Chapter 3

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 0 = c 1 e x c 2 e x . Then y (0) = c 1 + c 2 =0, y 0 (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 0 =4 c 1 e 4 x c 2 e x . Then y (0) = c 1 + c 2 , y 0 (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 0 = c 1 + c 2 (1+ln x ). Then y (1) = c 1 =3, y 0 (1) = c 1 + c 2 = 1 so that c 1 and c 2 = 4. The solution is y x 4 x ln x . 4. From y = c 1 + c 2 cos x + c 3 sin x we find y 0 = c 2 sin x + c 3 cos x and y 0 = c 2 cos x c 3 sin x . Then y ( π )= c 1 c 2 =0 , y 0 ( π c 3 , y 0 ( 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 0 c 2 x . Then y (0) = c 1 , y 0 (0)=2 c 2 · 0 = 0 and y 0 (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 , y 0 c 2 · 0=0so c 1 = 0 and c 2 is arbitrary. Two solutions are y = x 2 and y x 2 . 7. From x (0) = x 0 = c 1 we see that x ( t x 0 cos ωt + c 2 sin and x 0 ( t x 0 sin + c 2 ω cos . Then x 0 (0) = x 1 = c 2 ω implies c 2 = x 1 .Thu s x ( t x 0 cos + x 1 ω sin ωt. 8. Solving the system x ( t 0 c 1 cos 0 + c 2 sin 0 = x x 0 ( t 0 c 1 ω sin 0 + c 2 ω cos 0 = x 1 for c 1 and c 2 gives c 1 = ωx 0 cos 0 x 1 sin 0 ω and c 2 = x 1 cos 0 + 0 sin 0 ω . Thus x ( t 0 cos 0 x 1 sin 0 ω cos + x 1 cos 0 + 0 sin 0 ω sin = x 0 (cos cos 0 + sin sin 0 )+ x 1 ω (sin cos 0 cos sin 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 , y 0 (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 0 (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 0 = 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 (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) y (0) = c 1 y ( π )= c 1 e π = 1, which is not possible. (c) y (0) = c 1 =1 , y ( π/ 2) = c 2 e 2 = 1 so that c 1 = 1 and c 2 = e 2 . The solution is y = e x cos x + e 2 e x sin x . (d) 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) y ( 1) = c 1 + c 2 +3=0, y (1) = c 1 + c 2 + 3 = 4, which is not possible. (b) y (0) = c 1 · 0+ c 2 · 0 + 3 = 1, which is not possible. (c) y (0) = c 1 · 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 . Solutions are y = c 1 x 2 ( c 1 +3) x 4 +3. (d) y (1) = c 1 + c 2 +3=3, y (2)=4 c 1 +16 c 2 + 3 = 15 so that c 1 = 1 and c 2 = 1. The solution is y = x 2 + x 4 15. Since ( 4) x + (3) x 2 + (1)(4 x 3 x 2 ) = 0 the functions are linearly dependent.
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Chapter 3 - 3 Higher-Order Differential Equations Exercises...

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