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373_pdfsam_math 54 differential equation solutions odd

# 373_pdfsam_math 54 differential equation solutions odd - D...

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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 find that r = 0, 1, and 1. 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, we see that r = 0 and r = 1 are 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 x . 31. Writing this equation in operator form yields
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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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