## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 14

#### Answer

$$y= \frac{1}{\sec x +\tan x}\left( x-\cos x +C\right).$$

#### Work Step by Step

This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ \int \sec x dx}=e^{\ln(\sec x +\tan x)}=\sec x +\tan x$$ where we used the fact that $\int \sec xdx=\ln(\sec x+\tan x)$. Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=\frac{1}{\sec x +\tan x}\left( \int (\sec x +\tan x)\cos xdx +C\right)\\ &= \frac{1}{\sec x +\tan x}\left( \int (1 +\sin x) dx +C\right)\\ &= \frac{1}{\sec x +\tan x}\left( x-\cos x +C\right)\\ \end{align} where we used integration by parts. Thus, the general solution is: $$y= \frac{1}{\sec x +\tan x}\left( x-\cos x +C\right).$$

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