# 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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Answer

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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 \\

\]

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.

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.

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