Aldo Ashley

2022-10-25

Solve this differential equation
$\left(2x+y\right)dx+\left(x-2y\right)dy=0$
as an exact differential equation and I know it's exact because I solve the equaliy
$\frac{\mathrm{\partial }\left(2x+y\right)}{\mathrm{\partial }y}=1$
and
$\frac{\mathrm{\partial }\left(x-2y\right)}{\mathrm{\partial }x}=1$
so following the steps to solve this kind of equations i have:
${x}^{2}+{g}^{\prime }\left(y\right)=x-2y$
and
${g}^{\prime }\left(y\right)=\frac{x-2y}{{x}^{2}}$
to be honest I have many doubts what are the next steps so if you can guide me I'll apreciate

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Alexandria Rubio

Expert

Step 1
The solution is $F\left(x,y\right)=C$ such that
$\frac{\mathrm{\partial }F}{\mathrm{\partial }x}=2x+y$
$\frac{\mathrm{\partial }F}{\mathrm{\partial }y}=x-2y$
Treating y as constant, integrate the first equation with respect to x
$F\left(x,y\right)=\int \left(2x+y\right)\phantom{\rule{thinmathspace}{0ex}}dx={x}^{2}+xy+g\left(y\right)$
Step 2
Treating x as constant, take the partial w.r.t. y of the above expression
$\frac{\mathrm{\partial }F}{\mathrm{\partial }y}=x+{g}^{\prime }\left(y\right)=x-2y$
This gives
${g}^{\prime }\left(y\right)=-2y⇒g\left(y\right)=-{y}^{2}$
$F\left(x,y\right)={x}^{2}+xy-{y}^{2}$
so the solution is
${x}^{2}+xy-{y}^{2}=C$

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