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 pm
Find 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.
Amjad
May 26 @ 12:10 pm
yy’+xy^2=x
Solve it