Solve 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 separating the variables method.

Given differential equation of the form

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

This differential equation 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 separating variable technique the terms dy and dx are placed in the numerator with their respective variables.
Now integration 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}

Which is the required solution of the given differential equation.