# Question #1eb38

Mar 18, 2017

$y = {e}^{- k x} \left(x + C\right)$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} + k y = {e}^{- k x}$

We apply an Integrating Factor (IF): ${e}^{\int k \mathrm{dx}} = {e}^{k x}$

(Note that here the integration constant can be left out as it will cancel when we multiply both sides by the IF.)

Applying the IF to both sides:

${e}^{k x} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{k x} k y = {e}^{k x} {e}^{- k x}$

$\implies \left({e}^{k x} \setminus y\right) ' = 1$

$\implies {e}^{k x} \setminus y = \int \mathrm{dx} = x + C$

$\implies y = {e}^{- k x} \left(x + C\right)$