## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 27

#### Answer

$$y(x)=\frac{1}{m+n} e^{m x}+C e^{-n x} ,\ \ \ n\neq -m$$ $$y= xe^{-nx}+Ce^{-nx},\ \ \ \ n=-m$$

#### Work Step by Step

Given$$y^{\prime}+n y=e^{mx}$$ This is a linear equation with $p(x) =n, \ \ q(x) =e^{mx}$, so \begin{align*} \mu(x)&=e^{\int p(x)dx}\\ &=e^{\int nd x}\\ &=e^{nx} \end{align*} Then \begin{align*} y\mu(x) &=\int \mu(x)q(x)dx\\ e^{nx} y &=\int e^{nx}e^{mx}dx\\ &=\int e^{(n+m)x}dx\\ &=\frac{1}{n+m}e^{(n+m)x}+C \end{align*} Then if $n\neq-m$ $$y(x)=\frac{1}{m+n} e^{m x}+C e^{-n x}$$ If $n=-m$ $$e^{nx} y = x+C\ \ \ \to \ \ y= xe^{-nx}+Ce^{-nx}$$

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