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**Unformatted text preview: **y p ( t ) A . Substitution into the equation yields ( A ) + 2( A ) + 5( A ) = 10 5 A = 10 A = 2 , and so, on [0 , 3 / 2], a general solution to the original equation is y 1 ( t ) = ( c 1 cos 2 t + c 2 sin 2 t ) e t + 2 . We nd c 1 and c 2 by substituting this function into the initial conditions. 0 = y 1 (0) = c 1 + 2 , 0 = y 1 (0) = c 1 + 2 c 2 c 1 = 2 , c 2 = 1 y 1 ( t ) = (2 cos 2 t + sin 2 t ) e t + 2 . 206...

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