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# Solve the differential equation:$\left( {x{y^2} + x} \right)dx + \left( {{x^2}y + y} \right)dy = 0$.

Last updated date: 27th Mar 2023
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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:
$\Rightarrow x\left( {{y^2} + 1} \right)dx = - y\left( {{x^2} + 1} \right)dy, \\ \Rightarrow \dfrac{x}{{\left( {{x^2} + 1} \right)}}dx = - \dfrac{y}{{\left( {{y^2} + 1} \right)}}dy, \\ \Rightarrow \dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx = - \dfrac{{2y}}{{\left( {{y^2} + 1} \right)}}dy \\$
$\Rightarrow \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}dx} = - \int {\dfrac{{2y}}{{\left( {{y^2} + 1} \right)}}dy,}$
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 $\left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right) = \pm {e^C}$