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3 Higher-Order Differential Equations Exercises 3.1 1. From y = c1ex +c2ex we find y = c1ex c2ex. Then y(0) = c1 +c2 = 0, y(0) = c1 c2 = 1 so that c1 = 1/2 and c2 = 1/2. The solution is y = 1 2 ex 1 2 ex. 2. From y = c1e4x + c2ex we find y = 4c1e4x c2ex. Then y(0) = c1 + c2 = 1, y(0) = 4c 1 c2 = 2 so that c1 = 3/5 and c2 = 2/5. The solution is y = 3 5 e4x + 2 5 ex. 3. From y = c1x + c2x ln x we find y = c1 + c2(1 + lnx). Then y(1) = c1 = 3, y(1) = c1 + c2 = 1 so that c1 = 3 and c2 = 4. The solution is y = 3x 4x ln x. 4. From y = c1 + c2 cos x + c3 sin x we find y = c2 sin x + c3 cos x and y = c2 cos x c3 sin x. Then y() = c1 c2 = 0, y() = c3 = 2, y() = c2 = 1 so that c1 = 1, c2 = 1, and c3 = 2. The solu y = 1 cos x 2 sin x. 5. From y = c1 + c2x2 we find y = 2c2x. Then y(0) = c1 = 0, y(0) = 2c2 0 = 0 and y( 0) = 1 is not ossible. Since a2(x) = x is 0 at x = 0, Theorem 3.1 is not violated. 6. In this case we have y(0) = c1 = 0, y(0) = 2c2 0 = 0 so c1 = 0 and c2 is arbit rary. Two solutions are y = x2 and y = 2x2. 7. From x(0) = x0 = c1 we see that x(t) = x0 cos t + c2 sin t and x(t) = x0 sin t + c 2 cos t. Then x(0) = x1 = c2 implies c2 = x1/. Thus x(t) = x0 cos t + x1 sin t. 8. Solving the system x(t0) = c1 cos t0 + c2 sin t0 = x x (t0) = c1 sin t0 + c2 cos t0 = x1 for c1 and c2 gives c1 = x0 cos t0 x1 sin t0 and c2 = x1 cos t0 + x0 sin t0 . Thus x(t) = x0 cos t0 x1 sin t0 cos t + x1 cos t0 + x0 sin t0 sin t = x0(cos t cos t0 + sin t sin t0) + x1 (sin t cos t0 cos t sin t0) = x0 cos (t t0) + x1 sin (t t0). 9. Since a2(x) = x 2 and x0 = 0 the roblem has a unique solution for < x < 2. 10. Since a0(x) = tan x and x0 = 0 the roblem has a unique solution for /2 < x < / 2. 11. We have y(0) = c1 + c2 = 0, y(1) = c1e + c2e1 = 1 so that c1 = e/ e2 1

and c2 = e/ e2 1 . The solution is y = e (ex ex) / e2 1 . 12. In this case we have y(0) = c1 = 1, y(1) = 2c2 = 6 so that c1 = 1 and c2 = 3. The solution is y = 1+3x2. 77 Exercises 3.1 13. From y = c1ex cos x + c2ex sin x we find y = c1ex(sin x + cos x) + c2ex(cos x + sin x). (a) We have y(0) = c1 = 1, y(0) = c1+c2 = 0 so that c1 = 1 and c2 = 1. The solutio n is y = ex cos xex sin x. (b) We have y(0) = c1 = 1, y() = c1e = 1, which is not ossible. (c) We have y(0) = c1 = 1, y(/2) = c2e/2 = 1 so that c1 = 1 and c2 = e/2. The soluti on is y = ex cos x + e/2ex sin x. (d) We have y(0) = c1 = 0, y() = c1e = 0 so that c1 = 0 and c2 is arbitrary. Soluti ons are y = c2ex sin x, for any real numbers c2. 14. (a) We have y(1) = c1 + c2 + 3 = 0, y(1) = c1 + c2 + 3 = 4, which is not oss ible. (b) We have y(0) = c1 0 + c2 0 + 3 = 1, which is not ossible. (c) We have y(0) = c1 0 + c2 0 + 3 = 3, y(1) = c1 + c2 + 3 = 0 so that c1 is arb itrary and c2 = 3 c1. Solutions are y = c1x2 (c1 + 3)x4 + 3. (d) We have y(1) = c1 + c2 + 3 = 3, y(2) = 4c1 + 16c2 + 3 = 15 so that c1 = 1 and c2 = 1. The solution is y = x2 + x4 + 3. 15. Since (4)x + (3)x2 + (1)(4x 3x2) = 0 the functions are linearly de endent. 16. Since (1)0 + (0)x + (0)ex = 0 the functions are linearly de endent. A simila r argument shows that any set of functions containing f(x) = 0 will be linearly de endent. 17. Since (1/5)5 + (1) cos2 x + (1) sin2 x = 0 the functions are linearly de ende nt. 18. Since (1) cos 2x + (1)1 + (2) cos2 x = 0 the functions are linearly de endent . 19. Since (4)x + (3)(x 1) + (1)(x + 3) = 0 the functions are linearly de endent. 20. From the gra hs of f1(x) = 2+x and f2(x) = 2+ x we see that the functions are linearly inde endent since they cannot be multi les of each other. 21. The functions are linearly inde endent since W 1 + x, x, x2 = 1 +x x x2 1 1 2x 0 0 2 = 2 = 0. 22. Since (1/2)ex + (1/2)ex + (1) sinh x = 0 the functions are linearly de endent. 23. The functions satisfy the differential equation and are linearly inde endent since W

are linearly inde endent

are linearly inde endent + c2ex sin 2x. are linearly inde endent

ex/2, xex/2 = ex = 0 for < x < . The general solution is y = c1ex/2 + c2xex/2. 27. The functions satisfy the differential equation and are linearly inde endent since W x3, x4 = x6 = 0 for 0 < x < . The general solution is y = c1x3 + c2x4. 28. The functions satisfy the differential equation and are linearly inde endent since W (cos(ln x), sin(ln x)) = 1/x = 0 for 0 < x < . The general solution is y = c1 cos(ln x) + c2 sin(ln x). 29. The functions satisfy the differential equation and are linearly inde endent since W x, x 2, x 2 ln x = 9x 6 = 0 for 0 < x < . The general solution is y = c1x + c2x 2 + c3x 2 ln x. 30. The functions satisfy the differential equation and are linearly inde endent since W(1, x, cos x, sin x) = 1 for < x < . The general solution is y = c1 + c2x + c3 cos x + c4 sin x. 31. (a) From the gra hs of y1 = x3 and y2 = x 3 we see that

e 3x, e4x = 7ex = 0 for < x < . The general solution is y = c1e 3x + c2e4x. 24. The functions satisfy the differential equation and since W(cosh 2x, sinh 2x) = 2 for < x < . The general solution is y = c1 cosh 2x + c2 sinh 2x. 78 Exercises 3.1 25. The functions satisfy the differential equation and since W (ex cos 2x, ex sin 2x) = 2e2x = 0 for < x < . The general solution is y = c1ex cos 2x 26. The functions satisfy the differential equation and since W

the functions are linearly inde endent since they cannot be multi les of each other. It is easily shown that y1 = x3 solves x2y 4xy + 6y = 0. To show that y2 = x 3 is a solution let y2 = x3 for x 0 and let y2 = x3 for x < 0. (b) If x 0 then y2 = x3 and W(y1, y2) = x3 x3 3x2 3x2 = 0. If x < 0 then y2 = x3 and W(y1, y2) = x3 x3 3x2 3x2 = 0. This does not violate Theorem 3.3 since a2(x) = x2 is zero at x = 0. 79 Exercises 3.1 (c) The functions Y1 = x3 and Y2 = x2 are solutions of x2y 4xy + 6y = 0. They are l inearly inde endent since W x3, x2 = x4 = 0 for < x < . (d) The function y = x3 satisfies y(0) = 0 and y(0) = 0. (e) Neither is the general solution since we form a general solution on an inter val for which a2(x) = 0 for every x in the interval. 32. By the su er osition rinci le, if y1 = ex and y2 = ex are both solutions of a homogeneous linear differential equation, then so are 1 2 (y1 + y2) = ex + ex 2 = cosh x and 1 2 (y1 y2) = ex ex 2 = sinh x. 33. Since ex3 = e3ex = (e5e2)ex = e5ex+2, we see that ex3 is a constant multi le of e x+2 and the functions are linearly de endent. 34. Since 0y1 + 0y2 + + 0yk + 1yk+1 = 0, the set of solutions is linearly de end ent. 35. The functions y1 = e2x and y2 = e5x form a fundamental set of solutions of t he homogeneous equation, and y = 6ex is a articular solution of the nonhomogeneous equation. 36. The functions y1 = cos x and y2 = sin x form a fundamental set of solutions of the homogeneous equation, and y = x sin x + (cos x) ln(cos x) is a articular solution of the nonhomogeneous equation. 37. The functions y1 = e2x and y2 = xe2x form a fundamental set of solutions of the homogeneous equation, and y = x2e2x + x 2 is a articular solution of the nonhomogeneous equation. 38. The functions y1 = x1/2 and y2 = x1 form a fundamental set of solutions of the homogeneous equation, and

y = 1 15x2 1 6x is a articular solution of the nonhomogeneous equation. 39. (a) We have y 1 = 6e2x and y 1 = 12e2x, so y 1 6y

1 + 5y 1 = 12e2x 36e2x + 15e2x = 9e2x. Also, y 2 = 2x + 3 and y 2 = 2, so y 2 6y

2 + 5y 2 = 2 6(2x + 3) + 5(x2 + 3x) = 5x2 + 3x 16. (b) By the su er osition rinci le for nonhomogeneous equations a articular sol ution of y 6y + 5y = 5x2 + 3x 16 9e2x is y = x2 + 3x + 3e2x. A articular solution of the second equ ation is y = 2y 2 1 9y 1 = 2x2 6x 1 3e2x. 40. (a) y 1 = 5 (b) y 2 = 2x (c) y = y 1 + y 2 = 5 2x (d) y = 1 2y 1 2y 2 = 5 2 + 4x 80 Exercises 3.2 Exercises 3.2 In Problems 1 8 we use reduction of order to find a second solution. In Problems 9 16 we use formula (5) from the text. 1. Define y = u(x)e2x so y = 2ue2x + u e2x, y = e2xu + 4e2xu + 4e2xu, and y 4y + 4y = 4e2xu = 0. Therefore u = 0 and u = c1x + c2. Taking c1 = 1 and c2 = 0 we see that a second so lution is y2 = xe2x. 2. Define y = u(x)xex so y = (1 x)e xu + xe xu

, y = xe xu + 2(1 x)e xu (2 x)e xu, and y + 2y + y = e x(xu + 2u ) = 0 or u + 2 x u = 0. If w = u we obtain the first order equation w + 2 x w = 0 which has the integrating factor e2 dx/x = x2. Now d dx [x2w] = 0 gives x2w = c. Therefore w = u = c/x2 and u = c1/x. A second solution is y2 = 1 x xe x = e x. 3. Define y = u(x) cos 4x so y = 4u sin 4x + u cos 4x, y = u cos 4x 8u sin 4x 16u cos 4x and y + 16y = (cos 4x)u 8(sin 4x)u = 0 or u 8(tan 4x)u = 0. If w = u we obtain the first order equation w8(tan 4x)w = 0 which has the integrati ng factor e 8 tan 4xdx = cos2 4x. Now d dx [(cos2 4x)w] = 0 gives (cos2 4x)w = c. Therefore w = u = c sec2 4x and u = c1 tan 4x. A second solution is y2 = tan 4x c

os 4x = sin 4x. 4. Define y = u(x) sin 3x so y = 3u cos 3x + u sin 3x, y = u sin 3x + 6u cos 3x 9u sin 3x, and y + 9y = (sin 3x)u + 6(cos 3x)u = 0 or u + 6(cot 3x)u = 0. If w = u we obtain the first order equation w + 6(cot 3x)w = 0 which has the integ rating factor e6 cot 3xdx = sin2 3x. Now d dx [(sin2 3x)w] = 0 gives (sin2 3x)w = c. Therefore w = u = c csc2 3x and u = c1 cot 3x. A second solution is y2 = cot 3x s in 3x = cos 3x. 5. Define y = u(x) cosh x so y = u sinh x + u cosh x, y = u cosh x + 2u sinh x + u cosh x and y y = (cosh x)u + 2(sinh x)u = 0 or u + 2(tanh x)u = 0. 81 Exercises 3.2 If w = u we obtain the first order equation w +2(tanh x)w = 0 which has the integr ating factor e2 tanhxdx = cosh2 x. Now d dx [(cosh2 x)w] = 0 gives (cosh2 x)w = c. Therefore w = u = c sech2 x and u = c1 tanh x. A second solution is y2 = tanh x c osh x = sinh x. 6. Define y = u(x)e5x so y = 5e5xu + e5xu , y = e5xu + 10e5xu + 25e5xu and

y 25y = e5x(u + 10u ) = 0 or u + 10u = 0. If w = u we obtain the first order e