## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 13

#### Answer

$$y=e^{- x }\left( \frac{1}{2}e^{ 2x }+C\right) .$$

#### Work Step by Step

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

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