Therefore f x and g x are linearly independent on 1 1

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Unformatted text preview: are linearly independent on ( 1, 1). b) Solution: We have g 0 (x) = 2x, while f 0 ( x) = ( 2x, for x > 0 , 2x, for x < 0 and at x = 0, we have f (0 + h) h! 0 h f 0 (0) = lim f (0) h |h| = lim |h| = 0. h! 0 h h! 0 = lim Now, W (x) = f g 0 f 0 g . When x > 0, this is 2x3 2x3 = 0. When x < 0, it is 2x3 + 2x3 = 0. When x = 0, it is also 0, so the Wronskian is zero for all x 2 ( 1, 1). Since they are linearly independent functions, Proposition 3.26 of the Course Notes guarantees that if f and g were solutions to a 2nd-order homogeneous linear DE, then their Wronskian would not be zero. Therefore they cannot be solutions to such an equation....
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This document was uploaded on 02/09/2014.

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