# What is the general solution of the differential equation y' - (2xy)/(x^2+1) = 1?

##### 1 Answer
Sep 27, 2017

$y = \left({x}^{2} + 1\right) \arctan \left(x\right) + C \left({x}^{2} + 1\right)$

#### Explanation:

We have:

$y ' - \frac{2 x y}{{x}^{2} + 1} = 1$ ..... [A}

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

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

As the equation is already in this form, then the integrating factor is given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus - \frac{2 x}{{x}^{2} + 1} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(- \int \setminus \frac{2 x}{{x}^{2} + 1} \setminus \mathrm{dx}\right)$
 \ \ = exp( - ln(x^2+1)
$\setminus \setminus = \exp \left(\ln \left(\frac{1}{{x}^{2} + 1}\right)\right)$
$\setminus \setminus = \frac{1}{{x}^{2} + 1}$

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

$\left(\frac{1}{{x}^{2} + 1}\right) \setminus y ' - \left(\frac{1}{{x}^{2} + 1}\right) \frac{2 x y}{{x}^{2} + 1} = \left(\frac{1}{{x}^{2} + 1}\right) 1$

$\therefore \frac{d}{\mathrm{dx}} \left(\left(\frac{1}{{x}^{2} + 1}\right) \setminus y\right) = \frac{1}{{x}^{2} + 1}$

Which we can directly integrate to get:

$\frac{y}{{x}^{2} + 1} = \int \setminus \frac{1}{{x}^{2} + 1} \setminus \mathrm{dx}$

The RHS is a standard integral, so integrating we get:

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

Leading to the General Solution:

$y = \left({x}^{2} + 1\right) \arctan \left(x\right) + C \left({x}^{2} + 1\right)$