# Question #a071a

Sep 1, 2017

$\mu = - \sin x \cdot {e}^{\cos} x$

#### Explanation:

First manipulate the equation back into standard form, $\frac{\mathrm{dy}}{\mathrm{dx}} + p \left(x\right) y = q \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} + y \sin x = {e}^{x}$
$\therefore p \left(x\right) = \sin x \mathmr{and} q \left(x\right) = {e}^{x}$
$\mu = {e}^{\int \left(p \left(t\right)\right) \mathrm{dt}}$
$\therefore \mu = {e}^{\int \left(\sin x\right) \mathrm{dx}}$
$= {e}^{\cos} x$
$= - \sin x \cdot {e}^{\cos} x$ (multiply by the derivative of the power)