Chapter 3 - 3 Higher-Order Differential Equations Exercises...

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3 Higher-Order Differential Equations Exercises 3.1 1. From y = c 1 e x + c 2 e x we find y = c 1 e x c 2 e x . Then y (0) = c 1 + c 2 = 0, y (0) = c 1 c 2 = 1 so that c 1 = 1 / 2 and c 2 = 1 / 2. The solution is y = 1 2 e x 1 2 e x . 2. From y = c 1 e 4 x + c 2 e x we find y = 4 c 1 e 4 x c 2 e x . Then y (0) = c 1 + c 2 = 1, y (0) = 4 c 1 c 2 = 2 so that c 1 = 3 / 5 and c 2 = 2 / 5. The solution is y = 3 5 e 4 x + 2 5 e x . 3. From y = c 1 x + c 2 x ln x we find y = c 1 + c 2 (1 + ln x ). Then y (1) = c 1 = 3, y (1) = c 1 + c 2 = 1 so that c 1 = 3 and c 2 = 4. The solution is y = 3 x 4 x ln x . 4. From y = c 1 + c 2 cos x + c 3 sin x we find y = c 2 sin x + c 3 cos x and y = c 2 cos x c 3 sin x . Then y ( π ) = c 1 c 2 = 0, y ( π ) = c 3 = 2, y ( π ) = c 2 = 1 so that c 1 = 1, c 2 = 1, and c 3 = 2. The solution is y = 1 cos x 2sin x . 5. From y = c 1 + c 2 x 2 we find y = 2 c 2 x . Then y (0) = c 1 = 0, y (0) = 2 c 2 · 0 = 0 and y (0) = 1 is not possible. Since a 2 ( x ) = x is 0 at x = 0, Theorem 3 . 1 is not violated. 6. In this case we have y (0) = c 1 = 0, y (0) = 2 c 2 · 0 = 0 so c 1 = 0 and c 2 is arbitrary. Two solutions are y = x 2 and y = 2 x 2 . 7. From x (0) = x 0 = c 1 we see that x ( t ) = x 0 cos ωt + c 2 sin ωt and x ( t ) = x 0 sin ωt + c 2 ω cos ωt . Then x (0) = x 1 = c 2 ω implies c 2 = x 1 . Thus x ( t ) = x 0 cos ωt + x 1 ω sin ωt. 8. Solving the system x ( t 0 ) = c 1 cos ωt 0 + c 2 sin ωt 0 = x x ( t 0 ) = c 1 ω sin ωt 0 + c 2 ω cos ωt 0 = x 1 for c 1 and c 2 gives c 1 = ωx 0 cos ωt 0 x 1 sin ωt 0 ω and c 2 = x 1 cos ωt 0 + ωx 0 sin ωt 0 ω . Thus x ( t ) = ωx 0 cos ωt 0 x 1 sin ωt 0 ω cos ωt + x 1 cos ωt 0 + ωx 0 sin ωt 0 ω sin ωt = x 0 (cos ωt cos ωt 0 + sin ωt sin ωt 0 ) + x 1 ω (sin ωt cos ωt 0 cos ωt sin ωt 0 ) = x 0 cos ω ( t t 0 ) + x 1 ω sin ω ( t t 0 ) . 9. Since a 2 ( x ) = x 2 and x 0 = 0 the problem has a unique solution for −∞ < x < 2. 10. Since a 0 ( x ) = tan x and x 0 = 0 the problem has a unique solution for π/ 2 < x < π/ 2. 11. We have y (0) = c 1 + c 2 = 0, y (1) = c 1 e + c 2 e 1 = 1 so that c 1 = e/ ( e 2 1 ) and c 2 = e/ ( e 2 1 ) . The solution is y = e ( e x e x ) / ( e 2 1 ) . 12. In this case we have y (0) = c 1 = 1, y (1) = 2 c 2 = 6 so that c 1 = 1 and c 2 = 3. The solution is y = 1 + 3 x 2 . 77
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Exercises 3.1 13. From y = c 1 e x cos x + c 2 e x sin x we find y = c 1 e x ( sin x + cos x ) + c 2 e x (cos x + sin x ). (a) We have y (0) = c 1 = 1, y (0) = c 1 + c 2 = 0 so that c 1 = 1 and c 2 = 1. The solution is y = e x cos x e x sin x . (b) We have y (0) = c 1 = 1, y ( π ) = c 1 e π = 1, which is not possible. (c) We have y (0) = c 1 = 1, y ( π/ 2) = c 2 e π/ 2 = 1 so that c 1 = 1 and c 2 = e π/ 2 . The solution is y = e x cos x + e π/ 2 e x sin x . (d) We have y (0) = c 1 = 0, y ( π ) = c 1 e π = 0 so that c 1 = 0 and c 2 is arbitrary. Solutions are y = c 2 e x sin x , for any real numbers c 2 . 14. (a) We have y ( 1) = c 1 + c 2 + 3 = 0, y (1) = c 1 + c 2 + 3 = 4, which is not possible. (b) We have y (0) = c 1 · 0 + c 2 · 0 + 3 = 1, which is not possible. (c) We have y (0) = c 1 · 0 + c 2 · 0 + 3 = 3, y (1) = c 1 + c 2 + 3 = 0 so that c 1 is arbitrary and c 2 = 3 c 1 .
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