384_pdfsam_math 54 differential equation solutions odd

384_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: Chapter 6 So, v1 (x) = v2 (x) = v3 (x) = g (x)W1 (x) dx = W [x, x2 , x3 ](x) g (x)W2 (x) dx = W [x, x2 , x3 ](x) g (x)W3 (x) dx = W [x, x2 , x3 ](x) x−4 x4 1 dx = − 2 + c1 , 3 2x 4x −4 3 1 x (−2x ) dx = 3 + c2 , 3 2x 3x −4 2 1 x (x ) dx = − 4 + c3 , 3 2x 8x where c1 , c2 , and c3 are constants of integration. Substitution back into (6.24) yields yp (x) = − 1 + c1 x + 4x2 1 1 1 + c1 x + c2 x2 + c3 x3 . + c2 x2 + − 4 + c3 x3 = − 3 3x 8x 24x Since {x, x2 , x3 } is a fundamental solution set for the homogeneous equation, taking c1 , c2 , and c3 to be arbitrary constants, we obtain a general solution to the original nonhomogeneous equation. That is, y (x) = − 1 + c1 x + c2 x2 + c3 x3 . 24x 9. To find a particular solution to the nonhomogeneous equation, we will use the method of variation of parameters. We must first calculate the four determinants W [ex , e−x , e2x ](x), W1 (x), W2 (x), W3 (x). Thus, we have ex W [ex , e−x , e2x ](x) = e−x ex 0 W1 (x) = 1 e2x = (−1)3−1 0 −e−x 2e2x e−x 4e2x e2x = (−1)3−2 ex e x e−x e2x = −4e2x + 2e2x + e2x + e2x − 2e2x − 4e2x = −6e2x , ex −e−x 2e2x e−x 4e2x e−x e2x −e−x 2e2x e2x 2e 2x = 2ex + ex = 3ex , ex 0 W2 (x) = ex 0 2e2x ex 1 4e2x ex e−x 0 −e −x −x = − 2e3x − e3x = −e3x , W3 (x) = e e x x 0 1 = (−1)3−3 ex e−x e ex −e−x = −1 − 1 = −2. 380 ...
View Full Document

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.

Ask a homework question - tutors are online