373_pdfsam_math 54 differential equation solutions odd

373_pdfsam_math 54 differential equation solutions odd - (...

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Exercises 6.3 29. In operator form, the equation becomes ( D 3 2 D 2 + D ) [ z ]= D ( D 1) 2 [ z ]= x e x . (6.12) Solving the corresponding auxiliary equation, r ( r 1) 2 = 0, we Fnd that r =0 ,1 ,and1 . Thus z h ( x )= C 1 + C 2 e x + C 3 xe x is a general solution to the homogeneous equation associated with the original equation. To annihilate the right-hand side in (6.12), we apply the operator D 2 ( D 1) to this equation. Thus we obtain D 2 ( D 1) D ( D 1) 2 [ z ]= D 2 ( D 1) [ x e x ] D 3 ( D 1) 3 =0 . Solving the corresponding auxiliary equation, r 3 ( r 1) 3 =0 ,weseethat r =0and r =1are its roots of multiplicity three. Hence, a general solution is given by z ( x )= c 1 + c 2 x + c 3 x 2 + c 4 e x + c 5 xe x + c 6 x 2 e x . This general solution, when compared with z h ( x ), gives z p ( x )= c 2 x + c 3 x 2 + c 6 x 2 e
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Unformatted text preview: ( D 3 + 2 D 2 9 D 18 ) [ y ] = 18 x 2 18 x + 22 . (6.13) Since, D 3 + 2 D 2 9 D 18 = D 2 ( D + 2) 9( D + 2) = ( D + 2) ( D 2 9 ) = ( D + 2)( D 3)( D + 3) , (6.13) becomes ( D + 2)( D 3)( D + 3)[ y ] = 18 x 2 18 x + 22 . The auxiliary equation in this problem is ( r + 2)( r 3)( r + 3) = 0 with roots r = 2, 3, and 3. Hence, a general solution to the corresponding homogeneous equation has the form y h ( x ) = c 1 e 2 x + c 2 e 3 x + c 3 e 3 x . 369...
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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.

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