Unformatted text preview: side, yields D 3 ± ( D + 1) 2 + 1 ²± ( D + 1) 2 + 1 ² [ y ] = D 3 ± ( D + 1) 2 + 1 ²³ e − x cos x + x 2 ´ = 0 ⇒ D 3 ³ ( D + 1) 2 + 1 ´ 2 [ y ] = 0 . (6.11) The corresponding auxiliary equation, r 3 [( r + 1) 2 + 1] 2 = 0 has a root r = 0 of multiplicity three and double roots r = − 1 ± i . Therefore, a general solution to (6.11) is given by y ( x ) = ( c 1 cos x + c 2 sin x ) e − x + ( c 3 cos x + c 4 sin x ) xe − x + c 5 x 2 + c 6 x + c 7 . Since y ( x ) = y h ( x ) + y p ( x ), we conclude that y p ( x ) = ( c 3 cos x + c 4 sin x ) xe − x + c 5 x 2 + c 6 x + c 7 . 368...
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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