Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 17


$$y(x)=x^{x}+C x^{x} e^{-x}$$

Work Step by Step

Given $$y^{\prime}-(\ln x) y=x^{x}$$ This a linear equation with $p(x) = -\ln x$ and $q(x) =x^x$; then \begin{align*} \mu(x) &= e^{\int p(x)dx}\\ &=e^{\int -\ln xdx}\\ &=e^{-[x\ln x-x]}\\ &=e^{\ln x^{-x} +x}\\ &=x^{-x}e^x \end{align*} Hence, the general solution is \begin{align*} \mu(x) y &=\int \mu (x)xq(x)dx\\ x^{-x}e^xy&=\int x^{-x}e^x x^{x}dx\\ &=\int x^{-x}e^x x^{x}dx\\ &=\int e^{x}dx\\ &=e^{x}+C \end{align*} Hence $$y(x)=x^{x}+C x^{x} e^{-x} $$
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