**Unformatted text preview: **= 3 B cos 3 t-3 A sin 3 t and y 00 p =-9 A cos 3 t-9 B sin 3 t . Adding everything up, we get y 00 p + 2 y p + y p = (-8 A + 6 B ) cos 3 t + (-6 A-8 B ) sin 3 t . This must all equal sin 3 t , so we have (-8 A +6 B ) = 0 and (-6 A-8 B ) = 1. From (-8 A +6 B ) = 0, we get B = 4 3 A . Substituting into the equation (-6 A-8 B ) = 1, we have-6 A-32 3 A = 1, or-50 3 A = 1, so A =-3 50 . Since B = 4 3 A , we have B =-4 50 . . Hence y p =-3 50 cos 3 t-4 50 sin 3 t , and y ( t ) = C 1 e-t + C 2 te-t-3 50 cos 3 t-4 50 sin 3 t. 1...

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- Spring '07
- lee
- Heterogeneity, Homogeneity