Unformatted text preview: r 2 − 5 r + 6 = ( r − 2)( r − 3) = 0. Therefore, the solution to the homogeneous equation associated with the original problem is u h ( x ) = c 1 e 2 x + c 2 e 3 x . Since a general solution to this original problem is given by u ( x ) = u h ( x ) + u p ( x ) = c 1 e 2 x + c 2 e 3 x + u p ( x ) and since u ( x ) must be of the form u ( x ) = c 1 e 2 x + c 2 e 3 x + c 3 cos 2 x + c 4 sin 2 x + c 5 , we see that u p ( x ) = c 3 cos 2 x + c 4 sin 2 x + c 5 . 23. The function g ( x ) = e 3 x − x 2 is annihilated by the operator A := D 3 ( D − 3). Applying the operator A to both sides of the diFerential equation given in this problem yields A [ y − 5 y + 6 y ] = A ± e 3 x − x 2 ² = 0 366...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Summation, Homogeneity, D2

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