# What is the particular solution of the differential equation? : y'+4xy=e^(-2x^2) with y(0)=-4.3

Sep 8, 2017

$y = \left(x - 4.3\right) {e}^{- 2 {x}^{2}}$

#### Explanation:

We have:

$y ' + 4 x y = {e}^{- 2 {x}^{2}}$ ..... [A]

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

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

The given equation is already in standard form, so the integrating factor is given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus 4 x \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(2 {x}^{2}\right)$
$\setminus \setminus = {e}^{2 {x}^{2}}$

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

$2 {e}^{2 {x}^{2}} y ' + 8 x {e}^{2 {x}^{2}} y = 2 {e}^{2 {x}^{2}} {e}^{- 2 {x}^{2}}$

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

Which we can now "seperate the variables" to get:

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

Which is trivial to integrate giving the General Solution:

$2 {e}^{2 {x}^{2}} y = 2 x + C$

Applying the initial condition we get:

$2 {e}^{0} \left(- 4.3\right) = 0 + C \implies C = - 8.6$

Giving the Particular Solution:

$2 {e}^{2 {x}^{2}} y = 2 x - 8.6$
$\therefore {e}^{2 {x}^{2}} y = x - 4.3$
$\therefore {e}^{2 {x}^{2}} {e}^{- 2 {x}^{2}} y = \left(x - 4.3\right) {e}^{- 2 {x}^{2}}$
$\therefore y = \left(x - 4.3\right) {e}^{- 2 {x}^{2}}$