# Solve the Differential Equation  (1+x^2)y'+4xy=(1+x^2)^-2 ?

Nov 9, 2017

$y = \arctan \frac{x}{1 + {x}^{2}} ^ 2 + \frac{C}{1 + {x}^{2}} ^ 2$

#### Explanation:

We have:

$\left(1 + {x}^{2}\right) y ' + 4 x y = {\left(1 + {x}^{2}\right)}^{-} 2$ ..... [A]

We can rearrange [A] as follows:

$\left(1 + {x}^{2}\right) y ' + 4 x y = \frac{1}{1 + {x}^{2}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{4 x}{1 + {x}^{2}} y = \frac{1}{1 + {x}^{2}} ^ 3$ ..... [B]

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

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

So we form an Integrating Factor;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus \frac{4 x}{1 + {x}^{2}} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(2 \setminus \int \setminus \frac{2 x}{1 + {x}^{2}} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(2 \setminus \ln \left(1 + {x}^{2}\right)\right)$
$\setminus \setminus = \exp \left(\ln {\left(1 + {x}^{2}\right)}^{2}\right)$
$\setminus \setminus = {\left(1 + {x}^{2}\right)}^{2}$

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

${\left(1 + {x}^{2}\right)}^{2} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{4 x {\left(1 + {x}^{2}\right)}^{2}}{1 + {x}^{2}} y = {\left(1 + {x}^{2}\right)}^{2} / {\left(1 + {x}^{2}\right)}^{3}$
$\therefore \frac{d}{\mathrm{dx}} \left({\left(1 + {x}^{2}\right)}^{2} \setminus y\right) = \frac{1}{1 + {x}^{2}}$

Which we can directly integrate to get:

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

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

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

Leading to the General Solution:

$y = \arctan \frac{x}{1 + {x}^{2}} ^ 2 + \frac{C}{1 + {x}^{2}} ^ 2$