# Hw5 - 3.2.3 Consider = e-2t = te-2t = = e-2t-2t 1)e-2t te-2t-2e-2t = e-4t W = Wronskian 3.2.27 Suppose P(x)y Q(x)y R(x)y = 0 is exact Then(by

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Unformatted text preview: 3.2.3 Consider = e-2t , = te-2t . = - = e-2t (-2t + 1)e-2t - te-2t (-2e-2t ) = e-4t W {, } = Wronskian 3.2.27 Suppose P (x)y + Q(x)y + R(x)y = 0 is exact. Then (by definition), there exists a function f (x) such that 0 = P (x)y + Q(x)y + R(x)y = P (x)y + [f (x)y] . Expanding the right side, we see P (x)y + Q(x)y + R(x)y = P (x)y + (f (x) + P (x))y + f (x)y. Thus we must have Q(x) = f (x) + P (X) and R(x) = f (x). Differentiating the first equation, we have Q (x) = f (x) + P (x). Substituting in R(x) = f (x), we conclude Q (x) = R(x) + P (x), or P (x) - Q (x) + R(x) = 0. 1 ...
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## This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.

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Hw5 - 3.2.3 Consider = e-2t = te-2t = = e-2t-2t 1)e-2t te-2t-2e-2t = e-4t W = Wronskian 3.2.27 Suppose P(x)y Q(x)y R(x)y = 0 is exact Then(by

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