Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises - Page 626: 24

Answer

$y=x\ln\left(\frac{1}{C-\ln x}\right)$

Work Step by Step

Let $v=\frac{y}{x}$. Then, $y=xv$ $y'=v+v'x$ Substituting into the original differential equation, $xy'=y+xe^{y/x}$ $x(v+v'x)=xv+xe^v$ $xv+v'x^2=xv+xe^v$ $v'x^2=xe^v$ $v'x=e^v$ Solve for $v$: $\frac{dv}{dx}\cdot x=e^v$ (Separate the variables) $e^{-v}dv=\frac{1}{x}dx$ (Integrate) $\int e^{-v}dv=\int \frac{1}{x} dx$ $-e^{-v}=\ln x-C$ $e^v=\frac{1}{C-\ln x}$ (Take the logarithm) $v=\ln\left(\frac{1}{C-\ln x}\right)$ (Back Substitute $v=\frac{y}{x}$) $\frac{y}{x}=\ln\left(\frac{1}{C-\ln x}\right)$ $y=x\ln\left(\frac{1}{C-\ln x}\right)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.