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: 23


$$ y(x)=-\cos x+ \sin x$$

Work Step by Step

Given$$(\sin x) y^{\prime}=(\cos x) y+1, \quad y\left(\frac{\pi}{4}\right)=0$$ Then $$y^{\prime}-(\cot x) y=\csc x$$ This is a linear equation with $p(x) =-\cot x,\ \ q(x) =\csc x$, so \begin{align*} \mu(x)&=e^{\int p(x)dx}\\ &=e^{\int -\cot x dx}\\ &=e^{-\ln \sin x}\\ &=e^{\ln \csc x}\\ &=\csc x \end{align*} Then \begin{align*} y\mu(x) &=\int \mu(x)q(x)dx\\ \csc xy &=\int \csc^2 xdx\\ &=-\cot x+C \end{align*} Then $$(\csc x) y=-\cot x+C\ \ \to \ \ \ y(x)=-\cos x+c \sin x$$ Since $y(\pi/4)=0 $, then $C=1$, and hence $$ y(x)=-\cos x+ \sin x$$
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