# CSM_Chapters16.pdf

September 14, 2017 | Author: Clau Amaiia | Category: Zero Of A Function, Initial Condition, Equations, Analysis, Calculus

#### Description

Chapter 16

Higher-Order Differential Equations 16.1

Exact First-Order Equations

1. Since Py = 0 = Qx , the equation is exact. 3 fx = 2x + 4, f = x2 + 4x + g(y), fy = g 0 (y) = 3y − 1, g(y) = y 2 − y 2 3 The solution is x2 + 4x + y 2 − y = C. 2 2. Since Py = 1 and Qx = −1, the equation is not exact. 3. Since Py = 4 = Qx , the equation is exact. 5 fx = 5x + 4y, f = x2 + 4xy + g(y), fy = 4x + g 0 (y) = 4x − 8y 3 , g(y) = −2y 4 2 5 2 The solution is x + 4xy − 2y 4 = C. 2 4. Since Py = cos y − sin x = Qx , the equation is exact. fx = sin y−y sin x, f = x sin y+y cos x+g(y), fy = x cos y+cos x+g 0 (y) = cos x+x cos y−y, 1 g(y) = − y 2 2 1 The solution is x sin y + y cos x − y 2 = C. 2 5. Since Py = 4xy = Qx , the equation is exact. fx = 2y 2 x − 3, f = y 2 x2 − 3x + g(y), fy = 2yx2 + g 0 (y) = 2yx2 + 4, g(y) = 4y The solution is y 2 x2 − 3x + 4y = C.   y  1 1 3 6. − 4x + 3y sin 3x dx + 2y − + cos 3x dy = 0. Since Py = 2 + 3 sin 3x and Qx = 2 x x x 1 − 3 sin 3x, the equation is not exact. x2 7. (x2 − y 2 )dx + (x2 − 2xy)dy = 0. Since Py = −2y and Qx = 2x − 2y, the equation is not exact. 288

16.1. EXACT FIRST-ORDER EQUATIONS 8.

289

y 1 dx + (ln x − 1)dy = 0. Since Py = = Qx , the equation is exact. fy = x x y y 0 ln x − 1, f = y ln x − y + g(x), fx = + g (x) = 1 + ln x + , g 0 (x) = 1 + ln x, g(x) = x ln x x x The solution is y ln x − y + x ln x = C. 

1 + ln x +

9. (y 3 − y 2 sin x − x)dx + (3xy 2 + 2y cos x)dy = 0. Since Py = 3y 2 − 2y sin x = Qx , the equation is exact. fx = y 3 − y 2 sin x − x, f = xy 3 + y 2 cos x − 21 x2 + g(y), fy = 3xy 2 + 2y cos x + g 0 (y) = 3xy 2 + 2y cos, g(y) = 0 The solution is xy 3 + y 2 cos x − 21 x2 = C 10. Since Py = 3y 2 = Qx , the equation is exact. fx = x3 + y 3 , f = 3xy 2 + g 0 (y) = 3xy 2 , g(y) = 0 1 The solution is x4 + xy 3 = C. 4

1 4 x + xy 3 + g(y), fy = 4

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 = 3x2 y + ey , f = x3 y + xey + g(y), fy = x3 + xey + g 0 (y) = x3 + xey − 2y, g(y) = −y 2 The solution is x3 y + xey − y 2 = 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 0 (y) = −x, g(y) = 0 The solution is 2xex − 2ex − yx + 2x3 = C.     3 3 14. 1 − + y dx + 1 − + x dy = 0. Since Py = 1 = Qx , the equation is exact. x y 3 3 3 fx = 1 − + y, f = x − 3 ln |x| + xy + g(y), fy = x + g 0 (y) = 1 − + x, g 0 (y) = 1 − , x y y g(y) = y − 3 ln |y| The solution is x − 3 ln |xy| + xy + y = C. 15. Since Py = 3x2 y 2 = Qx , the equation is exact. 1 1 1 fy = x3 y 2 , f = x3 y 3 + g(x), fx = x2 y 3 + g 0 (x) = x2 y 3 − , g 0 (x) = − = 2 3 1 = 9x 1 + 9x2 1 1 1 x 1 1 , g(x) = − tan−1 = − tan1 3x − 9 1/9 + x2 9 1/3 1/3 3 1 1 The solution is x3 y 3 − tan−1 3x = C or x3 y 3 = tan −13x = C1 . 3 3 16. 2ydx − (5y − 2x)dy = 0. Since Py = 2 = Qx , the equation is exact. 5 fx = 2y, f = 2xy + g(y), fy = 2x + g 0 (y) = −5y + 2x, g(y)= − y 2 2 5 The solution is 2xy − y 2 = C. 2 17. Since Py = sin x cos y = Qx , the equation is exact. fy = cos x cos y, f = cos x sin y +g(x), fx = − sin x sin y +g 0 (x) = tan x−sin x sin y, g 0 (x) = tan x, g(x) = ln | sec x| The solution is cos x sin y + ln | sec x| = C or cos x sin y − ln | cos x| + C.

290

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS 2

2

18. (2y sin x cos x − y + 2y 2 exy )dx + (sin2 x + 4xyexy − x)dy = 0. Since Py = 2 sin x cos x − 2 2 2 1 + 4xy 3 exy + 4yexy = Qz , the equation is exact. fx = 2y sin x cos x − y + 2y 2 exy = 1 2 2 y sin 2x − y + 2y 2 exy , f = − y cos 2x − xy + 2exy + g(y), 2 2 2 1 1 fy = − cos 2x − x + 4xyexy + g 0 (y) = − (1 − 2 sin2 x) − x + 4xyexy + g 0 (y) 2 2 2 2 1 = − + sin2 x − x + 4xyexy + g 0 (y) = sin2 x + 4xyexy − x 2 1 1 g 0 (y) = , g(y) = y 2 2 1 1 2 The solution is − y cos 2x − xy + 2exy + y = C. 2 2 19. Since Py = 4t3 − 1 = Qt , the equation is exact. ft = 4t3 y − 15t2 − y, f = t4 y − 5t3 − yt + g(y), fy = t4 − t + g 0 (y) = t4 + 3y 2 − t, g 0 (y) − 3y 2 , g(y) = y 3 . The solution is t4 y − 5t3 − yt + y 3 = C. y 2 − t2 (t2 + y 2 ) + y(2y) = (t2 + y 2 )2 (t2 + y 2 )2 −2t , the equation is not exact. and Qy = 2 (t + y 2 )2

20. Since Py = −

21. Since Py = 2(x + y) = Qx , the equation is exact. 1 fx = (x+y)2 = x2 +2xy+y 2 , f = x3 +x2 y+xy 2 +g(y), fy = x2 +2xy+g 0 (y) = 2xy+x2 −1 3 1 g 0 (y) = −1, g(y) = −y A family of solutions is x3 + x2 y + xy 2 − y = C. Substituting x = 1 3 1 4 and y = 1 we obtain + 1 + 1 − 1 = = C. The solution subject to the given condition is 3 3 1 3 4 2 2 x + x y + xy − y = . 3 3 22. Since Py = 1 = Qx , the equation is exact. fx = ex + y, f = ex + xy + g(y), fy = x + g 0 (y) = 2 + x + yey , g 0 (y) = 2 + yey Using integration 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 the given 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 0 (y) = 6y + 4t − 1, g 0 (y) = 6y − 1, g(y) = 3y 2 − y A family of solutions is 4ty + t2 − 5t + 3y 2 − y = C. Substituting t = −1 and y = 2 we obtain −8 + 1 + 5 + 12 − 2 = 8 = C. The solution subject to the given condition is 4ty + t2 − 5t + 3y 2 − y = 8. 24. Since Py = 2y cos x − 3x2 = Qx , the equation is exact. fx = y 2 cos x − 3x2 y − 2x, f = y 2 sin x − x3 y − x2 + g(y), fy = 2y sin x − x3 + g 0 (y) = 2y sin x − x3 + ln y, g 0 (y) = ln y, g(y) = y ln y − y A family of solutions is y 2 sin x − x3 y − x2 + y ln y − y = C. Substituting x = 0 and y = e we obtain e − e = 0 = C. The solution subject to the given condition is y 2 sin x − x3 y − x2 + y ln y − y = 0.

16.2. HOMOGENEOUS LINEAR EQUATIONS

291

25. We want Py = Qx or 3y 2 + 4kxy 3 = 3y 2 + 40xy 3 . Thus, 4k = 40 and k = 10. 26. We want Py = Qx or 18xy 2 − sin y = 4kxy 2 − sin y. Thus 4k = 18 and k = 92 . 27. We need Py = Qx , so we must have 1 xy xe

+

∂M = exy + xyexy + 2y − ∂y

1 x2 .

This gives M (x, y) =

(yx − 1)exy y + y 2 − 2 + g(x) for some function g. x x

28. We need Py = Qx , so we must have x1/2 y −1/2 +

∂N x . This gives N (x, y) = = 12 x−1/2 y −1/2 − 2 ∂x (x + y 2 )2

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

∂ ∂  4 [µ(x, y)M (x, y)] = xy = 4xy 3 ∂y ∂y  ∂ ∂  2 3 [µ(x, y)N (x, y)] = 2x y + 3y 5 − 20y 3 = 4xy 3 ∂z ∂x 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 y 3 xydx + (2x2 + 3y 2 − 20)dy = 0, then xydx + (2x2 + 3y 2 − 20)dy = 0, provided y 6= 0. Therefore, to solve the original DE, we solve xy 4 dx + 2x2 y 3 + 3y 5 − 20y 3 dy = 0. fx = xy 4 , f = 12 x2 y 4 + g(y), fy = 2x2 y + g 0 (y) = 2x2 y 3 + 3y 5 − 20y 3 , g 0 (y) = 3y 5 − 20y , g(y) = 21 y 6 − 5y 4 , f = 21 x2 y 4 + 12 y 6 − 5y 4 . The solution is therefore 21 x2 y 4 + 12 y 6 − 5y 4 = C.

29. Let µ(x, y = y 3 . Then

1 dy − g(x)dx = 0. Since g is a function of x h(y) only and h is a function of y only, we have Py = Qx = 0.

30. True; a separable equation can be written as

16.2

Homogeneous Linear Equations

1. 3m2 − m = 0 =⇒ m(3m − 1) = 0 =⇒ m = 0, 1/3; y + C1 + C2 ex/3 2. 2m2 + 5m = 0 =⇒ m(2m + 5) = 0 =⇒ m = 0, −5/2; y = C1 + C2 e−5x/2 3. m2 − 16 = 0 =⇒ m2 = 16 =⇒ m = −4, 4; y = C1 e−4x + C2 e4x √ √ √ √ 4. m2 − 8 = 0 =⇒ m2 = 8 =⇒ m = −2 2, 2 2; y = C1 e−2 2x + C2 e2 2x 5. m2 + 9 = 0 =⇒ m2 = −9 =⇒ m = −3i, 3i; y = C1 cos 3x + C2 sin 3x 1 1 6. 4m2 + 1 = 0 =⇒ m2 = −1/4 =⇒ m = −i/2, i/2; y = C1 cos x + C2 sin x 2 2 7. m2 − 3m + 2 = 0 =⇒ (m − 1)(m − 2) = 0 =⇒ m = 1, 2; y = C1 ex + C2 e2x 8. m2 − m − 6 = 0 =⇒ (m + 2)(m − 3) = 0 =⇒ m = −2, 3; y = C1 e−2x + C2 33x 9. m2 + 8m + 16 = 0 =⇒ (m + 4)2 = 0 =⇒ m = −4, −4; y = C1 e−4x + C2 xe−4x 10. m2 − 10m + 25 = 0 =⇒ (m − 5)2 = 0 =⇒ m = 5, 5; y = C1 e5x + C2 xe5x

292

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

11. m2 + 3m − 5 = 0 =⇒ m = −3/2 ± 12. m2 + 4m − 1 = 0 =⇒ m = −2 ±

29/2; y = C1 e(−3/2−

5; y = C1 e(−2−

5)x

29/2)x

+ C2 e(−3/2+ √

+ C2 e(−2+

29/2)x

5)x

13. 12m2 − 5m − 2 = 0 =⇒ (3m − 2)(4m + 1) = 0 =⇒ m = −1/4, 2/3; y = C1 e−x/4 + C2 e2x/3 14. 8m2 + 2m − 1 = 0 =⇒ (4m − 1)(2m + 1) = 0 =⇒ m = −1/2, 1/4; y = C1 e−x/2 + C2 ex/4 15. m2 − 4m + 5 = 0 =⇒ m = 2 ± i; y = e2x (C1 cos x + C2 sin x) √ 16. 2m − 3m + 4 = 0 =⇒ m = 3/4 ± ( 23/4)i; y = e3x/4 2

! √ 23 23 C1 cos x + C2 sin x 4 4 √

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

2

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

−x/2



√ ! 2 2 C1 cos x + C2 sin x 3 3 1 1 C1 cos x + C2 sin x 2 2



19. 9m2 + 6m + 1 = 0 =⇒ (3m + 1)2 = 0 =⇒ m = −1/3, −1/3; y = C1 e−x/3 + C2 xe−x/3 20. 15m2 − 16m − 7 = 0 =⇒ (3m + 1)(5m − 7) = 0 =⇒ m = −1/3, 7/5; y = C1 e−x/3 + C2 e7x/5 21. m2 + 16 = 0 =⇒ m2 = −16 =⇒ m = ±4i; y = C1 cos 4x + C2 sin 4x; y 0 = −4C1 sin 4x + C2 cos 4x Using y(0) = 2 we obtain 2 = C1 . Using y 0 (0) = −2 we obtain −2 = 4C2 or C2 = −1/2. The 1 solution is y = 2 cos 4x − sin 4x. 2 22. m2 − 1 = 0 =⇒ m2 = 1 =⇒ m = ±1; y = C1 ex + C2 e−x ; y 0 = C1 ex − C2 e−x . Using y(0) = Y 0 (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 = C1 e−5x + C2 e−x ; y 0 = −5C1 e−5x − C2 e−x . Using y(0) = 0 and y 0 (0) = 3 we obtain the system C1 + C2 = 0, 3 3 − 5C1 − C2 = 3. Thus, C1 = −3/4 and c2 = 3/4. The solution is y = − e−5x + e−x . 4 4 24. m2 − 8m + 17 = 0 =⇒ m = 4 ± i; y = e4x (C1 cos x + C2 sin x); y 0 = e4x [(4C1 + C2 ) cos x + (−C1 + 4C2 ) sin x] . Using y(0) = 4 and y 0 (0) = −1 we obtain the system C1 = 4, 4C1 + C2 = −1. Thus, C1 = 4 and C2 = −17. The solution is y = e4x (4 cos x − 17 sin x). 1 1 1 25. 2m2 − 2m + 1 = 0 =⇒ m = 1/2 ± (1/2)i; y = ex/2 (C1 cos x + C2 sin x); y 0 = ex/2 [ (C1 + 2 2 2 1 1 1 C2 ) cos x− (C1 −C2 ) sin x]. Using y(0) = −1 and y 0 (0) = 0 we obtain the system C1 = −1, 2 2 2   1 1 1 1 x/2 C1 + C2 = 0. Thus, C1 = −1 and C2 = 1. The solution is y = e sin x − cos x . 2 2 2 2

16.2. HOMOGENEOUS LINEAR EQUATIONS

293

26. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; y = C1 ex + C2 xex ; y 0 = (C1 + C2 )ex + C2 xex . Using y(0) = 5 and y 0 (0) = 10 we obtain the system C1 = 5, C1 + C2 = 10. Thus, C1 = C2 = 5. The solution is y = 5ex + 5xex . √ √ √ 7 7 −x/2 2 (C1 cos x + C2 sin x); 27. m + m + 2 = 0 =⇒ m = −1/2 ± ( 7/2)i; y = e 2 2 " # √ √ √ √ 1 7 7 7 7 1 y 0 = e−x/2 (− C1 + C2 ) cos x + (− C1 − C2 ) sin x . 2 2 2 2 2 2 √ 1 7 C2 = 0. Thus, C1 = C2 = 0. Using y(0) = y 0 (0) = 0 we obtain the system C1 = 0, − C1 + 2 2 The solution is y = 0. 28. 4m2 − 4 − 3 = 0 =⇒ (2m − 3)(2m + 1) = 0 =⇒ m = −1/2, 3/2; y = C1 e−x/2 + C2 e3x/2 ; 1 3 y 0 = − C1 e−x/2 + C2 e3x/2 . Using y(0) = 1 and y 0 (0) = 5 we obtain the system C1 +C2 = 1, 2 2 3 7 11 1 − C1 + C2 = 5. Thus, c1 = −7/4 and C2 = 11/4. The solution is y = − e−x/2 + e3x/2 . 2 2 4 4 29. m2 −3m+2 = 0 =⇒ (m−1)(m−2) = 0 =⇒ m = 1, 2; y = C1 ex +C2 e2x ; y 0 = C1 ex +2C2 e2x . Using y(1) = 0 and y 0 (1) = 1 we obtain the system eC1 + e2 C2 = 0, eC1 + 2e2 C2 = 1. Thus, C1 = −e−1 and C2 = e−2. The solution is y = −ex−1 + e2x−2 . 0 30. m2 +1 = 0 =⇒ m2 = −1 =⇒ m = ±i; y = C1 cos x+C2 sin x; x+C2 cos x. Using √ y = C1 sin√ 1 3 3 1 y(π/3) = 0 and y 0 (π/3) = 2 we obtain the system C1 + C2 = 0, − C1 + C2 = 2. 2 2 2 2 √ √ Thus, C1 = − 3 and C2 = 1. The solution is y = − 3 cos x + sin x.

31. The auxiliary equation is (m − 4)(m + 5) = m2 + m − 20 = 0. The differential equation is y 00 + y 0 − 20y = 0. 32. The auxiliary equation is [(m − 3) − i] [(m − 3) + i] = (m − 3)2 − i2 = m2 − 6m + 10 = 0. The differential equation is y 00 = 6y 0 + 10y = 0. 33. The auxiliary equation is m2 +1 = 0, so m = ±i. The general solution is y = C1 cos x+C2 sin x. The boundary conditions yield y(0) = C1 = 0, y(π) = −C1 = 0, so y = C2 sin x. 34. The general solution is y = C1 cos x + C2 sin x. 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 cos x + C2 sin x. The boundary conditions yield y 0 (0) = C2 =  0 1 0, y 2 = −C1 = 2, so y = −2 cos x. 36. The auxiliary equation is m2 − 1 = 0, so m = ±1. The general solution is y = C1 ex + C2 e−x . The boundary conditions yield y(0)  = C1 + C2  = 1,  y(1) = C1e + C2 e−1 = −1, or C1 = −1 −1 −1 − e e+1 −1 − e e+1 and C2 = , so y = ex + e−x. e − e−1 e − e−1 e − e−1 e − e−1 37. The auxiliary equation is m2 − 2m + 2 = 0, so m = 1 ± i. The general solution is y = ex (C1 cos x + C2 sin x) . The boundary conditions yield y(0) = C1 = 1 and y(π) = −eπ C1 = −1, which is a contradiction. No solution.

294

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

38. The general solution is y = ex (C1 cos x + C2 sin x) . The boundary conditions yield y(0) =  C1 = 1 and y (π/2) = C2 eπ/2 = 1, so y = ex cos x + e−π/2 sin x 39. The auxiliary equation is m2 − 4m + 4 = 0, so m = 2 is a repeated root. The general solution is y = C1 e2x + C2 xe2x . The boundary conditions yield y(0) = C1 = 0 and y(1) = C2 e2 = 1, so y = xe−2 e2x = xe2(x−1) . 40. The general solution is y = C1 e2x +C2 xe2x . The boundary conditions yield y 0 (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 1 and 4±i. The general solution of the differential equation is y = C1 ex +e4x (C2 cos x+C3 sin x). 42. Assuming a solution of the form y = emx we obtain the auxiliary equation m3 +6m2 +m−34 = 0. Since y1 = e−4x cos x is a solution, we know that m1 = −4 + i is a root of the auxiliary equation. Using the fact that complex roots of real polynomial equations occur in conjugate pairs we have that m2 = −4−i is also a root. Thus [m−(−4+i)][m−(−4−i)] = m2 +8m+17 is a factor of the auxiliary equation and we can write it as m3 +6m2 +m−34 = (m2 +8m+17)(m− 2) = 0. The general solution of the differential equation is y = C1 e2x +e−4x (C1 cos x+C2 sin x). 43. y 0 = memx , y 00 = m2 emx , y 000 = m3 emx ; m3 emx − 4m2 emx − 5memx = 0 =⇒ (m3 − 4m2 − 5m)emx = 0 =⇒ m3 − 4m2 − 5m = 0 =⇒ m(m − 5)(m + 1) = 0 =⇒ m = 0, −1, 5; y = C1 + C2 e−x + C3 e5x 44. y 0 = memx , y 00 = m2 emx , y 000 = m3 emx ; m3 emx + 3m2 emx − 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 = C1 e−3x + C2 e−2x + C3 e2x 45. Case 1: λ = −α2 < 0 Auxiliary equation is m2 − α2 = 0, so m = ±α and general solution is y = C1 eαx + C2 e−αx . Boundary conditions yield y(0) = C1 +C2 = 0 and y(1) = C1 eα +C2 e−α = 0, or C1 = C2 = 0. So Case 1 yields no nonzero solutions. Case 2: λ = 0 Auxiliary equation is m2 = 0, so m = 0 is a repeated root and general solution is y = C1 +C2 x. Boundary conditions yield y(0) = C1 = 0 and y(1) = C2 = 0. So Case 2 yields no nonzero solutions. Case 3: λ = α2 > 0 Auxiliary equation is m2 + α2 = 0, so m = ±αi and the general solution is y = C1 cos αx + C2 sin x. Boundary conditions yield y(0) = C1 = 0 and y(1) = C2 sin α = 0. Hence, nonzero solutions exist only when sin α = 0, which implies α = ±nπ so that λ = n2 π 2 for n = 1, 2, 3, . . . (n = 0 is excluded since that would give λ = 0).

16.3. NONHOMOGENEOUS LINEAR EQUATIONS

295

4 4 46. (a) If the earth has density ρ then M = ρ πR3 and Mr = ρ πr3 , so that M/Mr = R3 /r3 3 3 and Mr = r3 M/R3 . Then F = −k

r3 M m/R3 mM Mr m = −k = −k 3 r. r2 r2 R

d2 r mM d2 r kM d2 r = −k r =⇒ + r = 0 =⇒ + ω2 r = dt2 R3 dt2 R3 dt2 0whereω 2 = kM/R3 . Since kmM/R2 = mg we have ω 2 = kM/R3 = g/R.

(b) Since a = d2 r/dt2 ,F = ma = m

(c) The general solution of the differential equation in part (b) is r(t) = c1 cos ωt + c2 sin ωt. The initial conditions r(0) = R and r0 (0) = 0 imply c1 = R and c2 = 0. Then r(t) = R cos ωt. The mass oscillates back and forth from one side of the earth to the other with a 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 = C1 e−3x + C2 e3x ; yp = A, yp0 = yp00 = 0; −9A = 54 =⇒ A = −6; yp = −6; y = C1 e−3x + C2 e3x − 6 2. 2m2 − 7m + 5 = 0 =⇒ (2m − 5)(m − 1) = 0 =⇒ m = 1, 5/2; yc = C1 ex + C2 e5x/2 ; yp = 29 A, yp0 = yp00 = 0; 5A = −29 =⇒ A = −29/5; y = C1 ex + C2 e5x/2 − 5 3. m2 + 4m + 4 = 0 =⇒ (m + 2)2 = 0 =⇒ m = −2, −2; yc = C1 e−2x + C2 xe−2x ; yp = Ax + B, yp0 = A, yp00 = 0; 4A + 4(Ax + B) = 2x + 6 =⇒ 4Ax + 4(A + B) = 2x + 6 Solving 4A = 2, 4A + 4B = 6, we obtain A = 1/2 and B = 1. Thus, y = C1 e−2x + C2 xe−2x + 1 x + 1. 2 4. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; yc = C1 ex + C2 xex ; yp = Ax3 + Bx2 + Cx + d, yp0 = 3Ax2 + 2Bx + C, yp00 = 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 + 4x Solving 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 = C1 ex + C2 xex + x3 + 6x2 + 22x + 32. 5. m2 + 25 = 0 =⇒ m = ±5i; yc = C1 cos 5x + C2 sin 5x; yp = A sin x + B cos x, yp0 = A cos x − B sin x, yp00 = −A sin x − B cos x; −A sin x − B cos x + 25(A sin x + B cos x) = 6 sin x =⇒ 1 24A sin x + 24B cos x = 6 sin x; A = 1/4, B = 0; y = C1 cos 5x + C2 sin 5x + sin x 4 6. m2 − 4 = 0 =⇒ m = −2,

2;

yc = C1 e−2x + C2 e2x ;

16Ae4x − 4Ae4x = 7e4x =⇒ 12Ae4x = 7e4x

yp0 = 4Ae4x , yp00 = 7 =⇒ A = 7/12; y = C1 e−2x + C2 e2x + e4x 12 yp = Ae4x ,

296

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

7. m2 − 2m − 3 = 0 =⇒ (m − 3)(m + 1) = 0 =⇒ m = −1, 3; yc = C1 e−x + C2 e3x yp = Ae2x + Bx3 + Cx2 + Dx + E, yp0 = 2Ae2x + 3Bx2 + 2Cx + D, yp00 = 4Ae2x + 6Bx + 2C (4Ae2x + 6Bx + 2C) − 2(2Ae2x + 3Bx2 + 2Cx + D) − 3(Ae2x + Bx3 + Cx2 + Dx + E) = 4e2x +2x3 =⇒ −3Ae2x 3Bx3 +(−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, we obtain A = −4/3, B = −2/3, C = 4/3, D = −28/9, and E = 80/27. Thus, 2 4 28 80 4 y = C1 e−x + C2 e3x − e2x − x3 + x2 − x + . 3 3 3 9 27 √ √ √ 1 3 8. m2 + m + 1 = 0 =⇒ m = − ± i; yc = e−x/2 (C1 cos 3x/2 + C2 sin 3x/2) 2 2 yp = Ax2 ex + Bxex + Cex + D, yp0 = Ax2 ex + (2A + B)xex + (B + C)ex yp00 = Ax2 ex + (4A + B)xex + (2A + 2B + C)ex  2 x    Ax (4A + B)xex + (2A+ 2B + C)ex + Ax2 ex + (2A + B)xex + (B + C)ex  e2 + + Ax ex + Bxex + Cex + D = x2 ex +3 =⇒ 3Ax2 ex +(6A+3B)xex (2A+3B +3C)ex +D = x2 ex + 3 Solving 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

√ 1 2 4 3x/2 + C2 sin 3x/2) + x2 ex − xex + ex + 3. 3 3 9

9. m2 − 8m + 25 = 0 =⇒ m = 4 ± 3i; yc = e4x (C1 cos 3x + C2 sin 3x); yp = Ae3x + B sin 2x + C cos 2x, yp0 = 3Ae3x + 2B cos 2x − 2C sin 2x, yp00 = 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 2x Solving 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 3x 96 126 e + sin 2x − cos 2x. 10 697 697

10. m2 − 5m + 4 = 0 =⇒ (m − 1)(m − 4) = 0 =⇒ m = 1, 4; yc = C1 ex + C2 e4x yp = A sinh 3x + B cosh 3x, yp0 = 3A cosh 3x + 3B sinh 3x, yp00 = 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 3x Solving 13A − 15B = 2, −15A + 13B = 0, we obtain A = −13/28 and B = −15/28. Thus, y + C1 ex + C2 e4x −

13 15 sinh 3x − cosh 3x. 28 28

11. m2 − 64 = 0 =⇒ m = −8, 8; yc = C1 e−8x + C2 e8x ; yp = A, yp0 = yp00 = 0 1 − 64A = 16 =⇒ A = −1/4; y = C1 e−8x + C2 e8x − , y 0 = −8C1 e−8x + 8C2 e8x . 4 1 Using y(0) = 1 and y 0 (0) = 0 we obtain C1 + C2 − = 1, −8C1 + 8C2 = 0, or C1 = C2 = 5/8. 4 5 5 1 Thus, y = e−8x + e8x − . 8 8 4

16.3. NONHOMOGENEOUS LINEAR EQUATIONS

297

12. m2 + 5m − 6 = 0 =⇒ (m + 6)(m − 1) = 0 =⇒ m = −6, 1; yc = C1 e−6x + C2 ex ; yp = Ae2x , yp0 = 2Ae2x , yp00 = 4Ae2x ; Ae2x + 5(2Ae2x ) − 6(Ae2x ) = 10e2x =⇒ 8Ae2x = 10e2x =⇒ 5 5 A = 5/4; y = C1 e−6x + C2 ex + e2x , y 0 = −6C1 e−6x + C2 ex + e2x . 4 2 5 5 9 Using y(0) = 1 and y 0 (0) = 0 we obtain C1 + C2 + = 1, −6C1 + C2 + = 0, or C1 = 4 2 28 4 9 −6x 4 x 5 2x and C2 = − . Thus, y = e − e + e . 7 28 7 4 cos x sin x =1 13. m2 + 1 = 0 =⇒ m = −i; i; yc = C1 cos x + C2 sin x; W = − sin x cos x u01 = − sin x sec x = − tan x, u1 = ln | cos x|; u02 = cos x sec x = 1, u2 = x yp = cos x ln | cos x| + x sin x; y = C1 cos x + C2 sin x + cos x ln | cos x| + x sin x cos x sin x 2 =1 14. m + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W = − sin x cos x sin2 x 1 − cos2 x u01 = − sin x tan x = − =− = − sec x + cos x, u1 = − ln | sec x + tan x| + sin x cos x cos x u02 = cos x tan x = sin x; u2 = − cos x yp = − cos x ln | sec x + tan x| + sin x cos x − sin x cos x = − cos x ln | sec x + tan x| y = C1 cos x + C2 sin x − cos x ln | sec x + tan x| cos x sin x 2 =1 15. m + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W = − sin x cos x 1 1 1 u01 = − sin2 x, u1 = − x + sin x cos x; u02 = sin x cos x, u2 = sin2 x 2 2 2 1 1 1 yp = − x cos x + sin x cos2 x + sin3 x 2 2 2 1 1 1 y = C1 cos x + C2 sin x − x cos x + sin x cos2 x + sin3 x 2 2 2 1 1 1 2 = C1 cos x + C2 sin x − x cos x + sin x(cos x + sin2 x) = C1 cos x + C3 sin x − x cos x 2 2 2 cos x sin x =1 16. m2 + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W = − sin x cos x 2 0 2 0 u1 = − sin x sec x tan x = − tan x = 1−sec x, u1 = x−tan x; u2 = cos x sec x tan x = tan x u2 = − ln | cos x|; yp = cos x(tan x − x) = x cos x − sin x − sin x ln | cos x| y = C1 cos x + C2 sin x + x cos x − sin x − sin x ln | cos x| = C1 cos x + C3 sin x + x cos x − sin x ln | cos x| cos x sin x 2 =1 17. m + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W = − sin x cos x u01 = − sin x cos2 x u1 = 31 cos3 x; u02 = cos x cos2 x = cos x−cos x sin2 x, u2 = sin x− 13 sin3 x yp = 31 cos4 x + sin2 x − 13 sin4 x = sin2 x + 13 (cos2 x − sin2 x) = sin2 x + 13 cos 2x 1 1 1 1 y = C1 cos x + C2 sin x + sin2 x + cos 2x = C1 cos x + C2 sin x + − cos 2x + cos 2x 3 2 2 3 1 1 = C1 cos x + C2 sin x + − cos 2x 2 6 cos x sin x 2 =1 18. m + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W = − sin x cos x

298

19.

20.

21.

22.

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS u01 = − sin x sec2 x = − tan x sec x, u1 = − sec x; u02 = cos x sec2 x = sec x; u2 = ln | sec x + tan x| yp = − cos x sec x + sin x ln | sec x + tan x| = −1 + sin x ln | sec x + tan x| y = C1 cos x + C2 sin x − 1 + sin x ln | sec x + tan x| −x e ex 2 −x x =2 m − 1 = 0 =⇒ m = −1, 1; yc = C1 e + C2 e ; W = −e−x ex 1 1 1 1 u01 = ex cosh x = − (e2x + 1), u1 = − e2x − x; 2 4 8 4 1 1 1 1 u02 = e−x cosh x = (1 + e−2x ) u2 = x − e−2x , 2 4 4 8 1 1 1 1 1 1 1 1 yp = e−x (− e2x − x) + ex ( x − e−2x = − ex − xe−x + xex − d−x 8 4 4 8 8 4 4 8 1 x 1 −x 1 = − e − e + x sinh x 8 8 2 1 1 1 1 y = C1 e−x + C2 ex − ex − e−x + x sinh x = C3 e−x + C4 ex + x sinh x 8 8 2 2 −x e ex 2 −x x =2 m − 1 = 0 =⇒ m = −1, 1; yc = C1 e + C2 e ; W = −e−x ex 1 1 1 1 u01 = − ex sinh 2x = − (e3x − e−x ), u1 = − e3x − e−x 2 4 12 4 1 1 1 1 u02 = d−x sinh 2x = (ex − e−3x ), u2 = ex + e−3x 2 4 4 12 1 1 1 1 1 1 1 1 yp = e−x (− e3x − e−x ) + ex ( ex + e−3x ) = − e2x − e−2x + e2x + e−2x 12 4 4 12 12 4 4 12 1 1 1 = e2x − e−2x = sinh 2x 6 6 3 1 −x x y = C1 e + C2 e + sinh 2x 3 e−2x e2x 2 −2x 2x m − 4 = 0 =⇒ m = −2, 2; yc = C1 e + C2 e ; W = =4 −2e−2x 2e2x  2x  Z x 4t 2x 4x 1 e 1e 1 e 1 e 1 1 u01 = − e2x ( )=− , u1 = − dt; u02 = e−2x = , u2 = ln |x| 4 x 4 x 4 x0 t 4 x 4x 4 Z Z x 4t 1 2x 1 e 1 1 −2x x e4t dt + e ln |x|; y = C1 e−2x + C2 e2x − e2x dt + e2x ln |x| yp = − e 4 4 4 4 x0 t x0 t e−3x e3x m2 − 9 = 0 =⇒ m = −3, 3; yc = C1 e−3x + C2 e3x ; W = =6 −3x −3e 3e3x 1 9x 3 3 9x 3 1 u01 = − e3x ( 3x = − x, u1 = − x2 ; u02 = e−3x ( 3x ) = xe−6x 6 e 2 4 6 e 2 Z

3 −6x xe dx Integration by parts 2 1 1 = − xe−6x − e−6x 4 24 3 1 1 3 1 yp = − x2 e−3x − xe−3x − e−3x ; y = C3 e−3x + C2 e3x − x2 e−3x − xe−3x 4 4 24 4 4 u2 =

16.3. NONHOMOGENEOUS LINEAR EQUATIONS

299

23. m2 + 3m + 2 = 0 =⇒ (m = −2, −1; yc = C1 e−2x + C2 e−x + 2)(m + 1) = 0 =⇒ m −x −x e−2x 1 e e2x e −3x ; u01 = − −3x W = =− −2x −x = e x −2e −e e 1+e 1 + ex Z 2x e dx v = 1 + ex , dv = ex dx, ex = v − 1 u1 = − 1 + ex Z v−1 =− dv = −v + ln |v| = −1 − ex + ln(1 + ex ) v 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 = C1 e−2x + C2 e−x − e−2x − e−x + e−2x ln(1 + ex ) + e−x ln(1 + ex ) = C3 e−2x + C4 e−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 = C1 ex + C2 e2x ; ex e2x e2x 1 2x e3x 3x 0 =− W = x 2x = e ; u1 = − 3x e x e 2e e 1+e 1 + ex Z 2x e u1 = − dx v = 1 + ex , dv = ex dx, ex = v − 1 1 + ex Z v−1 dv = −v + ln |v| = −1 − ex + ln(1 + ex ) =− v 1 e3x ex u02 = 3x ex = , u2 = ln(1 + ex ) x e 1+e 1 + ex yp = ex [−1 − ex + ln(1 + ex )] + e2x ln(1 + ex ) = −ex − e2x + ex ln(1 + ex ) + e2x ln(1 + ex ) y = C1 ex + C2 e2x − ex − e2x + ex ln(1 + ex ) + e2x ln(1 + ex ) = C3 ex + C4 e2x + ex ln(1 + ex ) + e2x ln(1 + ex )

−2x 25. m2 + 3m + 2 = 0 =⇒ (m + C2 e−x + 2)(m + 1) = 0 =⇒ m = −2, −1; yc = C1 e −x e−2x 1 e −3x ; u01 = − −3x e−x sin ex = −e2x sin ex W = −2x −x = e −2e −e e Z

u1 = −

e2x sin ex dx

Integration by parts

= ex cos ex − sin ex 1 u02 = −3x e−2x sin ex = ex sin ex , u2 = − cos ex e yp = e−2x (ex cos ex −sin ex )+e−x (− cos ex ) = −e−2x sin ex ; y = C1 e−2x +C2 e−x −e−2x sin ex

26. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; yc = C1 ex + C2 xex ; ex xex = e2x ; u01 = − 1 xex ex tan−1 x = −x tan−1 x W = x x e xe + ex e2x

300

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS Z u1 = −

x tan−1 xdx

Integration by parts

1 1 1 = − x2 tan−1 x − tan−1 x + x 2 2 2 1 1 x x −1 −1 0 u2 = 2x e e tan x = tan x, u2 = x tan−1 x − ln(1 + x2 ) e    2  1 1 1 1 2 −1 −1 x x yp = e − x tan x − tan x + x + xe x tan−1 x − ln(1 + x2 ) 2 2 2 2 1 2 x 1 1 1 = x e tan−1 x − ex tan−1 x + xex − xex ln(1 + x2 ) 2 2 2 2 1 x 1 1 2 x −1 x x y = C1 e + C3 xe + x e tan x − e tan−1 x − xex ln(1 + x2 ) 2 2 2 27. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; yc = C1 ex + C2 xex ; ex 1 ex x 1 xex 2x 0 x =− , u1 = − ln(1 + x2 ); W = x x x = e ; u1 = − 2x xe 2 2 e xe + e e 1+x 1+x 2 1 x ex 1 −1 0 u2 = 2x e = , u2 = tan x e 1 + x2 1 + x2 1 1 yp = − ex ln(1 + x2 ) + xex tan−1 x; y = C1 ex + C2 xex − ex ln(1 + x2 ) + xex tan−1 x 2 2 28. m2 − 2m + 2 = 0 =⇒ m = 1 ± i; yc = ex (C1 cos x + C2 sin x); x x e cos x e sin x = e2x W = −ex sin x + ex cos x ex cos x + ex sin x 1 1 u01 = − 2x ex sin x ex sec x = − tan x, u1 = ln | cos x|; u02 = 2x ex sec x = 1, u2 = x e e yp = ex cos x ln | cos x| + xex sin x; y = ex (C1 cos x + C2 sin x) + ex cos x ln | cos x| + xex sin x 29. m2 + 2m + 1 = 0 =⇒ (m + 1) 2 = 0 =⇒ m = −1, −1; yc = C1 e−x + C2 xe−x e−x 1 xe−x −2x ; u01 = − −2x xe−x e−x ln x = −x ln x W = −x −x −x = e −e −xe + e e Z u1 = −

x ln xdx

Integration by parts

1 2 1 2 x − x ln x 4 2 1 u02 = −2x e−x e−x ln x = ln x, u2 = x ln x − x e   1 2 1 2 1 3 −x yp = e x − x ln x + xe−x (x ln x − x) = x2 e−x ln x − x2 e−x 4 2 2 4 1 3 y = C1 e−x + C2 xe−x + x2 e−x ln x − x2 e−x 2 4 =

30. m2 + 10m + 25 = 0 =⇒ (m + 5)2 = 0 =⇒ m = −5, −5; yc = C1 e−5x + C2 xe−5x e−5x 1 e−10x e−5x xe−5x −10x W = ; u01 = −10x xe−5x 2 = − −5x −5x −5x = e −5e −5xe +e e x x Z −5x Z x −5t −10x −5x e 1 e e e u1 = − dx = − dt; u02 = −10x e−5x 2 = 2 , x t e x x x 0 Z x −5t Z −5x Z x −5t Z x −5x e e e e −5x −5x u2 = dx = dt; yp = −e dt + xe dt 2 2 x2 t t x0 x0 x0 t

16.3. NONHOMOGENEOUS LINEAR EQUATIONS y = C1 e−5x + C2 xe−5x − e−5x

Z

x

x0

e−5t dt + xe−5x t

301 Z

x

x0

e−5t dt t2

2 x/2 31. 4m2 − 4m + 1 = 0 =⇒ (2m − 1) + C2 xex/2 = 0 =⇒ m = 1/2, 1/2; yc = C1 e x/2 x/2 e xe 1 x/2 −x 1 1 x 0 W = 1 x/2 1 x/2 (2e + x) = −2xe−3x/2 − x2 e−x/2 x/2 = e ; u1 = − x xe e e 4 4 xe +e 2Z 2 Z 1 u1 = −2 xe−3x/2 dx − x2 e−x/2 dx Integration by parts 4 4 8 1 = xe−3x/2 + e−3x/2 + x2 e−x/2 + 2xe−x/2 + 4e−x/2 3 9 2 1 1 1 u02 = x ex/2 (2e−x + x) = 2e−3x/2 − xe−x/2 eZ 4 Z 4 1 −3x/2 −x/2 u2 = 2 e dx + xe dx Integration by parts 4 1 4 = − e−3x/2 − xe−x/2 − e−x/2 3 2 8 1 x/2 4 −3x/2 yp = e ( xe + e−3x/2 + x2 e−x/2 + 2xe−x/2 + 4e−x/2 ) 3 9 2 4 −3x/2 1 −x/2 8 x/2 + xe (− e − xe − e−x/2 ) = e−x + x + 4 3 2 9 8 −x x/2 x/2 y = C1 e + C2 xe + e +x+4 9

32. 4m2 − 4m + 1 = 0 =⇒ (2m − 1) 2 = 0 =⇒ m = 1/2, 1/2; yc = C1 ex/2 + C2 xex/2 √ ex/2 xex/2 1 x/2 ex/2 √ x 1 − x2 0 x 0 2 1−x = − u1 = W = 1 x/2 1 x/2 = e ; u1 = − x xe e e 4 4 xe + ex/2 2 2 √ 1 1 ex/2 √ 1 − x2 (1 − x2 )3/2 ; u02 = x ex/2 1 − x2 = 12 e 4 4 Z 1 p 2 u2 = 1 − x dx Trig substitution 4 1 p 1 = sin−1 x + x 1 − x2 8 8 √ 1 1 1 x/2 e (1 − x2 )3/2 + xex/2 sin−1 x + x2 ex/2 1 − x2 yp = 12 8 8 √ 1 1 1 x/2 2 3/2 x/2 x/2 y = C1 e + C2 xe + e (1 − x ) + xex/2 sin−1 x + x2 ex/2 1 − x2 12 8 8 −x e ex 33. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1 e−x + C2 ex ; W = =2 −e−x ex 1 1 u01 = ex xex = − xe2x 2 Z 2 1 2x u1 = − xe dx Integration by parts 2 1 1 = e2x − xe2x 8 4 1 1 1 1 1 1 1 1 1 u02 = e−x xex = x, u2 = x2 ; yp = e−x ( e2x − xe2x ) + ex ( x2 ) = ex − xex + x2 ex 2 2 4 8 4 4 8 4 4

302

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS 1 1 1 1 1 y = C1 e−x + C3 ex − xex + x2 ex ; y 0 = −C1 e−x + C3 ex − ex + xex + x2 ex 4 4 4 4 4 1 0 Using y(0) = 1 and y (0) = 0 we have C1 + C3 = 1, −C1 + C3 − = 0, or C1 = 3/8 and 4 3 5 1 1 C3 = 5/8. Thus, y = e−x + ex − xex + x2 ex . 8 8 4 4

34. 2m2 + m − 1 = 0 =⇒ (2m − 1)(m + 1) = 0 =⇒ m = −1, 1/2; yc = C1 e−x + C2 ex/2 e−x ex/2 3 (x + 1) 1 1 2 = − (xex + ex ), u1 = − xex W = 1 x/2 = e−x/2 ; u01 = − −x/2 ex/2 −e−x 2 2 3 3 e 3e 2 2 1 −x (x + 1) 0 u2 = −x/2 e = e−x/2 (x + 1) 2 3 3eZ 1 e−x/2 (x + 1)dx Integration by parts u2 = 3 2 = − xe−x/2 − 2e−x/2 3 2 1 −x yp = e (− xex ) + ex/2 (− xe−x/2 − 2e−x/2 ) = −x − 2 3 3 1 y = C1 e−x + C2 ex/2 − x − 2; y 0 = −C1 e−x + C2 ex/2 − 1 2 1 Using y(0) = 1 and y 0 (0) = 0 we obtain C1 + C2 − 2 = 1, −C1 + C2 − 1 = 0, or C1 = 1/3 2 1 8 and C2 = 8/3. Thus, y = e−x + ex/2 − x − 2. 3 3 x x ln x 1 0 1 4 00 =x 35. y − y + 2 y = ln x; yc = C1 x + C2 x ln x; W = 1 1 + ln x x x x 1 4 4 4 4 4 1 u01 = − (x ln x)( ln x) = − (ln x)2 , u1 = − (ln x)3 ; u02 = (x)( ln x) = ln x, x x x 3 x x x 4 2 2 u2 = 2(ln x)2 ; yp = − x(ln x)3 + 2x(ln x)3 = x(ln x)3 ; y + C1 x + C2 x ln x + x(ln x)3 3 3 3 2 x 4 0 6 1 x3 00 2 3 = x4 ; 36. y − y + 2 y = 3 ; yc = C1 x + C2 x ; W = 2x 3x2 x x  x    1 1 1 1 1 1 1 1 u01 = − 4 (x3 ) = − 4 ; u1 = 3 ; u02 = 4 (x2 ) = 5 , u2 = − 4 ; 3 3 x 3x x x x 4x  x  x  1 1 1 2 3 yp = x +x − 4 = 3x3 4x 12x 1 2 3 y = C1 x + C2 x + 12x 37. Writing the differential equation in the form d2 C/dx2 − (1/λ2 )C = −C(∞)/λ2 we see that the auxiliary equation is m2 − 1/λ2 = 0. Thus, Cc = c1 ex/λ + c2 e−x/λ . Using undetermined coefficients with Cp = A we find that A = C(∞). Then C(x) = c1 ex/λ + c2 e−x/λ + C(∞). Since C(0) = c1 +c2 +C(∞) = 0 and lim C(x) = C(∞) we see that c1 = 0 and c2 = −C(∞). x→∞

Thus, C(x) = C(∞)(1 − e−x/λ ). 38. If yc is the complementary function and yp is a particular solution, we have ayc00 + byc0 + cyc = 0

and ayp00 + byp0 + cyp = g(x).

16.4. MATHEMATICAL MODELS

303

Therefore, letting y = yc + yp , we have ay 00 = by 0 + cy 0 = a(yc + yp )00 + b(yc + yp )0 + c(yc + yp ) = ayc00 + ayp00 + byc0 + byp0 + cyc + cyp = [ayc00 + byc0 + cyc ] + [ayp00 + byp0 + cyp ] = ayp00 + byp0 + 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

5 5 . Therefore, yp = xex . 2 2 (c) The auxiliary equation is m2 + 2m − 3 = 0, so m = −3 or m = 1. This gives yc = C1 e−3x + C2 ex . Therefore, the general solution is which gives A =

5 y = yc + yp = C1 e−3x + C2 ex + xex 2 40. The auxiliary equation is m2 − 1 = 0, so m = ±1. This gives yc = C1 e−x + C2 ex . We look for a particular solution of the form yp = Axex + B(x − 2)e−x − Axex − Bxex = e−x − ex . Equating coefficients of xex , ex , xe−x , and e−x , we get A − A = 0, 2A = −1, B − B = 0, −2B = 1, 1 1 1 1 which gives A = − , B = − . Therefore, yp = − xex − xe−x and the general solution is 2 2 2 2 1 1 y = yc + yp = C1 e−x + C2 ex − xex − xe−x 2 2

16.4

Mathematical Models

1. A weight of 4 pounds is pushed up 3 feet above the equilibrium position. At t = 0 it is given an 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 = 0 it is released from rest. 1 1 3. Using m = W/g = 8/32 = 1/4, the initial value problem is x00 + x = 0; x(0) = , x0 (0) = 4 2 3 1 . The auxiliary equation is m2 + 1 = 0, so m = ±2i and x = C1 cos 2t + C2 sin 2t, 2 4 3 x0 = −2C1 sin 2t + 2C2 cos 2t. Using the initial condition, we obtain C1 = 1/2 and C2 = . 4 1 3 The equation of motion is x(t) = cos 2t + sin 2t. 2 4

304

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

4. From Hooke’s law we have 24 = k(1/3), so k = 72. Using m = W/g = 24/32 = 3/4, the initial 3 3 value problem is x00 +72x = 0; x(0) = −3, x0 (0) = 0. The auxiliary equation is m2 +72 = √ √ √ √ √ 4 √ √ 4 0, so m = ±4 6i and x = C1 cos 4 6t+C2 sin 4 6t, x0 = −4 6C1 sin 4 6+4 6C2 cos 4 6t. √ 1 Using the initial conditions, we obtain C1 = −1/4 and C2 = 0. Thus, x(t) = − cos 4 6t. 4 5. From Hooke’s law we have 400 = k(2), so k = 200. The initial value problem is 50x00 + 200x = 0; x(0) = 0, x0 (0) = −1 = . The auxiliary equation is 50m2 + 200 = 0, so m = ±2i and x = C1 cos 2x + C2 sin 2x, x0 = −2C1 sin 2x + 2C2 cos 2x. Using the initial conditions, we obtain C1 = 0 and C2 = −5. Thus, x(1) = −5 sin 2x. 1 00 x + 4x = 0; x(0) = 6. Using m = W/g = 2/32 = 1/16, the initial value problem is 16 2 4 , x0 (0) = − . The auxiliary equation is m2 /16 + 4 = 0, so m = ±8i and x = C1 cos 8x + 3 3 C2 sin 8x, x0 = −8C1 sin 8x + 8C2 cos 8x. Using the initial conditions, we obtain C1 = 2/3 1 2 and C2 = −1/6. Thus, x(t) = cos 8t − sin 8t. 3 6 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 downward speed of 1 ft/s. A damping force numerically equal to the instantaneous velocity acts on the system. 1 00 x + x0 + 2x = 0; x(0) = 8 −1, x0 (0) = 8. The auxiliary equation is m2 /8 + m + 2 = 0 or (m + 4)2 = 0, so m = −4, −4 and x = C1 e−4t + C2 te−4t , x0 = (C2 − 4C1 )e−4t − 4C2 te−4t . Using the initial conditions, we obtain C1 = −1 and C2 = 4. Thus, x(t) = −e−4t +4te−4t . Solving x(t) = −e4t +4te−4t = 0, we see that the weight passes through the equilibrium position at t = 1/4s. To find the maximum displacement we solve x0 (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 at t = 1/2s.

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

10. From Hooke’s law we have 40(980) = k(10), so k = 3920. The initial value problem is 40x00 + 560x0 + 3920x = 0; x(0) = 0, x0 (0) − 2. The auxiliary equation is 40m2 + 560m + 3920 = 0 or m2 + 14m + 98 = 0, so m = −7 ± 7i and x = e−7t (C1 cos 7t + C2 sin 7t), x0 = −7(C1 + C2 )e−7t sin 7t − 7(C1 − C2 )e−7t cos 7t. Using the initial conditions, we obtain 2 C1 = 0 and C2 = 2/7. Thus, x(t) = e−7t sin 7t. 7 11. From Hooke’s law we have 10 = k(7 − 5), so k = 5. Using m = W/g = 8/32 = 1/4, the 1 1 initial value problem is x00 + x0 + 5x = 0; x(0) = ; x0 (0) = 1. The auxiliary equation is 4 2 m2 /4+m+5 = 0 or m2 +4m+20 = 0, so m = −2±4i. Thus, x = e−2t (C1 cos 4t+C2 sin 4t) and x0 = −2(C1 − 2C2 )e−2t cos 4t − 2(2C1 + C2 )e−2t sin 4t. Using the initial conditions, we obtain 1 1 1 = C1 and 1 = −2( − 2C2 ), so C1 = 1/2 and C2 = 1/2. Therefore x(t) = e−2t (cos 4t + 2 2 2 sin 4t).

16.4. MATHEMATICAL MODELS

305

12. From Hooke’s law we have 24 = k(4), so k = 6. Using m = W/g = 24/32 = 3/4, the initial 3 value problem is x00 + βx0 + 6x = 0; x(0) = 0, x0 (0) = −2. The auxiliary equation 4 p 3 is m2 + βm + 6 = 0. Using the quadratic formula, m = (−β ± β 2 − 18/(3/2). When 4 √ √ 2 2p 2 2 2p 2 β > 18 = 3 2, we have m1 = − β + β − 18 and m2 = − β − β − 18. Thus, 3 3 3 3 √ 2 √ 2 x(t) = C1 e−2βt/3+2t β −18/3 + C2 e−2βt/3−2t β −18/3   p   p   2 2 −2βt/3 2 2 =e C3 cosh β − 18 t + C4 sinh β − 18 t 3 3 see Example 5 in Section 16.2. 2p 2 β − 18t). The velocity is 3 2p 2 2p 2 2p 2 2β x0 (t) = C4 e−2βt/3 sinh( β − 18C4 e−2βt/3 cosh( β − 18t) − β − 18t). 3 3 3 3 p 2p 2 From x0 (0) = −2 we obtain −2 = β − 18C4 or C4 = −3/ β 2 − 18. Therefore, 3 2p 2 −3 e−2βt/3 sinh( β − 18t). x(t) = p 3 β 2 − 18 From x(0) = 0 we obtain C3 = 0 so that x(t) = C4 e−2βt/3 sinh(

13. From Hooke’s law we have 10 = k(2), so k = 5. Using m = W/g = 10/32 = 5/16, the 5 00 5 2 differential equation is x + βx0 + 5 = 0. The auxiliary is m + βm + 5 = 0 Using the 16 16 p 2 quadratic formula, m = (−β ± β − 25/4)/(5/8). For β > 0 the motion is (a) overdamped when β 2 − 25/4 > 0 or β > 5/2 (b) critically damped when β 2 − 25/4 = 0 or β = 5/2 (c) underdamped when β 2 − 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 initial value problem is x00 + 8x0 + 16x = 8 sin 4t; x(0) = x0 (0) = 0. The auxiliary equation is m2 + 8m + 16 = (m + 4)2 = 0 so m = −4, −4, and xc = C1 e−4t + C2 te−4t . Using xp = A sin 4t + B cos 4t we find A = 0 and B = −1/4. Thus, x(t) = C1 e−4t + C2 te−4t −

1 cos 4t and x0 (t) = −4C1 e−4t − 4C2 te−4t + C2 e−4t + sin 4t. 4

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

306

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

15. The initial value problem is x00 + 8x0 + 16x = e−t sin 4t; x(0) = x0 (0) = 0. Using xp = Ae−t sin 4t + Be−t cos 4t we find A = −7/625 and B = −24/625. Thus, x(t) = C1 e−4t + C2 te−4t − x0 (t) = −4C1 e−4t − 4C2 te−4t + C2 e−4t −

7 −t 24 −t e sin 4t − e cos 4t, 625 625

28 −t 7 −t 96 −t e cos 4t + e sin 4t + e sin 4t 625 625 625

24 −t e cos 4t. 625 Using the initial conditions, we obtain C1 = 24/625 and C2 = 100/625. Thus, +

x(t) =

24 −4t 100 −4t 7 −t 24 −t e te e sin 4t − e cos 4t. + − 625 625 625 626

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 is m2 + 9 = 0, so m = ±3i and xc = C1 cos 3t + C2 sin 3t. Using variation of parameters 5 5 sin 3t, so xp = − t cos 3t + 6 18 5 x(t) = C1 cos 3t + C3 sin 3t − t cos 3t 6 5 5 x0 (t) = −3C1 sin 3t + 3C2 cos 3t + t sin 3t − cos 3t. 2 6 Using the initial conditions, we obtain C1 = 2 and C2 = 5/18. Thus, x(t) = 2 cos 3t + 5 5 sin 3t − t cos 3t. The spring-mass system is in pure resonance. 18 6 17. The DE describing charge is .05q 00 + 2q 0 + 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). The 0 initial conditions yield q(0) = C1 = 5 and i(0)  = q (0) = −20C1+ 40C2 = 0, which gives 5 C1 = 5 and C2 = 25 . Therefore q(t) = e−20t 5 cos 40t + sin 40t , and q(0.01) = 4.568C. 2 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 41 q 00 +20q 0 +300q = 0. The auxiliary equation is 41 m2 +20m+300 = 0, so m = −20 or m = −60. The general solution is q(t) = C1 e−20t + C2 e−60t . The initial conditions yield q(0) = C1 + C2 = 4 and i(0) = q 0 (0) = −20C1 − 60C2 = 0, which give C1 = 6 and C2 = −2. Therefore, q(t) = 6e−20t − 2e−60t . The charge is never equal to zero. 5 00 q + 10q 0 + 30q = 300. The auxiliary equation is 53 m2 + 10m + 30 = 0. so 3 m = −3 ± 3i. This gives qc = e−3t (C1 cos 3t + C2 sin 3t). Assume a particular solution of the form qp = A. Substituting into the DE, we have 30A = 300 so that A = 10 and therefore qp = 10. Thus, the general solution is q = qc + qp = e−3t (C1 cos 3t + C2 sin 3t) + 10. The initial conditions yield q(0) = C1 +10 = 0 and i(0) = q 0 (0) = −3(C1 −C2 ) = 0, which gives C1 = −10 and C2 = −10. Therefore, q(t) = e−3t (−10 cos 2t − 10 sin 3t) + 10 − 10 − 10e−3t (cos 3t + sin 3t), i(t) = q 0 (t) = 60e−3t sin 3t. The charge q(t) attains a maximum of 10.432 C at t = π3 .

19. The DE is

16.5. POWER SERIES SOLUTIONS

307

20. The DE is q 00 +100q 0 +2500q = 30. The auxiliary equation is m2 +100m+250000, so m = −50 is a repeated root. This gives qc = C1 e−50t + C2 te−50t . Assume a particular solution of the 3 form qp = A. Substituting into the DE, we have 2500A = 30 so that A = 250 and therfore 3 3 −50t −50t + C2 te + 250 . The general solution qp = 250 . The general solution is q = qc + qp = C1 e 3 is q = qc + qp = C1 e−50t + C2 te−50t + 250 . The initial conditions yield q(0) = C1 = 0 and 3 0 i(0) = q (0) = −50C! + C2 = 2 which give C1 = 0 and C2 = 2. Therefore, q(t) = 2te−50t + 250 1 0 −50t and i(t) = q (t) = (2 − 100t)e . The charge q(t) attains a maximum of 0.0267 C at t = 50 s. 21.

16.5 1.

∞ X

Power Series Solutions n(n − 1)cn xn−2 +

cn xn =

n=0

n=2

|

∞ X

{z

k=n−2

∞ X

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

k=0

∞ X

ck xk

k=0

} =

∞ X

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

k=0

ck , k = 0, 1, 2, . . . (k + 2)(k + 1)ck+2 + ck = 0; ck+2 = − (k + 2)(k + 1) c0 c0 c1 c1 c2 c0 c0 c2 = − = − , c 3 = − = − , c4 = − = = , 2 2! 3·2 3! 4 cos 3 4 · 3 · 2! 4! c1 c1 c4 c0 c0 c3 = = , c6 = − =− =− , c5 = − 5·4 5 · 4 · 3! 5! 6·5 6 · 5 · 4! 6! c5 c1 c1 c7 = − =− =− 7 · 6 · 5! 7!   7 · 6  1 4 1 6 1 5 1 7 1 2 1 3 y = c0 1 − x + x − x + · · · + c1 x − x + x − x + · · · 2! 4! 6! 3! 5! 7! ∞ ∞ X X 1 1 x2n + c1 (−1)n x2n+1 = c0 (−1)n (2n)! (2n + 1)! n=0 n=0 2.

∞ X

n(n − 1)cn xn−2 −

n=2

|

∞ X

cn xn =

n=0

{z

k=n−2

∞ X

(k + 2)(k + 1)ck+2 xk −

k=0

∞ X

ck xk

k=0

} =

∞ X

[(k + 2)(k + 1)ck+2 − ck ]xk = 0

k=0

c0 ck (k + 2)(k + 1)ck+2 − ck = 0; ck+2 = , k = 0, 1, 2, . . . ; c2 = , (k + 2)(k + 1) 2! c1 c1 c2 c0 c3 c1 c4 c0 c5 c1 c3 = = , c4 = = , c5 = = , c6 = = , c7 = = 3 ·2 3! 4·3 4! 6·5 6! 7!  5 · 4 5! 7 · 6 1 1 1 1 1 1 y = c0 1 + x2 + x4 + x6 + · · · + c1 x + x3 + x5 + x7 + · · · 2! 4! 6! 3! 5! 7! ∞ ∞ X 1 X 1 = c0 x2n + c1 x2n+1 (2n)! (2n + 1)! n=0 n=0

308

3.

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS ∞ X

n(n − 1)cn xn−2 −

ncn xn−1 =

n=1

n=2

|

∞ X

{z

k=n−2

|

}

∞ X

(k + 2)(k + 1)ck+2 xk −

k=0

{z

∞ X

(k + 1)ck+1 xk

k=0

}

k=n−1

=

∞ X

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

k=0

c1 c1 ck+1 , k = 0, 1, 2, . . . ; c2 = = , (k + 2)(k + 1)ck+2 − (k + 1)ck+1 = 0; ck+2 = (k + 2) 2 2! c2 c1 c3 c1 c3 = = , c4 = = , 3 3! 4 4!   P∞ 1 1 2 1 3 y = c0 + c1 x + x + x + · · · = c0 + c1 n=1 xn 2! 3! n!

4.

∞ X

n(n − 1)cn x

n−2

ncn x

n−1

∞ X

=2

n=1

n=2

|

∞ X

{z

k=n−2

}

|

k

(k + 2)(k + 1)ck+2 x +

k=0

{z

∞ X

(k + 1)ck+1 xk

k=0

}

k=n−1

=

∞ X

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

k=0

ck+1 2(k + 2)(k + 1)ck+2 + (k + 1)ck+1 = 0; ck+2 = − , k = 0, 1, 2, . . . ; 2(k + 2) c1 c1 c2 c1 c3 c1 c2 = − =− , c3 = − = 2 , c4 = − =− 3 , 2·2 2 · 2! 2·3 2 · 3! 2 · 4! 2 · 4 1 1 1 2 x + 2 x3 − 3 x4 + · · · y = c0 + c1 x − 2 · 2! 2 · 3! 2 · 4!

5.

∞ X

n(n − 1)cn xn−2 − x

n=2

|

∞ X

cn xn =

n=0

{z

k=n−2

}

|

{z

k=n+1

∞ X

(k + 2)(k + 1)ck+2 xk −

∞ X

ck−1 xk

k=1

k=1

} = 2c2 +

∞ X

[(k + 2)(k + 1)ck+2 − ck−1 ]xk = 0

k=0

ck−1 c2 = 0; (k + 2)(k + 1)ck+2 − ck−1 = 0; ck+2 = , k = 1, 2, 3, . . . ; (k + 2)(k + 1) c0 c2 c3 c0 c4 c1 c3 = , c5 = = 0, c6 = = , c7 = = 3·2 5·4 6·5 6·5·3·2 7·6 7·6·4·3 c5 c6 c0 c7 c1 c9 = = 0, c9 = = c9 = , c10 = = 8 ·7 9·8 9·8·6·5·3·2 10 ·9 10 · 9 · 7 · 6 · 4 · 3 1 3 1 1 y = c0 1 + x + x6 + x9 + · · · 3·2 6·5·3·2 9·8·6·5·3·2   1 4 1 1 7 10 + c1 x + x + x + x + ··· 4·3 7·6·4·3 10 · 9 · 7 · 6 · 4 · 3

16.5. POWER SERIES SOLUTIONS

6.

∞ X

n(n − 1)cn xn−2 +x2

∞ X

cn xn =

n=0

n=2

{z

|

k=n−2

309 ∞ X

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

k=0

∞ X

ck−2 xk

k=2

| {z }

}

k=n+2

= 2c2 + +c3 x +

∞ X

[(k + 2)(k + 1)ck+2 + ck−2 ]xk = 0

k=2

ck−2 , k = 2, 3, 4, . . . ; c2 = c3 = 0; (k + 2)(k + 1)ck+2 + ck−2 = 0; ck+2 = − (k + 2)(k + 1) c0 c1 c2 c3 c4 c0 c4 = − , c5 = − , c6 = − = 0, c7 = − , c8 = − = 4·3 5·4 7·6 7·6 8·7 8·7·4·3 c5 c1 c6 c7 c9 = − = , c10 = − = 0, c11 − = 0, 9·8 9·8·5·4 10 · 9 11 · 10 c8 c0 c9 c1 c12 = − =− , c13 = − =− , 12 · 11 12 · 11 · 8 · 7 · 4 · 3 13 · 12 13 · 12 · 9 · 8 · 5 · 4    1 1 1 1 4 x + x8 − · · · c1 x − x5 + x9 − · · · y = c0 1 − 4·3 8·7·4·3 5·4 9·8·5·4

7.

∞ X

n(n − 1)cn xn−2 −2x

n=2

∞ X

cn xn−1 +

n=1

|

{z

k=n−2

∞ X

cn x n

n=0

} =

∞ X

k

(k + 2)(k + 1)ck+2 x −

k=0

∞ X k=1

= c0 + 2c2 +

∞ X

k

kck x +

∞ X

ck xk

k=0

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

k=1

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

c0 2

c1 c1 3c2 3c0 (2k − 1)ck , k = 1, 2, 3, . . . ; c3 = = , c4 = =− , (k + 2)(k + 1) 3·2 3! 4·3 4! 5c3 5c1 7c4 7 · 3c0 9c5 9 · 5c1 c5 = = , c6 = = , c7 = = 5 · 4 5! 6·5 6! 7 · 6 7!    1 2 3 4 7·3 6 5 9·5 7 1 3 y = c0 1 − x − x − x − · · · + c1 x + x + x 5 + x + ··· 2! 4! 6! 3! 5! 7! ck+2 =

8.

∞ X

n(n − 1)cn xn−2 −x

n=2

|

∞ X

ncn xn−1 + 2

n=1

{z

k=n−2

∞ X

cn xn

n=0

} =

∞ X

(k + 2)(k + 1)ck+2 xk −

k=0

= 2c0 + 2c2 +

∞ X k=1

∞ X

kck xk + 2

∞ X

ck x k

k=0

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

k=1

2c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 − (k − 2)ck = 0; c2 = −c0 (k − 2)ck c1 c1 c3 c1 ck+2 = , k = 1, 2, 3, . . . ; c3 = − = − , c4 = 0, c5 = =− , (k + 2)(k + 1) 3·2 3! 5·4 5!

310

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS 3c5 3c1 5c7 5 · 3c1 c6 = c8 = c10 = 0, c7 = =− , c9 = =− 7 · 6 7! 9 · 8 9!   1 5 3 7 5·3 9 1 3 2 x + ··· y = c0 (1 − x ) + c1 1 − x − x − x − 3! 5! 7! 9!

9.

∞ X

n(n − 1)cn xn−2 + x2

ncn xn−1 + x

n=1

n=2

|

∞ X

{z

k=n−2

}

| =

∞ X

cn x n

n=0

{z

k=n+1 ∞ X

}

|

{z

k=n+1

}

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

k=0

∞ X k=2

= 2c2 + (6c3 + c0 )x +

∞ X

∞ X

(k − 1)ck−1 xk +

ck−1 xk

k=1

[(k + 2)(k + 1)ck+2 + kck−1 ]xk = 0

k=2

c0 2c2 = 0, 6c3 + c0 = 0 (k + 2)(k + 1)ck+2 + kck−1 = 0; c2 = 0, c3 = − 3·2 kck−1 2c1 4c3 4c0 ck+2 = − , k = 2, 3, 4, . . . ; c4 = − , c5 = 0, c6 = − = (k + 2)(k + 1) 4·3 6·5 6·5·3·2 5c4 5 · 2c1 7c6 7 · 4c0 c7 = − = , c8 = c11 = c14 = · · · = 0, c9 = − =− 7·6 7·6·4·3 9·8 9·8·6·5·3·2 8c7 8 · 5 · 2c1 c10 = − =− 10 · 9 · 7 · 6 · 4 · 3     10 · 9 22 4 52 · 22 7 82 · 52 · 22 10 1 3 42 6 72 · 42 9 x · · · + c1 x − x + x − x + ··· y = c0 1 − x + x − 3! 6! 9! 4 7! 10!

10.

∞ X

n(n − 1)cn xn−2 +2x

n=2

|

∞ X

ncn xn−1 + 2

n=1

{z

k=n−2

∞ X

cn x n

n=0

} =

∞ X

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

k=0

= 2c2 + 2c0 +

∞ X k=1

∞ X

kck xk + 2

∞ X

ck xk

k=0

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

k=1

2c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 + 2(k + 1)ck = 0; c2 = −c0 2c1 2c2 2c0 2c3 22 c1 ck , k = 1, 2, 3, . . . ; c3 = − , c4 = − = , c5 = − = , ck+2 = − k+2 3 4 4 5 5·3 2 3 3 2c4 2 c0 2c5 2 c1 2c6 2 c0 c6 = − =− , c7 = − =− , c8 = − = 6·4 7 7·5·3 8·6·4 6  8  2 4 22 6 23 2 3 22 5 23 2 8 7 y = c0 1 − x + x − x + x + · · · +c1 x − x + x − x + ··· 4 6·4 8·6·4 3 5·3 7·5·3

16.5. POWER SERIES SOLUTIONS

11. (x − 1)

∞ X

n(n − 1)cn xn−2 +

n=2

∞ X

311 ncn xn−1

n=1

=

∞ X

n(n − 1)cn xn−1 −

∞ X

n(n − 1)cn xn−2 +

n=2

n=2

{z

| =

∞ X

}

k=n−1

(k + 1)kck+1 xk −

k=1

ncn xn−1

n=2

|

{z

k=n−2

∞ X

∞ X

}

|

{z

k=n−1 ∞ X

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

(k + 1)ck+1 xk

k=0

= c1 − 2c2 +

∞ X

}

k=0

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

k=1

c1 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 = 0; c2 = (k + 1)ck+1 2c2 c1 3c3 c1 , k = 1, 2, 3, . . . ; c3 = = , c4 = = k+ 2 3 4 4  3 ∞ X 1 n 1 2 1 3 1 4 x y = c0 + c1 x + x + x + x + · · · = c0 + c1 2 3 4 n n=1

c1 2

ck+2 =

12. (x + 2)

∞ X

n(n − 1)cn xn−2 + x

∞ X

ncn xn−1 −

=

∞ X

n(n − 1)cn xn−1 +2

=

∞ X

n(n − 1)cn xn−2 +

n=2

n=2

|

cn xn

n=0 ∞ X

n=1

n=2

∞ X

{z

}

k=n−1

(k + 1)kck+1 xk + 2

k=1

= 4c2 − c0 +

| ∞ X

∞ X n=1

{z

k=n−2

cn xn

n=0

∞ X

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

kck xk −

k=1

∞ X

ck xk

k=0

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

k=1

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

c0 4

(k + 1)kck+1 + (k − 1)ck kck+1 (k − 1)ck =− − , k = 1, 2, 3, . . . 2(k + 2)(k + 1) 2(k + 2) 2(k + 2)(k + 1) c2 c0 2c3 c2 c0 c0 c3 = − =− , c4 = − − = − =0 2 2·3 2·3·4 2·4 2·4·3 2·3·4 2 · 3 · 42 c9 4c5 c0 2c3 = , c6 = − −0=− c5 = 0 − 5 · 42 · 3 · 2 2·6  2·5·4  6 · 5 · 4 · 3 · 22 1 1 1 y = c0 1 + x2 − x3 + x5 − · · · + c1 x 2 4 4·3·2 5·4 ·3·2 ck+2 = −

∞ X

}

k=0 ∞ X

ncn xn −

312

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

13. (x2 − 1)

∞ X

n(n − 1)cn xn−2 + 4x

n=2

=

∞ X

∞ X

ncn xn−1 + 2

n=1 ∞ X

n(n − 1)cn xn −

n=2

∞ X

cn xn

n=0

n(n − 1)cn xn−2 +4

k(k − 1)ck xk −

k=2

∞ X

∞ X

ncn xn + 2

n=1

n=2

{z

| =

∞ X

cn xn

n=0

}

k=n−2

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

k=0

∞ X

kck xk + 2

k=1

= (2c0 − 2c2 ) + (2c1 + 4c1 − 6c3 )x +

∞ X

∞ X

∞ X

ck xk

k=0

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

k=2

=0 2c0 − 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 ,Pc6 = c4 = c0 , Pc7 = c5 = c1 ∞ ∞ y = c0 1 + x2 + x4 + · · · + c1 [x + x3 + x5 + · · · ] = c0 n=0 x2n + c1 n=0 x2n+1

14. (x2 + 1)

∞ X

n(n − 1)cn xn−2 − 6

∞ X

cn xn

n=0

n=2

=

∞ X

n(n − 1)cn xn +

n=2

∞ X

=

k=2

k

k(k − 1)ck x +

∞ X

∞ X

ncn xn

n=0

n=2

{z

| ∞ X

n(n − 1)cn xn−2 −6 k=n−2

} k

(k + 2)(k + 1)ck+2 x − 6

k=0

= (2c2 − 6c0 ) + (6c3 − 6c1 )x +

∞ X

ck x k

k=0 ∞ X

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

k=2

2c2 − 6c0 = 0; 6c3 − 6c1 = 0; (k − 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0; c2 = 3c0 , c3 = c1 ; −c2 c4 c0 (k − 3)ck , k = 2, 3, 4, . . . ; c4 = − = c0 , c5 = 0, c6 = − = − ck+2 = − k+1 3 5 5 3c6 3c0 5c8 5 · 3c0 c7 = c9 = c11 = · · · = 0, c8 = − = , c10 = − =− 7  7·5 9 9·7·5  3 6 2 4 3 y = c0 1 + 3x + x − x + · · · + c1 (x + x ) 5·3

16.5. POWER SERIES SOLUTIONS 15. (x2 + 2)

∞ X

n(n − 1)cn xn−2 + 3x

n=2

313 ∞ X

ncn xn−1 −

n=1

=

∞ X

n(n − 1)cn xn + 2

n=2

∞ X

cn x n

n=0

∞ X

n(n − 1)cn xn−2 +3

k(k − 1)ck xk + 2

{z

k=2

ncn xn −

∞ X

cn xn

n=0

}

k=n−2

∞ X

∞ X n=1

n=2

| =

∞ X

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

k=0

∞ X

kck xk −

k=1

= (4c2 − c0 ) + (12c3 + 3c1 − c1 )x +

∞ X

∞ X

ck xk

k=0

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

k=2

=0 c1 c0 , c3 = − ; 4 6 (k 2 + 2k − 1)ck 7c2 7 14c3 =− , k = 2, 3, 4, . . . ; c4 = − =− c0 , c5 = − = 2(k + 2)(k + 1) 2·4·3 4 · 4! 2·5·4

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

14 c1 2 · 5!

  23c4 23 · 7 34c5 34 · 14 7 4 23 · 7 6 1 c6 = − = 3 c0 , c7 = − =− c1 y = c0 1 + x2 − x + x − ··· + 2·6·5 2 · 6! 2 · 7 ·6 4 · 7! 4 4 · 4! 8 · 6!  14 5 34 · 14 7 1 3 c1 x − x + x − x + ··· 6 2 · 5! 4 · 7!

16. (x2 − 1)

∞ X

n(n − 1)cn xn−2 + x

∞ X

ncn xn−1 −

=

∞ X

n(n − 1)cn xn −

∞ X

| =

k=2

n(n − 1)cn xn−2 +

n=2

n=2

∞ X

cn xn

n=0

n=0

n=2

∞ X

k(k − 1)ck xk −

∞ X

∞ X

ncn xn −

}

k=n−2

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

k=0

= −(2c2 + c0 ) + (c1 − 6c3 − c1 )x +

cn x n

n=0

n=1

{z

∞ X

∞ X k=1

∞ X

kck xk −

∞ X

ck x k

k=0

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

k=2

=0 c0 , c3 = 0; 2 (k − 1)ck c2 c0 3c4 = , k = 2, 3, 4, . . . ; c4 = =− , c5 = c7 = c9 = · · · = 0, c6 = = k+2 4 4·2 6

2c2 + c0 = 0; −6c3 = 0; (k + 1)(k − 1)ck − (k + 2)(k + 1)ck+2 = 0; c2 = − ck+2 c0 − 4 · 22 y = c0

 1 2 1 4 1 6 1 − x − x − x − · · · + c1 x 2 8 16

314

17.

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS ∞ X

n(n − 1)cn xn−2 − (x + 1)

n=2

∞ X

ncn xn−1 −

n=1

=

∞ X

n(n − 1)cn xn−2 −

∞ X

ncn xn −

n=1

{z

| ∞ X

cn xn

n=0

n=2

=

∞ X

ncn xn−1 − {z

|

(k + 2)(k + 1)ck+2 xk −

k=0

∞ X

= 2c2 − c1 − c0 +

cn x n

}

k=n−1 ∞ X

kck xk −

k=1 ∞ X

∞ X n=0

n=1

}

k=n−2

∞ X

(k + 1)ck+1 xk −

k=0

∞ X

ck x k

k=0

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

k=1

c0 + c1 2 ck + ck+1 c1 + c2 c1 + c0 /2 + c1 /2 c0 + 3c1 ck+2 = , k = 1, 2, 3, . . . ; c3 = = = k+2 3 3 6 c2 + c3 c0 /2 + c1 /2 + c0 /6 + c1 /2 2c0 + 3c1 c4 = = = 4 4 12 c3 + c4 c0 /6 + c1 /2 + c0 /6 + c1 /4 4c0 + 9c1 c5 = = = 5 60 5    1 2 1 3 1 4 1 2 1 3 1 4 y = c0 1 + x + x + x + · · · + c1 x + x + x + x + · · · 2 6 6 2 2 4 2c2 − c1 − c0 = 0; (k + 2)(k + 1)ck+2 − (k + 1)ck+1 − (k + 1)ck = 0; c2 =

18.

∞ X

n(n − 1)cn xn−2 − x

n=2

∞ X

ncn xn−1 − (x + 2)

n=1

=

∞ X

=

∞ X

n(n − 1)cn x

n−2

∞ X

n

ncn x −

n=1

{z

(k + 2)(k + 1)ck+2 xk −

k=0

| ∞ X k=1

∞ X

∞ X

cn x

n+1

n=0

}

k=n−2

= 2c2 − 2c0 +

cn xn

n=0

n=2

|

∞ X

−2

∞ X

cn xn

n=0

{z

k=n+1 ∞ X

kck xk −

}

ck−1 xk − 2

k=1

∞ X

ck x k

k=0

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

k=1

2c2 − 2c0 = 0; (k + 2)(k + 1)ck+2 − (k + 2)ck − ck−1 = 0; c2 = 0 ck−1 c1 c0 c0 c1 ck ++ , k = 1, 2, 3, . . . ; c3 = + = + ck+2 = k+1 (k + 2)(k + 1) 2 3·2 3! 2 c2 c1 2c0 2c1 c3 c2 c1 c0 c0 11c0 3c1 c4 = + = + , c5 = + = + + = + 3 4·3 3! 4! 4 5·4 4·2 4! 5·4 5! 4! c4 c3 2c0 2c1 c1 c0 52c0 4c1 c6 = + = + + + = + 5 6 · 5 5 · 3! 5! 6 · 5 · 2 6 · 5 · 3! 6! 5!  1 3 2 4 11 5 1 3 2 4 3 5 2 y = c0 1 + x + x + x + x + · · · + c1 x + x + x + x + · · · 3! 3! 5! 2 4! 4!

16.5. POWER SERIES SOLUTIONS

19. (x − 1)

∞ X

n(n − 1)cn xn−2 − x

n=2

315

∞ X

ncn xn−1 +

n=1

=

∞ X

=

n(n − 1)cn xn−1 −

n(n − 1)cn xn−2 −

n=2

{z

}

k=n−1

∞ X

cn xn

n=0 ∞ X

n=2

|

∞ X

k=1

{z

k=n−2

∞ X

= c0 − 2c2 +

∞ X

cn xn

n=0

}

(k + 2)(k + 1)ck+2 xk −

k=0 ∞ X

ncn xn +

n=1

|

(k + 1)kck+1 xk −

∞ X

∞ X

kck xk +

k=1

∞ X

ck xk

k=0

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

k=1

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

1 c0 2

2c2 c0 (k + 1)kck+1 − (k − 1)ck , k = 1, 2, 3, . . . ; c3 = = (k + 2)(k + 1) 3·2 3·2 3 · 2c3 − c2 c0 − c0 /2 c0 c4 = = = 4 · 3 4 · 3 4 · 3·2     1 3 1 1 1 2 x + x4 + · · · + c1 x; y 0 = c0 x + x2 + · · · + c1 y = c0 1 + x + 2 3·2 4·3·2 2 Using the initial conditions, we obtain −2 = y(0) = c0 and 6 = y 0 (0) = c1 . The solution is ck+2 =

  1 3 1 1 1 4 1 x + x4 + · · · + 6x = −2 + 6x − x2 − x3 − x + ··· y = −2 1 + x2 + 2 3·2 4·3·2 3 4·3

20. (x − 1)

∞ X

n(n − 1)cn x

n−2

− 2x

n=2

∞ X

ncn x

n−1

+8

n=1

=

∞ X

=

∞ X

cn x n

n=0

n(n − 1)cn xn−2 −2

n=2

|

∞ X

∞ X

ncn xn + 8

n=1

{z

(k + 2)(k + 1)ck+2 xk − 2

k=0

= 2c2 + 8c0 +

cn xn

n=0

}

k=n−2

∞ X k=1

∞ X

∞ X

kck xk + 8

∞ X

ck xk

k=0

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

k=1

2c2 + 8c0 = 0; (k + 2)(k + 1)ck+2 − 2(k − 4)ck = 0; c2 = −4c0 2(k − 4)ck −2 · 3c1 −2 · 2c2 4 ck+2 = , k = 1, 2, 3, . . . ; c3 = = −c1 , c4 = = c0 (k + 2)(k + 1) 3·2 4·3 3 −2 · 1c3 1 2 · 0c4 c5 = = c1 , c 6 = = 0, c8 = c10 = c12 = · · · = 0 5·2  5 · 4  6·5  4 4 1 5 2 3 y = c0 1 − 4x + x + c1 x − x + x + · · · 3  10    16 1 0 3 2 4 y = c0 −8x + x + c1 1 − 3x + x + · · · 3 2

316

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS Using the initial conditions, we obtain 3 = y(0) = c0 and 0 = y 0 (0) = c1 . The solution is   4 4 2 y = 3 1 − 4x + x = 3 − 12x2 + 4x4 3

Chapter 16 in Review A. True/False 1. True 2. True. We know a general solution is y = Aex + Be−x . Now  x   x  e + e−x e − e−x C1 cosh x + C2 sinh x = C1 + C2 2 2     C1 C C2 C 1 2 x = + e + − e−x . 2 2 2 2 By varying C1 and C2 , we see that the two equations are different forms of the same general solution. 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 + C2 ex . Boundary conditions yield y(0) = C1 + C2 =  1 and y(1)  = C1 + C2 e = 0, which e −1 e 1 and C2 = . Therefore, y = − ex . give C1 = e−1 e−1 e−1 e−1 3. 10 = k(2.5) =⇒ k = 4 lb/ft; 32 = 4x =⇒ x = 8 ft 4. We have a repeated root m = −7. Therefore, y = C1 e−7x + C2 xe−7x . 5. yp = Ax2 + Bx + C + Dxe2x + Ee2x

CHAPTER 16 IN REVIEW

317

C. Exercises 1. Py = −6xy 2 sin y 3 = Qx , and the equation is exact. fx = 2x cos y 3 , f = x2 cos y 3 + g(y), fy = −3x2 y 2 sin y 3 + g 0 (y) = −1 − 3x2 y 2 sin y 3 , g 0 (y) = −1, g(y) = −y, f = x2 cos3 y − y. Therefore, the solution is x2 cos y 3 − y = C. 2. Py = 6y 2 = Qx , and the equation is exact. fx = 3x2 + 2y 3 , f = x3 + 2xy 3 + g(y), fy = 6xy 2 + g 0 (y) = 6xy 2 + y 2 , y3 y3 g 0 (y) = y 2 , g(y) = , f = x3 + 2xy 3 + . 3 3 3 y Therefore, the solution is x3 + 2xy 3 + = C. 3 3. Py = −2xy −5 = Qx , and the equation is exact. fx = 21 xy −4 , f = 14 x2 y −4 + g(y), fy = −x2 y −5 + g 0 (y) = 3y −3 − x2 y −5 g 0 (y) = 3y −3 , g(y) = − 32 y −2 , f = 14 x2 y −4 − 32 y −2 . Therefore, the general solution is 14 x2 y −4 − 32 y −2 = C. Since y(1) = 1, we have 41 (1)(1)− 32 (1) = C or C = − 54 . Thus, the solution is 41 x2 y −4 − 32 y −2 = − 54 . 4. Py = 2x + sin x = Qx and the equation is exact. 1 fx = y 2 + y sin x, f = xy 2 − y cos x + g(y), fy = 2xy − cos x + g 0 (y) = 2xy − cos x − 1+y 2, 1 , g(y) = tan−1 (y), f = xy 2 − y cos x + tan−1 (y). Therefore, the general g 0 (y) = 1 + y2 solution is xy 2 − y cos x + tan−1 (y) = C. Since y(0) = 1, we have −1 + π4 = C. Thus, the π solution is xy 2 − y cos x + tan−1 (y) = − 1 4 √ √ √ 5. m2 − 2m − 2 = 0 =⇒ m = 1 ± 3; y = C1 e(1− 3)x + C2 e(1+ 3)x √ √ √ 6. m2 − 8 = 0 =⇒ m = ±2 2; y = C1 e−2 2x + C2 e2 2x 7. m2 − 3m − 10 = 0 =⇒ (m − 5)(m + 2) = 0 =⇒ m = −2, 5; y = C1 e−2x + C2 e5x 8. 4m2 + 20m + 25 = 0 =⇒ (2m + 5)2 = 0 =⇒ m = −5/2, −5/2; y = C1 e−5x/2 + C2 xe−5x/2 1 x x 9. 9m2 + 1 = 0 =⇒ m = ± i; y = C1 cos + C2 sin 3 3 3 10. 2m2 − 5m = 0 =⇒ m(2m − 5) = 0 =⇒ m = 0, 5/2; y = C1 + C2 e5x/2 11. Letting y = ux we have dx − eu du = 0 x =⇒ ln |x| − eu = C1 =⇒ ln |x| − ey/x = C1 .

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

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. The general solution is y = C1 e−2x + C2 xe−2x . Initial conditions yield y(0) = C1 = −2 and y 0 (0) = −2C1 +C2 = 0 which give C1 = −2 and C2 = −4. The solution is y = −2e−2x −4xe−2x .

318

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

13. m2 − m − 12 = 0 =⇒ (m − 4)(m + 3) = 0 =⇒ m = −3, 4; yc = C1 e−3x + C2 e4x yp = Axe2x + Be2x , yp0 = 2Axe2x + (A + 2B)e2x ; yp00 = 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 = C1 e−3x + C2 e4x −

1 2x 13 2x xe − e . 10 100

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, we have 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 cos x + C2 sin x) x x e cos x e sin x = e2x W = x x x x −e sin x + e cos x e cos x + e sin x sin2 x cos2 x − 1 1 = = cos x−sec x, u = sin x−ln | sec x+tan x| u0 = − 2x ex sin x ex tan x = − e cos x cos x 1 v 0 = 2x ex cos x ex tan x = sin x, v = − cos x e yp = ex cos x(sin x − ln | sec x + tan x|) − ex sin x cos x = −ex cos x ln | sec x + tan x| y = ex (C1 cos x + C2 sin x) − ex cos x ln | sec x + tan x| −x e ex 2 −x x =2 16. m − 1 = 0 =⇒ m = −1, 1; yc = C1 e + C2 e ; W = −e−x ex 1 2ex e2x e3x u0 = − ex x =− x = − 2x −x −x 2Z e + e e +e e +1 e3x x x u=− dx t = e , dt = e dx e2x + 1  Z  Z 1 t2 dt = − 1 − =− dt = tan−1 t − t = tan−1 ex − ex t2 + 1 t2 + 1 1 2ex 1 ex v 0 = e−x x = = e + e−x ex + e−x e2x + 1 Z2 ex v= dx t = ex , dt = ex dx e2x + 1 Z dt = tan−1 t = tan−1 ex = t2 + 1 yp = e−x (tan−1 ex − ex ) + ex tan−1 ex = (ex + e−x ) tan−1 ex − 1 y = C1 e−x + C2 ex + (ex + e−x ) tan−1 ex − 1 cos x sin x =1 17. m2 + 1 = 0 =⇒ m = ±i; yc = C1 cos x + C2 sin x; W = − sin x cos x 1 u0 = − sin x sec3 x = − tan x sec2 x, u = − sec2 x; v 0 = cos x sec3 x = sec2 x, v = tan x 2

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319

1 1 sin2 x 1 yp = − cos x sec2 x + sin x tan x = sin x tan x − sec x = − 2 2 cos x 2 cos x 2 sin2 x − 1 sin2 x − cos2 x 1 1 = = = sin x tan x − cos x 2 cos x 2 cos x 2 2 1 1 0 y = C3 cos x + C2 sin x + sin x tan x, y = −C3 sin x + C2 cos x + sin x sec2 x + sin x 2 2 Using the initial conditions, we obtain C3 = 1 and C2 = 1/2. Thus,

1 2 cos2 x sin2 x 1 1 sin x + sin x tan x = + + sin x 2 2 2 cos x 2 cos x 2 cos2 x + 1 1 1 = + sin x = (sin x + cos x + sec x). 2 cos x 2 2

y = cos x +

18. The auxiliary equation is m2 +2m+2 = 0, so m = −1±i. Therefore, yc = e−x (C1 cos x + C2 sin x) . Assume a particular solution of the form yp = A. Substituting this into the DE, we have 2A = 1, or A = 21 . Therefore, the general solution is y = yc +yp = e−x (C1 cos x + C2 sin x)+ 21 . The initial conditions yield y(0) = C1 + 12 = 0 and y 0 (0) = −C1 + C2 = 1 which give C1 = − 21 and C2 = 12 . Thus, the solution is y = e−x − 21 cos x + 12 sin x + 12 .

19.

∞ X

n(n − 1)cn xn−2 + x

cn xn =

n=0

n=2

|

∞ X

{z

k=n−2

}

|

{z

k=n+1

∞ X

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

k=0

∞ X

ck−1 xk

k=1

} = 2c2 +

∞ X

[(k + 2)(k + 1)ck+2 + ck−1 ]xk = 0

k=1

ck−1 , k = 1, 2, 3, . . . c2 = 0; (k + 2)(k + 1)ck+2 + ck−1 = 0; ck+2 = − (k + 2)(k + 1) c0 c1 c3 c0 c3 = − , c4 = − , c5 = 0, c6 = − = , 3·2 4·3 6·5 6·5·3·2 c4 c1 c7 = − = 7·6 7·6·4·3 c6 c0 c7 c1 c8 = 0. c9 = − =− , c10 = − =− 9·8 9·8·6·5·3·2 10 · 9 10   ·9·7·6·4·3 1 3 1 1 y = c0 1 − x + x6 − x9 + · · · 3·2 6·5·3·2 9·8·6·5·3·2   1 4 1 1 7 10 + c1 x − x + x − x + ··· 4·3 7·6·4·3 10 · 9 · 7 · 6 · 4 · 3

320

CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS

20. (x − 1)

∞ X

n(n − 1)cn xn−2 + 3

n=2

∞ X

cn xn

n=1

=

∞ X

(n)(n − 1)cn xn−1 −

=

∞ X

n(n − 1)cn xn−2 +3

n=2

n=2

|

∞ X

{z

}

k=n−1

(k + 1)kck+1 xk −

k=1

= 3c0 − 2c2 +

| ∞ X

cn xn

n=0

{z

k=n−2

}

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

k=0 ∞ X

∞ X

∞ X

ck x k

k=0

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

k=1

3c0 ; 3c0 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck = 0; c2 = 2 kck+1 3ck c2 3c1 c0 c1 2c3 3c2 ck+2 = + , k = 1, 2, 3, . . . ; c3 = + = + , c4 = + = k + 2 (k + 2)(k + 1) 3 3·2 2 2 4 4·3 c0 c1 3c0 5c0 c1 + + = + , 4 4 8 8 4 3c4 3c3 3c0 3c1 3c0 3c1 9c0 9c1 c5 = + = + + + = + 5 5·4 8 20 40  40  20 40  1 3 1 4 3 3 1 3 5 4 y = c0 1 + x + x + x + · · · + c1 x + x + x + · · · 2 2 8 2 4 21. The differential equation is mx00 + 4x0 + 2x = 0. The solutions of the auxiliary equation are √ √ 1 1 (−4 ± 16 − 8m) = (−2 ± 4 − 2m). 2m m The motion will be non-oscillatory when 4 − 2m ≥ 0 or 0 < m ≤ 2. 22. Substituting xp = αA into the differential equation we obtain ω 2 αA = A, so α = 1/ω 2 and xp = A/ω 2 . 1 00 x + x0 + 3x = e−t ; x(0) = 2, 8 x0 (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, √ √ 8 x(t) = e−4t (C1 cos 2 2t + C2 sin 2 2t) + e−t 17

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

√ √ √ √ 8 and x0 (t) = e−4t [(2 2C2 − 4C1 ) cos 2 2t − (2 2C1 − 4C2 ) sin 2 2t] − e−t . 17 √ 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   √ √ 28 √ 8 26 cos 2 2t + 2 sin 2 2t + e−t . x(t) = e−4t 17 17 17

CHAPTER 16 IN REVIEW

321

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 mx00 + kxp= 0 then becomes x00 + (4g/3)x = 0. The p sin 2 g/3t. The initial conditions x(0) = 1 and solution is x(t) = C1 cos 2 g/3t + C2 √ x0 (0) = 2/3 imply C1 = 1 and C2 = 1/ 3g. (b) To find the maximum speed of the weight we compute r r r r g 4 g g 2 g 2p 0 3g + 1. x (t) = 2 sin 2 + cos 2 t and |x (t)| = 4 + = 3 3 3 3 3 9 3 0

25. The auxiliary equation is m2 /4 + m + 1 = 0 or (m + 2)2 = 0, so m = −2, −2 and x(t) = C1 e−2t + C2 te−2t and x0 (t) = −2C1 e−2t − 2C2 te−2t + C2 e−2t . Using the initial conditions, we obtain 4 = C1 and 2 = −2C1 +C2 . Thus, C1 = 4 and C2 = 10. Therefore x(t) = 4e−2t +10te−2t and x0 (t) = 2e−2t − 20te−2t . Setting x0 (t) = 0 we obtain the critical point t = 1/10. The maximum vertical displacement is x(1/10) = 5e−0.2 ≈ 4.0937.