370_pdfsam_math 54 differential equation solutions odd

370_pdfsam_math 54 differential equation solutions odd - r...

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Chapter 6 21. In operator form, the given equation can be written as ( D 2 5 D +6 ) [ u ]=cos2 x +1 . The function g ( x )=cos2 x +1 is a sum of two functions: cos 2 x is of the type (iii) on page 334 of the text with β = 2, and so it is innihilated by ( D 2 + 4); 1, as a constant, is annihilated by D . Therefore, the operator D ( D 2 + 4) innihilates the right-hand side, g ( x ). Applying this operator to both sides of the diFerential equation given in this problem yields D ( D 2 +4 )( D 2 5 D +6 ) [ u ]= D ( D 2 +4 ) [cos 2 x +1]=0 D ( D 2 +4 ) ( D 3)( D 2)[ u ]=0 . This last equation has the associated auxiliary equation r ( r 2 +4)( r 3)( r 2) = 0, which has roots r =2 ,3 ,0 , ± 2 i . Thus, a general solution to the diFerential equation associated with this auxiliary equation is u ( x )= c 1 e 2 x + c 2 e 3 x + c 3 cos 2 x + c 4 sin 2 x + c 5 . The homogeneous equation, u 0 5 u 0 +6
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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.

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