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Unformatted text preview: Review Problems
(e) Representing the given function as a linear combination, x2 − 2x + xe−x + (sin 2x) − (cos 3x), we ﬁnd an annihilator for each term. Thus, we have: x2 − 2x is annihilated by D 3 , xe−x sin 2x cos 3x is annihilated by [D − (−1)]2 = (D + 1)2 (by (ii), page 334) , is annihilated by D 2 + 22 = D 2 + 4 is annihilated by D 2 + 32 = D 2 + 9 (by (iii), page 334) , (by (iii), page 334) . Therefore, the product D 3 (D + 1)2 (D 2 + 4)(D 2 + 9) annihilates the given function. 9. A general solution to the corresponding homogeneous equation, x3 y − 2x2 y − 5xy + 5y = 0, is given by yh (x) = c1 x + c2 x5 + c3 x−1 . We now apply the variation of parameters method described in Section 6.4, and seek for a particular solution to the original nonhomogeneous equation in the form yp (x) = v1 (x)x + v2 (x)x5 + v3 (x)x−1 . Since (x) = 1, (x5 ) = 5x4 , (x) = 0 , (x5 ) = 20x3 , (x−1 ) = −x−2 , (x−1 ) = 2x−3 , the Wronskian W [x, x5 , x−1 ](x) and determinants Wk (x) given in (10) on page 340 of the text become x W [x, x5 , x−1 ](x) = 1 x5
3 x−1 = (x) 2x
−3 5x4 −x−2 5x4 −x−2 20x3 2x−3 − (1) x5 x−1 0 20x x5 5x
4 20x3 2x−3 = (x)(30x) − (−18x2 ) = 48x2 , W1 (x) = (−1)3−1 x−1 −x
−2 = −6x3 , 387 ...
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