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 - 1.8 Change of Variables - Problems - Page 79: 9

Answer

$y=x\tan (\ln|x|+C)$

Work Step by Step

Given: $$\frac{dy}{dx}=\frac{y^2+xy+x^2}{x^2}$$ $$\frac{dy}{dx}=(\frac{y}{x})^2+\frac{y}{x}+1$$ Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$ Subtituting: $$v+x\frac{dv}{dx}=v^2+v+1$$ $$x\frac{dv}{dx}=v^2+1$$ $$x\frac{dv}{v^2+1}=dx$$ $$\frac{dv}{v^2+1}=\frac{dx}{x}$$ Integrating left side: $$\int \frac{dv}{v^2+1}=\int \frac{dx}{x}+C$$ $$\arctan v=\ln|X| + C$$ $$\tan v=\tan (\ln|x|+C)$$ Subtitute when $y=vx \rightarrow y=x\tan (\ln|x|+C)$
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