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: 21

Answer

$\rightarrow y^2x=Ce^{\frac{4x}{y}}$

Work Step by Step

Given: $$2x(y+2x)y'=y(4x-y)$$ $$\frac{dy}{dx}=\frac{y(4x-y)}{2x(y+2x)}$$ Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$ Subtituting: $$x\frac{dv}{dx}+v=\frac{(4-v)v}{2(v+2)}$$ $$x\frac{dv}{dx}=\frac{(4-v)v}{2(v+2)}-v$$ $$x\frac{dv}{dx}=\frac{-3v^2}{2v+4}$$ $$\frac{-2(v+2)}{3v^2}dv=\frac{1}{x}dx$$ Integrating left sides: $\int \frac{-2(v+2)}{3v^2}dv=\int \frac{1}{x}dx$$ $$-\frac{2}{3}\int \frac{1}{v}dv-\frac{4}{3}\int \frac{1}{v^2}dv=\int \frac{1}{x}dx$ $$-\frac{2}{3}\ln v +\frac{4}{3v}=\ln x +\ln C$$ where $C$ is a constant of integration. $$\ln v^2 +\ln x^3 +\ln C=\ln(Cv^2x^3)=\frac{4}{v}$$ Subtitute when $y=vx $ $$\rightarrow \ln(Cy^2x)=\frac{4x}{y}$$ $$\rightarrow y^2x=Ce^{\frac{4x}{y}}$$
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