## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 10 - Introduction to Differential Equations - Chapter Review Exercises - Page 526: 23

#### Answer

$$y= \cos x \sin x -2\cos x .$$

#### Work Step by Step

This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ \int \tan x dx}=e^{-\ln \cos x}=\frac{1}{\cos x}$$ Now the general solution is \begin{align} y& =\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ & =\cos x \left( \int \cos x dx+C\right)\\ & =\cos x \left( \sin x+C\right)\\ & =\cos x \sin x +C\cos x \end{align} Since, $y(\pi)=2$, then $C=-2$ Thus, the general solution is: $$y= \cos x \sin x -2\cos x .$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.