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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 ] = cos 2 x + 1 . The function g ( x ) = cos 2 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 differential 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 differential 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 5 u + 6 u = 0, associated with the original problem, has as its corresponding auxiliary equation
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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 ² = 0 366...
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