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Unformatted text preview: Exercises 6.4
Therefore, we have yp (x) = v1 (x) + v2 (x) cos x + v3 (x) sin x = ln(sec x) + cos2 x + sin2 x − sin x ln(sec x + tan x) ⇒ yp (x) = ln(sec x) − sin x ln(sec x + tan x) + 1. Since y ≡ 1 is a solution to the homogeneous equation, we may choose yp (x) = ln(sec x) − sin x ln(sec x + tan x). Note: We left the absolute value signs oﬀ ln(sec x) and ln(sec x + tan x) because of the stated domain: 0 < x < π/2. 7. First, we divide the diﬀerential equation by x3 to obtain the standard form y − 3x−1 y + 6x−2 y − 6x−3 y = x−4 , x > 0, from which we see that g (x) = x−4 . Given that {x, x2 , x3 } is a fundamental solution set for the corresponding homogeneous equation, we are looking for a particular solution of the form yp (x) = v1 (x)x + v3 (x)x2 + v3 (x)x3 . Evaluating determinants W [x, x2 , x3 ](x), W1 (x), W2 (x), and W3 (x) yileds x x2 W [x, x2 , x3 ](x) = 0 2 x3
2 (6.24) 1 2x 3x =x x2 2x 3x2 2 x3
2 6x 6x − x2 x3 2 6x = 2x3 , W1 (x) = (−1)3−1 W [x2 , x3 ](x) = W2 (x) = (−1)
3−2 2x 3x x = x4 , = −2x3 , W [x, x ](x) = − 3 x3
2 1 3x W3 (x) = (−1)3−3 W [x, x2 ](x) = x x2 1 2x = x2 . 379 ...
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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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