = 4 lb/ft;32 = 4x =⇒ x = 8 ft

4. We have a repeated root m = −7. Therefore, y = C1e−7x + C2xe−7x.

5. yp = Ax2 +Bx+ C +Dxe2x + Ee2x


C. Exercises

1. Py = −6xy2 sin y3 = Qx, and the equation is exact.fx = 2x cos y3, f = x2 cos y3 + g(y), fy = −3x2y2 sin y3 + g�(y) = −1− 3x2y2 sin y3,g�(y) = −1, g(y) = −y, f = x2 cos3 y − y.Therefore, the solution is x2 cos y3 − y = C.

2. Py = 6y2 = Qx, and the equation is exact. fx = 3x2 + 2y3, f = x3 + 2xy3 + g(y), fy =6xy2 + g�(y) = 6xy2 + y2,

g�(y) = y2, g(y) =y3

3, f = x3 + 2xy3 +

y3

3.

Therefore, the solution is x3 + 2xy3 +y3

3= C.

3. Py = −2xy−5 = Qx, and the equation is exact.fx = 1

2xy−4, f = 1

4x2y−4 + g(y), fy = −x2y−5 + g�(y) = 3y−3 − x2y−5

g�(y) = 3y−3, g(y) = − 32y

−2, f = 14x

2y−4 − 32y

−2.Therefore, the general solution is 1

4x2y−4− 3

2y−2 = C. Since y(1) = 1, we have 1

4 (1)(1)−32 (1) =

C or C = − 54 . Thus, the solution is 1

4x2y−4 − 3

2y−2 = − 5

4 .

4. Py = 2x+ sinx = Qx and the equation is exact.fx = y2 + y sinx, f = xy2 − y cosx+ g(y), fy = 2xy − cosx+ g�(y) = 2xy − cosx− 1

1+y2 ,

g�(y) =1

1 + y2, g(y) = tan−1(y), f = xy2 − y cosx + tan−1(y). Therefore, the general

solution is xy2 − y cosx + tan−1(y) = C. Since y(0) = 1, we have −1 + π4 = C. Thus, the

solution is xy2 − y cosx+ tan−1(y) =π

4− 1

5. m2 − 2m− 2 = 0 =⇒ m = 1±√3; y = C1e(1−

√3)x + C2e(1+

√3)x

6. m2 − 8 = 0 =⇒ m = ±2√2; y = C1e−2

√2x + C2e2

√2x

7. m2 − 3m− 10 = 0 =⇒ (m− 5)(m+ 2) = 0 =⇒ m = −2, 5; y = C1e−2x + C2e5x

8. 4m2 + 20m+ 25 = 0 =⇒ (2m+ 5)2 = 0 =⇒ m = −5/2, −5/2; y = C1e−5x/2 + C2xe−5x/2

9. 9m2 + 1 = 0 =⇒ m = ±1

3i; y = C1 cos

x

3+ C2 sin

x

3

10. 2m2 − 5m = 0 =⇒ m(2m− 5) = 0 =⇒ m = 0, 5/2; y = C1 + C2e5x/2

11. Letting y = ux we have

(x+ uxeu)dx− xeu(udx+ xdu) = 0 =⇒ dx−xeudu = 0 =⇒ dx

x− eudu = 0

=⇒ ln |x|− eu = C1 =⇒ ln |x|− ey/x = C1.

Using y(1) = 0 we find C1 = −1. The solution of the initial-value problem is ln |x| = ey/x − 1.

12. The auxiliary equation is m = m2 + 4m + 4 = 0, so m = −2 is a repeated root. Thegeneral solution is y = C1e−2x + C2xe−2x. Initial conditions yield y(0) = C1 = −2 andy�(0) = −2C1+C2 = 0 which give C1 = −2 and C2 = −4. The solution is y = −2e−2x−4xe−2x.


13. m2 −m− 12 = 0 =⇒ (m− 4)(m+ 3) = 0 =⇒ m = −3, 4; yc = C1e−3x + C2e4x

yp = Axe2x +Be2x, y�p = 2Axe2x + (A+ 2B)e2x; y��p = 4Axe2x + 4(A+B)e2x

[4Axe2x+4(A+B)e2x]− [2Axe2x + (A+ 2B)e2x]− 12[Axe2x +Be2x]

= −10Axe2x + (3A− 10B)e2x = xe2x + ex

Solving −10A = 1, 3A− 10B = 1 we obtain A = −1/10 and B = −13/100. Thus,

y = C1e−3x + C2e

4x − 1

10xe2x − 13

100e2x.

14. The auxiliary equation is m2 + 4 = 0, so m = ±2i. Therefore, yc = C1 cos 2x + C2 sin 2x.Assume a particular solution of the form yp = Ax2 + Bx + C. Substituting into the DE, wehave

2A+Ax2 +Bx+ C = 16x2.

Equating coefficients, we get 2A+C = 0, B = 0, and A = 16. This gives C = −32. Therefore,yp = 16x2 − 32. The general solution is y = yc + yp = C1 cos 2x+ C2 sin 2x+ 16x2 − 32.

15. m2 − 2m+ 2 = 0 =⇒ m = 1± i; yc = ex(C1 cosx+ C2 sinx)

W =

��ex cosx ex sinx

−ex sinx+ ex cosx ex cosx+ ex sinx

�� = e2x

u� = − 1

e2xex sinx ex tanx = − sin2 x

cosx=

cos2 x− 1

cosx= cosx−secx, u = sinx−ln | secx+tanx|

v� =1

e2xex cosx ex tanx = sinx, v = − cosx

yp = ex cosx(sinx− ln | secx+ tanx|)− ex sinx cosx = −ex cosx ln | secx+ tanx|y = ex(C1 cosx+ C2 sinx)− ex cosx ln | secx+ tanx|

16. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1e−x + C2ex; W =

��e−x ex

−e−x ex

�� = 2

u� = −1

2ex

2ex

ex + e−x= − e2x

ex + e−x= − e3x

e2x + 1

u = −�

e3x

e2x + 1dx t = ex, dt = exdx

= −�

t2

t2 + 1dt = −

� �1− 1

t2 + 1

�dt = tan−1 t− t = tan−1 ex − ex

v� =1

2e−x 2ex

ex + e−x=

1

ex + e−x=

ex

e2x + 1

v =

�ex

e2x + 1dx t = ex, dt = exdx

=

�dt

t2 + 1= tan−1 t = tan−1 ex

yp = e−x(tan−1 ex − ex) + ex tan−1 ex = (ex + e−x) tan−1 ex − 1y = C1e−x + C2ex + (ex + e−x) tan−1 ex − 1

17. m2 + 1 = 0 =⇒ m = ±i; yc = C1 cosx+ C2 sinx; W =


�� = 1

u� = − sinx sec3 x = − tanx sec2 x, u = −1

2sec2 x; v� = cosx sec3 x = sec2 x, v = tanx


yp = −1

2cosx sec2 x+ sinx tanx = sinx tanx− 1

2secx =

sin2 x

cosx− 1

2 cosx

=2 sin2 x− 1

2 cosx=

sin2 x− cos2 x

2 cosx=

1

2sinx tanx− 1

2cosx

y = C3 cosx+ C2 sinx+1

2sinx tanx, y� = −C3 sinx+ C2 cosx+

1

2sinx sec2 x+ sinx

Using the initial conditions, we obtain C3 = 1 and C2 = 1/2. Thus,

y = cosx+1

2sinx+

1

2sinx tanx =

2 cos2 x

2 cosx+

sin2 x

2 cosx+

1

2sinx

=cos2 x+ 1

2 cosx+

1

2sinx =

1

2(sinx+ cosx+ secx).

18. The auxiliary equation ism2+2m+2 = 0, som = −1±i. Therefore, yc = e−x (C1 cosx+ C2 sinx) .Assume a particular solution of the form yp = A. Substituting this into the DE, we have2A = 1, or A = 1

2 . Therefore, the general solution is y = yc+yp = e−x (C1 cosx+ C2 sinx)+12 .

The initial conditions yield y(0) = C1+12 = 0 and y�(0) = −C1+C2 = 1 which give C1 = − 1

2and C2 = 1

2 . Thus, the solution is y = e−x�− 1

2 cosx+ 12 sinx

�+ 1

2 .

19.∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+x∞�

n=0

cnxn

� �� k=n+1

=∞�

k=0

(k + 2)(k + 1)ck+2xk +

∞�

k=1

ck−1xk

= 2c2 +∞�

k=1

[(k + 2)(k + 1)ck+2 + ck−1]xk = 0

c2 = 0; (k + 2)(k + 1)ck+2 + ck−1 = 0; ck+2 = − ck−1

(k + 2)(k + 1), k = 1, 2, 3, . . .

c3 = − c03 · 2 , c4 = − c1

4 · 3 , c5 = 0, c6 = − c36 · 5 =

c06 · 5 · 3 · 2 ,

c7 = − c47 · 6 =

c17 · 6 · 4 · 3

c8 = 0. c9 = − c69 · 8 = − c0

9 · 8 · 6 · 5 · 3 · 2 , c10 = − c710 · 9 = − c1

10 · 9 · 7 · 6 · 4 · 3y = c0

�1− 1

3 · 2x3 +

1

6 · 5 · 3 · 2x6 − 1

9 · 8 · 6 · 5 · 3 · 2x9 + · · ·

�

+ c1

�x− 1

4 · 3x4 +

1

7 · 6 · 4 · 3x7 − 1

10 · 9 · 7 · 6 · 4 · 3x10 + · · ·

�


20. (x− 1)∞�

n=2

n(n− 1)cnxn−2 + 3

∞�

n=1

cnxn

=∞�

n=2

(n)(n− 1)cnxn−1

� �� k=n−1

−∞�

n=2

n(n− 1)cnxn−2

� �� k=n−2

+3∞�

n=0

cnxn

=∞�

k=1

(k + 1)kck+1xk −

∞�

k=0

(k + 2)(k + 1)ck+2xk + 3

∞�

k=0

ckxk

= 3c0 − 2c2 +∞�

k=1

[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck]xk = 0

3c0 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck = 0; c2 =3c02

;

ck+2 =kck+1

k + 2+

3ck(k + 2)(k + 1)

, k = 1, 2, 3, . . . ; c3 =c23+

3c13 · 2 =

c02+c12, c4 =

2c34

+3c24 · 3 =

c04

+c14

+3c08

=5c08

+c14,

c5 =3c45

+3c35 · 4 =

3c08

+3c120

+3c040

+3c140

=9c020

+9c140

y = c0

�1 +

3

2x3 +

1

2x3 +

5

8x4 + · · ·

�+ c1

�x+

1

2x3 +

1

4x4 + · · ·

�

21. The differential equation is mx�� + 4x� + 2x = 0. The solutions of the auxiliary equation are

1

2m(−4±

√16− 8m) =

1

m(−2±

√4− 2m).

The motion will be non-oscillatory when 4− 2m ≥ 0 or 0 < m ≤ 2.

22. Substituting xp = αA into the differential equation we obtain ω2αA = A, so α = 1/ω2 andxp = A/ω2.

23. Using m = W/g = 4/32 = 1/8, the inital value problem is1

8x�� + x� + 3x = e−t; x(0) = 2,

x�(0) = 0. The auxiliary equation is m2/8 + m + 3 = 0. Using the quadratic formula, m =−4± 2

√2i. Thus, xc = e−4t(C1 cos 2

√2t+C2 sin 2

√2t). Using xp = Ae−t, we find A = 8/17.

Thus,

x(t) = e−4t(C1 cos 2√2t+ C2 sin 2

√2t) +

8

17e−t

and x�(t) = e−4t[(2√2C2 − 4C1) cos 2

√2t− (2

√2C1 − 4C2) sin 2

√2t]− 8

17e−t.

Using the initial conditions, we obtain 2 = C1 + 8/17 and 0 = 2√2C2 − 4C1 − 8/17. Then

C1 = 26/17 and C2 = 28√2/17 and

x(t) = e−4t

�26

17cos 2

√2t+

28

17

√2 sin 2

√2t

�+

8

17e−t.


24. (a) From k1 = 2W and k2 = 4W we find 1/k = 1/2W + 1/4W = 3/4W. Then k = 4W/3 =4mg/3. The differential equation mx�� + kx = 0 then becomes x�� + (4g/3)x = 0. Thesolution is x(t) = C1 cos 2

�g/3t + C2 sin 2

�g/3t. The initial conditions x(0) = 1 and

x�(0) = 2/3 imply C1 = 1 and C2 = 1/√3g.

(b) To find the maximum speed of the weight we compute

x�(t) = 2

�g

3sin 2

�g

3+

2

3cos 2

�g

3t and |x�(t)| =

�4g

3+

4

9=

2

3

�3g + 1.

25. The auxiliary equation is m2/4 + m + 1 = 0 or (m + 2)2 = 0, so m = −2, −2 and x(t) =C1e−2t +C2te−2t and x�(t) = −2C1e−2t − 2C2te−2t +C2e−2t. Using the initial conditions, weobtain 4 = C1 and 2 = −2C1+C2. Thus, C1 = 4 and C2 = 10. Therefore x(t) = 4e−2t+10te−2t

and x�(t) = 2e−2t − 20te−2t. Setting x�(t) = 0 we obtain the critical point t = 1/10. Themaximum vertical displacement is x(1/10) = 5e−0.2 ≈ 4.0937.

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Zill Calculo 4e Manual de Solucionario c16