Assume the hypothesis of Theorem 1 and assume that y 1 (x) and y 2 (x) are

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Assume the hypothesis of Theorem 1 and assume that y1(x) and y2(x) are both solutions to the linear first-order equation satisfying the initial condition y(x0) = y0.

a. Verify that y(x) = y1(x)– y2(x) satisfies the initial value problem:

y' + P(x) y = 0, y(x) = 0

b. For the integrating factor μ(x) defined by Equation (11.63), show that:

Equation 11.63

(x) = e/P(x) dx (11.63)

-(u(x) [y(x) - y(x)]) = 0 dx Conclude that u(x)[y (x) - y(x)] = constant.

c. From part (a), we have y1(x0)–y2(x0) =0. Since μ(x) > 0 forα 1(x) y2(x)≡ 0 on the interval; (α,β) Thus y1(x) = y2(x) for allα

Data from theorem 1

Suppose that P(x) and Q(x) are continuous functions over the interval a < x < . Then there is one and only

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A First Course In Mathematical Modeling

ISBN: 9781285050904

5th Edition

Authors: Frank R. Giordano, William P. Fox, Steven B. Horton

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