Unformatted text preview: = 3 B cos 3 t3 A sin 3 t and y 00 p =9 A cos 3 t9 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 A8 B ) sin 3 t . This must all equal sin 3 t , so we have (8 A +6 B ) = 0 and (6 A8 B ) = 1. From (8 A +6 B ) = 0, we get B = 4 3 A . Substituting into the equation (6 A8 B ) = 1, we have6 A32 3 A = 1, or50 3 A = 1, so A =3 50 . Since B = 4 3 A , we have B =4 50 . . Hence y p =3 50 cos 3 t4 50 sin 3 t , and y ( t ) = C 1 et + C 2 tet3 50 cos 3 t4 50 sin 3 t. 1...
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 Spring '07
 lee
 Heterogeneity, Homogeneity

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