# What is the general solution of the differential equation?  (x+1)y'+y= tan^-1(x) / (x+1)

Jan 10, 2018

$\left(1 + x\right) y = \int \setminus {\tan}^{-} 1 \frac{x}{1 + x} \setminus \mathrm{dx} + C$

#### Explanation:

We have:

$\left(x + 1\right) y ' + y = {\tan}^{-} 1 \frac{x}{x + 1}$

We can re-arrange this ODE as follows:

$\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{x + 1} y = {\tan}^{-} 1 \frac{x}{1 + x} ^ 2$ ..... 

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

We can readily generate an integrating factor when we have an equation of this form, given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus \frac{1}{1 + x} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(\ln \left(1 + x\right)\right)$
$\setminus \setminus = \left(1 + x\right)$

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

$\left(1 + x\right) \frac{\mathrm{dy}}{\mathrm{dx}} + y = {\tan}^{-} 1 \frac{x}{1 + x}$
$\frac{d}{\mathrm{dx}} \left(\left(1 + x\right) y\right) = {\tan}^{-} 1 \frac{x}{1 + x}$

(Which is incidentally the equation we started with). We can directly integrate to get:

$\left(1 + x\right) y = \int \setminus {\tan}^{-} 1 \frac{x}{1 + x} \setminus \mathrm{dx} + C$

And in terms of elementary functions this is as far we can get: