# What is the general solution of the differential equation  dy/dx +3y = e^(4x) ?

Oct 6, 2017

$y = \frac{1}{7} {e}^{4 x} + C {e}^{- 3 x}$

#### Explanation:

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{\mathrm{dy}}{\mathrm{dx}} + P \left(x\right) y = Q \left(x\right)$

Ther equation is already in standard form:

$\frac{\mathrm{dy}}{\mathrm{dx}} + 3 y = {e}^{4 x}$ ..... [A}

Then the integrating factor is given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus 3 \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(3 x\right)$
$\setminus \setminus = {e}^{3 x}$

And if we multiply the DE [A] by this Integrating Factor, $I$, we will have a perfect product differential;

$\frac{\mathrm{dy}}{\mathrm{dx}} + 3 y = {e}^{4 x}$

$\therefore {e}^{3 x} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {e}^{3 x} y = {e}^{3 x} {e}^{4 x}$

$\therefore \frac{d}{\mathrm{dx}} \left({e}^{3 x} y\right) = {e}^{7 x}$

Which we can directly integrate to get:

$\setminus \setminus \setminus \setminus \setminus {e}^{3 x} y = \frac{1}{7} {e}^{7 x} + C$

$\therefore y = \frac{1}{7} {e}^{7 x} {e}^{- 3 x} + C {e}^{- 3 x}$

$\therefore y = \frac{1}{7} {e}^{4 x} + C {e}^{- 3 x}$