 Questions & Answers    Question Answers

# Solve the differential equation:$\left( {x{y^2} + x} \right)dx + \left( {{x^2}y + y} \right)dy = 0$.  Answer Verified
Hint: Separate the terms with $x$ variable on one side and terms with $y$ variable on other side. And then solve the equation integrating both sides.

The given differential equation is $\left( {x{y^2} + x} \right)dx + \left( {{x^2}y + y} \right)dy = 0$. This can be simplified as:
$\Rightarrow x\left( {{y^2} + 1} \right)dx = - y\left( {{x^2} + 1} \right)dy, \\ \Rightarrow \dfrac{x}{{\left( {{x^2} + 1} \right)}}dx = - \dfrac{y}{{\left( {{y^2} + 1} \right)}}dy, \\ \Rightarrow \dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx = - \dfrac{{2y}}{{\left( {{y^2} + 1} \right)}}dy \\$
Integrating both sides, we’ll get:
$\Rightarrow \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx} = - \int {\dfrac{{2y}}{{\left( {{y^2} + 1} \right)}}dy,}$
We know that $\int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx} = \log \left| {{x^2} + 1} \right| + C$, Using this in the above equation, we’ll get:
$\Rightarrow \log \left| {{x^2} + 1} \right| = - \log \left| {{y^2} + 1} \right| + C, \\ \Rightarrow \log \left| {{x^2} + 1} \right| + \log \left| {{y^2} + 1} \right| = C, \\ \Rightarrow \log \left( {\left| {{x^2} + 1} \right|\left| {{y^2} + 1} \right|} \right) = C, \\ \Rightarrow \left| {\left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right)} \right| = {e^C}, \\ \Rightarrow \left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right) = \pm {e^C} \\$
Thus the solution of the differential equation is $\left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right) = \pm {e^C}$

Note: The method used in solving the above differential equation is called variable separation method i.e. keeping the terms containing the same variable on one side and terms having other variables on the other side. And then integrating on both the sides.
Bookmark added to your notes.
View Notes
Solve Separable Differential Equations  How to Solve Linear Differential Equation?  Exact Differential Equation  Homogeneous Differential Equation  Differential Equations For Class 12  Solution of Differential Equation  First Order Differential Equation  Order and Degree of Differential Equations  Second-Order Differential Equation  Differential Equations  