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: 18


$$ y= \frac{1}{2}\left(\sin x+\cos x\right)+Ce^{-x}$$

Work Step by Step

Given $$y^{\prime}+ y=\cos{x}$$ This a linear equation with $p(x) =1$ and $q(x) =\cos x$, then \begin{align*} \mu(x) &= e^{\int p(x)dx}\\ &=e^{\int dx}\\ &= e^x \end{align*} Hence the general solution is \begin{align*} \mu(x) y &=\int \mu (x)xq(x)dx\\ e^x y&=\int e^x \cos {x}dx \end{align*} Use $$\int e^{a u} \cos b u d u=\frac{e^{a u}}{a^{2}+b^{2}}(a \cos b u+b \sin b u)+C $$ We get \begin{align*} e^x y&=\int e^x \cos {x}dx \\ &= \frac{1}{2}e^x\left(\sin x+\cos x\right)+C \end{align*} Hence $$ y= \frac{1}{2}\left(\sin x+\cos x\right)+Ce^{-x}$$
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