Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 8 - Mathematical Modeling With Differential Equations - 8.4 First-Order Differential Equations And Applications - Exercises Set 8.4 - Page 592: 5


$$y=\frac{c}{\sqrt{x^2+1}} $$

Work Step by Step

Given $$(x^2+1)\frac{dy}{dx} +xy=0$$ Rewriting the equation $$ \frac{dy}{dx} +\frac{x}{x^2+1}y=0$$ Since \begin{align*} \mu(x)&=e^{\int \frac{x}{x^2+1}dx}\\ &=e^{\frac{1}{2}\ln(x^2+1) }\\ &=e^{\ln \sqrt{x^2+1}}\\ &= \sqrt{x^2+1} \end{align*} Then \begin{align*} y\mu(x)&=\int \mu(x) q(x)dx\\ \sqrt{x^2+1}y&=\int 0 dx\\ &=c \end{align*} Hence $$y=\frac{c}{\sqrt{x^2+1}} $$
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