## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 9: First-Order Differential Equations - Section 9.5 - Systems of Equations and Phase Planes - Exercises 9.5 - Page 557: 14

#### Answer

$y=2xe^{x}-e^{x}+ce^{-x}$

#### Work Step by Step

Here, we have $y'+y=4(xe^x)$ The integrating factor is: $I=e^{\int (1) dx}=e^{x}$ Now, $(e^{x})[y'+y]=4(xe^x)(e^{x})$ This implies that $\int [e^{x}y]' =\int 4xe^{2x} dx$ $\implies e^{x}y=(2x)(e^{2x})-e^{x}+c$ Hence, $y=(2x)(e^{x})-e^{x}+ce^{-x}$

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