# Solve dy/dx = 1+ 1/y^2 ?

Dec 6, 2017

$y - \arctan \left(y\right) = x + C$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{1}{y} ^ 2$

Which is a First Order Separable Ordinary Differential Equation so we can rearrange and "separate the variables":

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + {y}^{2}}{y} ^ 2$

$\implies \int \setminus {y}^{2} / \left(1 + {y}^{2}\right) \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$

We can manipulate the LHS integral:

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

$\therefore \int \setminus 1 - \frac{1}{1 + {y}^{2}} \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$

Which is now trivial to integrate giving us:

$y - \arctan \left(y\right) = x + C$

Which is the general implicit solution.