AEM_3e_Chapter03

# 17 from 3 m 3 10 m 2 15 m 4 0 we obtain m 1 3 and m 3

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17. From 3 m 3 + 10 m 2 + 15 m + 4 = 0 we obtain m = 1 / 3 and m = 3 / 2 ± ( 7 / 2) i so that y = c 1 e x/ 3 + e 3 x/ 2 c 2 cos 7 2 x + c 3 sin 7 2 x . 18. From 2 m 4 + 3 m 3 + 2 m 2 + 6 m 4 = 0 we obtain m = 1 / 2, m = 2, and m = ± 2 i so that y = c 1 e x/ 2 + c 2 e 2 x + c 3 cos 2 x + c 4 sin 2 x. 19. Applying D 4 to the differential equation we obtain D 4 ( D 2 3 D + 5) = 0. Then y = e 3 x/ 2 c 1 cos 11 2 x + c 2 sin 11 2 x y c + c 3 + c 4 x + c 5 x 2 + c 6 x 3 and y p = A + Bx + Cx 2 + Dx 3 . Substituting y p into the differential equation yields (5 A 3 B + 2 C ) + (5 B 6 C + 6 D ) x + (5 C 9 D ) x 2 + 5 Dx 3 = 2 x + 4 x 3 . Equating coeﬃcients gives A = 222 / 625, B = 46 / 125, C = 36 / 25, and D = 4 / 5. The general solution is y = e 3 x/ 2 c 1 cos 11 2 x + c 2 sin 11 2 x 222 625 + 46 125 x + 36 25 x 2 + 4 5 x 3 . 189

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CHAPTER 3 REVIEW EXERCISES 20. Applying ( D 1) 3 to the differential equation we obtain ( D 1) 3 ( D 2 D + 1) = ( D 1) 5 = 0. Then y = c 1 e x + c 2 xe x y c + c 3 x 2 e x + c 4 x 3 e x + c 5 x 4 e x and y p = Ax 2 e x + Bx 3 e x + Cx 4 e x . Substituting y p into the differential equation yields 12 Cx 2 e x + 6 Bxe x + 2 Ae x = x 2 e x . Equating coeﬃcients gives A = 0, B = 0, and C = 1 / 12. The general solution is y = c 1 e x + c 2 xe x + 1 12 x 4 e x . 21. Applying D ( D 2 + 1) to the differential equation we obtain D ( D 2 + 1)( D 3 5 D 2 + 6 D ) = D 2 ( D 2 + 1)( D 2)( D 3) = 0 . Then y = c 1 + c 2 e 2 x + c 3 e 3 x y c + c 4 x + c 5 cos x + c 6 sin x and y p = Ax + B cos x + C sin x . Substituting y p into the differential equation yields 6 A + (5 B + 5 C )cos x + ( 5 B + 5 C )sin x = 8 + 2sin x. Equating coeﬃcients gives A = 4 / 3, B = 1 / 5, and C = 1 / 5. The general solution is y = c 1 + c 2 e 2 x + c 3 e 3 x + 4 3 x 1 5 cos x + 1 5 sin x. 22. Applying D to the differential equation we obtain D ( D 3 D 2 ) = D 3 ( D 1) = 0. Then y = c 1 + c 2 x + c 3 e x y c + c 4 x 2 and y p = Ax 2 . Substituting y p into the differential equation yields 2 A = 6. Equating coeﬃcients gives A = 3. The general solution is y = c 1 + c 2 x + c 3 e x 3 x 2 . 23. The auxiliary equation is m 2 2 m + 2 = [ m (1 + i )][ m (1 i )] = 0, so y c = c 1 e x sin x + c 2 e x cos x and W = e x sin x e x cos x e x cos x + e x sin x e x sin x + e x cos x = e 2 x . Identifying f ( x ) = e x tan x we obtain u 1 = ( e x cos x )( e x tan x ) e 2 x = sin x u 2 = ( e x sin x )( e x tan x ) e 2 x = sin 2 x cos x = cos x sec x. Then u 1 = cos x , u 2 = sin x ln | sec x + tan x | , and y = c 1 e x sin x + c 2 e x cos x e x sin x cos x + e x sin x cos x e x cos x ln | sec x + tan x | = c 1 e x sin x + c 2 e x cos x e x cos x ln | sec x + tan x | . 24. The auxiliary equation is m 2 1 = 0, so y c = c 1 e x + c 2 e x and W = e x e x e x e x = 2 . 190
CHAPTER 3 REVIEW EXERCISES Identifying f ( x ) = 2 e x / ( e x + e x ) we obtain u 1 = 1 e x + e x = e x 1 + e 2 x u 2 = e 2 x e x + e x = e 3 x 1 + e 2 x = e x + e x 1 + e 2 x . Then u 1 = tan 1 e x , u 2 = e x + tan 1 e x , and y = c 1 e x + c 2 e x + e x tan 1 e x 1 + e x tan 1 e x .

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