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 - 1.12 Chapter Review - Additional Problems - Page 110: 22

Answer

$y=\frac{\ln (e^{2x}+1)-x+C}{1+e^{2x}}$

Work Step by Step

We are given: $$\frac{dy}{dx}+\frac{2e^{2x}}{1+e^{2x}}y=\frac{1}{e^{2x}-1}$$ Integrating factor: $$I=e^{\int \frac{2e^{2x}}{1+e^{2x}}dx}$$ To solve this, let $$u=1+e^{2x} \rightarrow du=2e^{2x}$$ then $$I=e^{\ln u}=e^{\ln (1+e^{2x})}=1+e^{2x}$$ The equation becomes: $$\frac{d}{dx}[y(1+e^{2x})]=\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{2x}}{e^{2x+1}}-\frac{1}{e^{2x+1}}$$ Integrating both sides: $$y(1+e^{2x})=\frac{1}{2}\ln (e^{2x}+1)-[-\frac{1}{2}\ln(e^{2x}+1)+x]+C=\ln (e^{2x}+1)-x+C$$ Hence general solution is $y=\frac{\ln (e^{2x}+1)-x+C}{1+e^{2x}}$
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