# I have the following equation (xy^{2}+x)dx+(yx^{2}+y)dy=0 and I am told it is separabl

I have the following equation
$\left(x{y}^{2}+x\right)dx+\left(y{x}^{2}+y\right)dy=0$
and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.
Let

$\int M.dx⇒\int x{y}^{2}+x={x}^{2}{y}^{2}+\frac{{x}^{2}}{2}+g\left(y\right)$
Partial of $\left({x}^{2}{y}^{2}+\frac{{x}^{2}}{2}+g\left(y\right)\right)⇒x{y}^{2}+{g\left(y\right)}^{\prime }$
${g\left(y\right)}^{\prime }=y$
$g\left(y\right)=\frac{{y}^{2}}{2}$
the general solution then is
$C={x}^{2}\frac{{y}^{2}}{2}+\frac{{x}^{2}}{2}+\frac{{y}^{2}}{2}$
Is this solution the same I would get if I had taken the Separate Equations route?
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Annie Levasseur
Basically, yes. You would write
$\int \frac{ydy}{{y}^{2}+1}=\int -\frac{xdx}{{x}^{2}+1}$
$\frac{1}{2}\mathrm{ln}\left({y}^{2}+1\right)=-\frac{1}{2}\mathrm{ln}\left({x}^{2}+1\right)+c$
Multiply both sides by 2 and exponentiate:
${y}^{2}+1=\frac{{e}^{2c}}{{x}^{2}+1}$
This is equivalent to your solution, with $C=\frac{{e}^{2c}-1}{2}$
###### Not exactly what you’re looking for?
Chanell Sanborn
Taking the other route you get ${y}^{\prime }y\left({x}^{2}+1\right)=-x{y}^{2}-x$.
Denote $z={y}^{2}$, hence ${z}^{\prime }=2y{y}^{\prime }$. So you get ${z}^{\prime }+\frac{2x}{{x}^{2}+1}z=-\frac{x}{{x}^{2}+1}$. This gives you $z=\frac{k\left(x\right)}{{x}^{2}+1}$nSKSubstituting, you have ${k}^{\prime }\left(x\right)=-x$, which implies $k\left(x\right)=-\frac{{x}^{2}}{2}+C$. Hence $z=-\frac{{x}^{2}}{2\left({x}^{2}+1\right)}+\frac{C}{{x}^{2}+1}$. Finally, ${y}^{2}=z$
From here you get $2{y}^{2}\left({x}^{2}+1\right)+{x}^{2}=C$, which is the answer you got.