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Unformatted text preview: Chapter 6
whose roots are r = 1, −1, and 3. Therefore, the functions y1 (t) = et , y2 (t) = e−t , and y3 (t) = e3t form a fundamental solution set. Substituting back et = x we ﬁnd that y1 (x) = et = x , y2 (x) = e−t = (et )
−1 = x−1 , y3 (x) = e3t = (et ) = x3 form a fundamental solution set for the homogeneous equation (6.25). Next, we apply the variation of parameters to ﬁnd a particular solution to the original equation. A particular solution has the form yp (x) = v1 (x)x + v2 (x)x−1 + v3 (x)x3 . (6.26) 3 To ﬁnd functions v1 (x), v2 (x), and v3 (x) we use formula (11) on page 340 of the text. We compute x W [x, x−1 , x3 ](x) = 0 x−1 2x−3 x3 =x x−1 x3
2 1 −x−2 3x2 6x −x−2 3x2 2x−3 x3 6x − x−1 x3 2x−3 6x = −16 , W1 (x) = (−1)3−1 W [x−1 , x3 ](x) = W2 (x) = (−1)
3−2 −x−2 3x2 x 1 3x x = 4x , W [x, x ](x) = − 3 = −2x3 , = −2x−1 . W3 (x) = (−1)3−3 W [x, x−1 ](x) = x−1 1 −x−2 Also, writing the given equation in standard form, y− 3 3 y + 3 y = x cos x, 2 x x we see that the nonhomogeneous term is g (x) = x cos x. Thus, by (11), v1 (x) = 382 1 x cos x(4x) dx = − −16 4 x2 cos x dx = − 12 x sin x + 2x cos x − 2 sin x + c1 , 4 ...
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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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