# Solving the Differential Equation (y^2+xy^2)y’=1

In this tutorial we shall solve a differential equation of the form $\left( {{y^2} + x{y^2}} \right)y’ = 1$, by using the separating the variables method.

Given the differential equation of the form

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

This differential equation can also be written as

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

Separating the variables, the given differential equation can be written as

${y^2}dy = \frac{1}{{1 + x}}dx\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Keep in mind that in the separating variable technique the terms $dy$ and $dx$ are placed in the numerator with their respective variables.

Now integrating both sides of the equation (i), we have
$\int {{y^2}dy = \int {\frac{1}{{1 + x}}dx} }$

Using the formulas of integration $\int {\frac{{f’\left( x \right)}}{{f\left( x \right)}}dx = \ln f\left( x \right)}$ and $\int {\frac{1}{x}dx = \ln x}$, we get
$\begin{gathered} \frac{{{y^3}}}{3} = \ln \left( {1 + x} \right) + \ln c \\ \Rightarrow \frac{{{y^3}}}{3} = \ln \left( {1 + x} \right)c \\ {y^3} = 3\left[ {\ln \left( {1 + x} \right)c} \right] \\ \end{gathered}$

This is the required solution of the given differential equation.