391_pdfsam_math 54 differential equation solutions odd

391_pdfsam_math 54 differential equation solutions odd -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online