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hw5solns

# hw5solns - Section 3.2 3{e2x xe2x By definition c1 e2x c2...

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Section 3.2 3. { e 2 x , xe 2 x } . By definition, c 1 e 2 x + c 2 xe 2 x = 0 e 2 x ( c 1 + c 2 x ) = 0 c 1 + c 2 x = 0 c 1 = c 2 = 0 linearly independent Using the Wronskian, det e 2 x xe 2 x 2 e 2 x e 2 x (1 + x ) x =0 = det 1 0 2 1 = 1 linearly independent 7. Using the Wronskian, W ( x ) = det sin 2 x cos 2 x 2 cos 2 x - 2 sin 2 x W (0) = det 0 1 2 0 = - 2 = 0 . Therefore { sin 2 x, cos 2 x } is linearly independent. 8. c 1 ( x 3 - 4) + c 2 x + c 3 3 x = 0. Grouping like terms and choosing c 1 = 0, c 2 = - 3, and c 3 = 1 shows the set is linearly dependent. 22. (a) On [0 , 1], | x | = x . Then c 1 x + c 2 x = 0 when c 1 = - c 2 shows the set is linearly dependent. (b) On [ - 1 , 0], | x | = - x . Then c 1 x + c 2 ( - x ) when c 1 = c 2 shows the set is linearly dependent. (c) On [ - 1 , 1], we need c 1 x + c 2 | x | = 0. This can only happen when c 1 = - c 2 and when c 1 = c 2 . These are both true only if c 1 = c 2 = 0. Thus the set is linearly independent on [ - 1 , 1]. (d) W ( x ) = det x - x 1 - 1 = 0 on [ - 1 , 0] W ( x ) = det x x 1 1 = 0 on [0 , 1] 29. There are two functions and the equation is 2nd order. We have ( e - 2 x ) - ( e - 2 x ) - 6( e - 2 x ) = 4 e - 2 x + 2 e - 2 x - 6 e - 2 x = 0 and ( e 3 x ) - ( e 3 x ) - 6( e 3

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hw5solns - Section 3.2 3{e2x xe2x By definition c1 e2x c2...

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