HW6 - Math 293 Fall 2005 Homework Solution 6 Section 3.2...

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Math 293 - Fall 2005 Homework Solution 6 - Section 3.2 Alexandre ROCH October 5, 2005 Problem. 5. (1)(17) + ( - 34)(cos 2 x ) + (17)(cos 2 x ) = 0, because 2 cos 2 x = 1 + cos 2 x Problem. 12. W = fl fl fl fl fl fl x cos(ln( x )) sin(ln( x )) 1 - sin(ln( x )) /x cos(ln( x )) /x 0 - cos(ln( x )) /x 2 + sin(ln( x )) /x 2 - sin(ln( x )) /x 2 - cos(ln( x )) /x 2 fl fl fl fl fl fl W = x - 2 [2 cos 2 (ln x ) + 2 sin 2 (ln x )] = 2 x - 2 is nonzero for x > 0 Problem. 18. y (0) = 1 = c 1 e x + c 2 e x cos x + c 3 e x sin x y 0 (0) = 0 = c 1 e x + c 2 ( e x cos x - e x sin x ) + c 3 ( e x sin x + e x cos x ) y 00 (0) = 0 = c 1 e x - 2 c 2 e x sin x + 2 c 3 e x cos x c 1 + c 2 = 1 , c 1 + c 2 + c 3 = 0 , c 1 + 2 c 3 = 0 c 1 = 2 , c 2 = - 1 , c 3 = - 1 y ( x ) = e x (2 - cos x - sin x ) Problem. 23. y (0) = 3 = c 1 e - x + c 2 e 3 x - 2 y 0 (0) = 11 = - c 1 e - x + 3 c 2 e 3 x c 1 + c 2 - 2 = 3 , - c 1 + 3 c 2 = 11 c 1 = 1 , c 2 = 4 y ( x ) = e - x + 4 e 3 x - 2 Problem. 30. We need to find a 1 x + a 2 x 2 = 0 x . 1 + a 2 a 1 x = 0 x = - a 1 a 2 1
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Hence it is not satisfied for all x. That is, x and x 2 are not linearly dependent. When the equation x 2 y 00 - 2 xy 0 + 2 y = 0 is rewritten in standard form y 00 + ( - 2 /x ) y 0 + (2 /x 2 ) y = 0 the coefficient functions
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