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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 ...
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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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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