Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.4 Separable Differential Equations - Problems - Page 43: 10

Answer

$y = k(16-x^2)^2 + 2$

Work Step by Step

Given that: $\frac{dy}{dx} = \frac{x^2y - 32}{16 - x^2} + 2$ Add terms on the right side into a single fraction and simplify: $$\frac{dy}{dx} = \frac{x^2y -32}{16 - x^2} + \frac{2(16 - x^2)}{16 - x^2}$$ $$\frac{dy}{dx} = \frac{x^2y -32 + 32 - 2x^2}{16 - x^2}$$ $$\frac{dy}{dx} = \frac{x^2y - 2x^2}{16 - x^2}$$ $$\frac{dy}{dx} = \frac{x^2(y - 2)}{16 - x^2}$$ Separate and integrate: $$\frac{dy}{y-2} = \frac{x^2}{16 - x^2}dx$$ $$\int{\frac{dy}{y-2}} = \int{\frac{x^2}{16 - x^2}}dx$$ $$\ln{|y-2|} = \int{\frac{x^2}{(4-x)(4+x)}}dx$$ $$\ln{|y-2|} = \int{\frac{2}{4-x}+\frac{2}{4+x}}dx$$ $$\ln{|y-2|} = 2\ln{|4-x|}+2\ln{|4+x|} +C$$ $$\ln{|y-2|} = \ln{(16-x^2)^2} +C$$ Solve for $y$: $$y-2 = e^{\ln{(16-x^2)^2} +C}$$ $$y-2 = (16-x^2)^2 \cdot e^C$$ Let $e^C = k$: $$y-2 = k(16-x^2)^2$$ $$y = k(16-x^2)^2 + 2$$
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