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

##### 1 Answer
Feb 7, 2018

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

#### Explanation:

We have:

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

This is a First Order Separable Ordinary Differential Equation. We can rewrite in the form:

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

So we can "separate the variables" to get:

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

Both integrals are standard calculus results , so integrating we get:

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

Leading to the General Solution:

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