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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 = 0 366...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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