AMath350.W13.A4.Sol

# Use the wronskian to verify that they are linearly

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Unformatted text preview: that they are linearly independent. Sol’n: We ﬁnd that W ( y 1 , y2 , y3 ) = x x3 x4 2 1 3x 4 x3 0 6x 12x2 x3 x4 3 x2 4 x3 = = x6 Since the diﬀerential equation satisﬁes the conditions of the Existence/Uniqueness Theorem for all x 6= 0 , and since W (x) 6= 0 for all such x, we can conclude that the given functions are linearly independent on (0, 1), and also on ( 1, 0). 6. Let y1 be the solution on (0, 1) of the IVP x2 y 00 + y 0 + xy = 0; y (1) = 1, y 0 (1) = 1 and let y2 be the solution on (0, 1) of the IVP x2 y 00 + y 0 + xy = 0; y (1) = 0, y 0 (1) = 1. a) Verify that {y1 , y2 } is a fundamental set of solutions of x2 y 00 +y 0 +xy = 0 on (0, 1) (that is, show that they are linearly independent on the interval). Sol’n: We know that y1 y2 W ( x) = 0 0 y1 y2 and so W (1) = y1 (1) y2 (1) 0 0 y1 (1) y2 (1) = 1 1 0 1 = 1 6= 0. Since y1 and y2 are solutions to a homogeneous linear diﬀerential equation, the fact that their Wronskian is nonzero at x = 1 means that their Wronskian is nonzero on the entire interval of existence (0, 1), and so we can conclude that they are linearly independent functions, and form a fundamental solution set. b) Let y3 be the solution o...
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## This document was uploaded on 02/09/2014.

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