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Unformatted text preview: Exercises 6.4
Therefore, r = 1, −2, and −2 are the roots of the auxiliary equation, and so the functions z1 = ex , z2 = e−2x , and z3 = xe−2x form a fundamental solution set. A particular solution then has the form zp (x) = v1 (x)z1 (x) + v2 (x)z2 (x) + v3 (x)z3 (x) = v1 (x)ex + v2 (x)e−2x + v3 (x)xe−2x . (6.23) To ﬁnd functions vj ’s we need four determinants, the Wronskian W [z1 , z2 , z3 ](x) and W1 (x), W2 (x), and W3 (x) given in (10) on page 340 of the text. Thus we compute ex W e ,e
x −2x e−2x xe−2x =e
−3x 1 1 xe−2x
−2x 1 x = 9e−3x , , xe −2x (x) = ex −2e−2x (1 − 2x)e−2x ex 4e−2x (4x − 4)e−2x e−2x −2e ex e
x −2x 1 −2 1 − 2x 4 4x − 4 W1 (x) = (−1)3−1 W e−2x , xe−2x (x) = W2 (x) = (−1)
3−2 (1 − 2x)e xe−2x
−2x = e−4x , W e , xe x −2x (x) = − ex e
x (1 − 2x)e e−2x
−2x = (3x − 1)e−x , W3 (x) = (−1)3−3 W ex , e−2x (x) = −2e = −3e−x . Substituting these expressions into the formula (11) on page 340 of the text, we obtain v1 (x) = v2 (x) = g (x)W1 (x) dx = W [ex , e−2x , xe−2x ] g (x)W2 (x) dx = W [ex , e−2x , xe−2x ] = v3 (x) = g (x)W3 (x) dx = W [ex , e−2x , xe−2x ] 1 9 e2x e−4x 1 dx = ex , −3x 9e 9 2x −x e (3x − 1)e dx 9e−3x x 7 − e4x , 12 144 e2x (−3e−x ) 1 dx = − e4x . −3x 9e 12 (3x − 1)e4x dx = Substituting these expressions into (6.23) yields zp (x) = 1 xx ee + 9 x 7 1 4x −2x 1 2x − e xe e. = e4x e−2x − 12 144 12 16 377 ...
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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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