Unformatted text preview: equation.) DiFerentiating, we ±nd y p ( t ) = Ate − 2 t + Bte t ⇒ y p ( t ) = A (1 − 2 t ) e − 2 t + B ( t + 1) e t ⇒ y p ( t ) = A (4 t − 4) e − 2 t + B ( t + 2) e t . Substitution into (4.15) yields − 3 Ae − 2 t + 3 Be t = 6 e − 2 t + 3 e t ⇒ A = − 2 , B = 1 . Thus a general solution to (4.15) is given by y ( t ) = y h ( t ) + y p ( t ) = c 1 e t + c 2 e − 2 t − 2 te − 2 t + te t . 252...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Trigraph, general solution, dt dt d2, d2 y dy, dt dt x2

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