# What is a solution to the differential equation dy/dx=y^2/x^2+y/x+1?

Jul 11, 2016

$y = x \tan \left(\ln x + C\right)$

#### Explanation:

This is a first order linear homogeneous equation. NB here homogeneous has its own meaning. it means that the equation can be written in form $y ' = f \left(x , y\right)$ and that $f \left(k x , k y\right) = f \left(x , y\right)$ for constant k.

so standard approach is to let $v = \frac{y}{x}$

so
$y = v \cdot x$
$y ' = v ' x + v$

thus, plugging this into the original.....

$v ' x + v = {v}^{2} + v + 1$

$v ' x = {v}^{2} + 1$

$\frac{v '}{{v}^{2} + 1} = \frac{1}{{v}^{2} + 1} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x}$

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

standard integral: ${\tan}^{- 1} v = \ln x + C$

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

$\left(\frac{y}{x}\right) = \tan \left(\ln x + C\right)$

$y = x \tan \left(\ln x + C\right)$