Chem Differential Eq HW Solutions Fall 2011 186

Chem Differential Eq HW Solutions Fall 2011 186 - A186...

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Unformatted text preview: A186 Appendix A Ordinary Differential Equations: Review of Concepts and Methods which implies (3). 49. 3 y + 13 y + 10 y = sin x, y1 = e−x . As in the previous exercise, let y1 = e−x , y = ve−x , y = v e−x − ve−x , y = v e−x − 2v e−x + ve−x . Then 3 y + 13 y + 10 y = sin x ⇒ 3(v e−x − 2v e−x + ve−x ) +13(v e−x − ve−x ) + 10ve−x = sin x ⇒ 3v + 7v = ex sin x 7 1 ⇒ v + v = ex sin x. 3 3 We now solve the first order o.d.e. in v : d 7x/3 v e dx e7x/3v v = 1 e7x/3 ex sin x 3 = 1 10x/3 sin x e 3 = 1 3 = 7 e7x/3 v + e7x/3v 3 1 e10x/3 10 ( sin x − cos) + C 3 ( 10 )2 + 1 3 3 = ex 9 (10 sin x − cos) + C. 109 3 1 10x/3 sin x dx e 3 (We used the table of integrals to evaluate the preceding integral. We will use it again below.) Integrating once more, y = 10 109 = v 10 ex 9 ex (sin x − cos x) − (cos x + sin x) + C 109 2 327 2 = vy1 = = − ex sin x dx − 9 327 ex cos x dx 10 9 (sin x − cos x) − (cos x + sin x) + Ce−x 218 654 13 7 cos x + sin x + Ce−x . 218 218 ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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