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**Unformatted text preview: **Chapter 6
To ﬁnd functions vj ’s we need four determinants, the Wronskian W [y1 , y2 , y3](x) and W1 (x), W2 (x), and W3 (x) given in (10) on page 340 of the text. Thus we compute e−x W e , e , xe
−x 2x 2x e2x 2e
2x xe2x (1 + 2x)e e2x 2e
2x 2x 11 =e e e xe2x
−x 2x 2x x = 9e3x , (x) = −e −x −1 2 1 + 2x 1 4 4 + 4x e−x 4e2x (4 + 4x)e2x W1 (x) = (−1)3−1 W e2x , xe2x (x) = W2 (x) = (−1)
3−2 2x (1 + 2x)e e−x
−x = e4x , = −(1 + 3x)ex , W e , xe −x 2x (x) = − xe2x (1 + 2x)e
2x −e e−x W3 (x) = (−1)3−3 W e−x , e2x (x) = e2x −e−x 2e2x = 3ex . Substituting these expressions into the formula (11) for determining vj ’s, we obtain v1 (x) = v2 (x) = v3 (x) = g (x)W1 (x) dx = W [e−x , e2x , xe2x ] g (x)W2 (x) dx = W [e−x , e2x , xe2x ] g (x)W3 (x) dx = W [e−x , e2x , xe2x ] e2x e4x 1 3x e, dx = 3x 9e 27 −e2x (1 + 3x)ex 1 dx = − 3x 9e 9 2x x e 3e x dx = , 3x 9e 3 x x2 , (1 + 3x) dx = − − 9 6 where we have chosen zero integration constants. Then formula (12), page 340 of the text, gives a particular solution yp (x) = 1 3x −x ee − 27 x x2 + 9 6 e2x + x 2x 1 2x xe2x x2 e2x xe = e− + . 3 27 9 6 Note that the ﬁrst two terms in yp (x) are solutions to the corresponding homogeneous equation. Thus, another (and simpler) answer is yp (x) = x2 e2x /6. 3. Let us ﬁnd a fundamental solution set for the corresponding homogeneous equation, z + 3z − 4z = 0. Factoring the auxiliary polynomial, r 3 + 3r 2 − 4, yields r 3 + 3r 2 − 4 = r 3 − r 2 + 4r 2 − 4 = r 2 (r − 1) + 4(r + 1)(r − 1) = (r − 1)(r + 2)2 . 376 ...

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