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.