# What is the general solution of the differential equation  dy/dx=(y+2x)^2+2 ?

Sep 27, 2017

$y = 2 \tan \left(2 x + A\right) - 2 x$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(y + 2 x\right)}^{2} + 2$ ..... [A]

We can perform a substitution, Let

$v = y + 2 x$

Then if we differentiate, we get:

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} + 2 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{dx}} - 2$

Substituting into the original DE $A$, we get:

$\frac{\mathrm{dv}}{\mathrm{dx}} - 2 = {v}^{2} + 2$
$\therefore \frac{\mathrm{dv}}{\mathrm{dx}} = {v}^{2} + 4$

This is now a First Order separable DE, so we can rearrange and seperate the variables, giving:

$\int \setminus \frac{1}{{v}^{2} + 4} \setminus \mathrm{dv} = \int \setminus \mathrm{dx}$

Which we can integrate, using standard integrals, to get:

$\frac{1}{2} \arctan \left(\frac{v}{2}\right) = x + C$

And, restoring the substitution:

$\frac{1}{2} \arctan \left(\frac{y + 2 x}{2}\right) = x + C$
$\arctan \left(\frac{y + 2 x}{2}\right) = 2 x + 2 C$
$\therefore \frac{y + 2 x}{2} = \tan \left(2 x + A\right)$
$\therefore y + 2 x = 2 \tan \left(2 x + A\right)$
$\therefore y = 2 \tan \left(2 x + A\right) - 2 x$