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: 15


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

Work Step by Step

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