# 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.

rakibul

October 1@ 7:17 pmFind the area of the quadrilateral circuit formed by the lines

6x

2 − 5xy − 6y

2 + 14x + 5y + 4 = 0 and x

2 − xy − 2y

2 − x − 4y − 2 = 0.