What is a solution to the differential equation (dy/dx) - y - e^(3x) = 0?

Jul 11, 2016

$y = \frac{1}{2} {e}^{3 x} + C {e}^{x}$

Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} - y - {e}^{3 x} = 0$

this is not separable so we use an Integrating Factor (IF)

$\frac{\mathrm{dy}}{\mathrm{dx}} - y = {e}^{3 x}$

$I F = {e}^{\int \left(- 1\right) \mathrm{dx}} = {e}^{-} x$

$\implies {e}^{-} x \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{-} x y = {e}^{2 x}$

Or $\frac{d}{\mathrm{dx}} \left({e}^{-} x y\right) = {e}^{2 x}$

${e}^{-} x y = \int \setminus {e}^{2 x} \setminus \mathrm{dx}$

${e}^{-} x y = \frac{1}{2} {e}^{2 x} + C$

$y = \frac{1}{2} {e}^{3 x} + C {e}^{x}$