Is it possible to solve a separable equation a different way and still arrive at the same answer? I have the following equation 
$$(xy^2 + x)dx + (yx^2 + y)dy=0$$ and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method. 
Let $M = xy^2 + x $ and $N = yx^2 + y$
$$My = 2xy \text{ and } Nx = 2xy $$
$$ \int M.dx ==> \int xy^2 + x = x^2y^2 + (x^2)/2 + g(y)$$
$$ \text{Partial of } (x^2y^2 + (x^2)/2 + g(y)) => xy^2 + g(y)'$$
$$g(y)' = y$$
$$g(y) = y^2/2$$
the general solution then is 
$$C = x^2y^2/2 + x^2/2 + y^2/2$$
Is this solution the same I would get if I had taken the Separate Equations route? 

Taking the other route you get $y'y(x^2+1)=-xy^2-x$. Denote $z=y^2$, hence $z'=2yy'$. So you get $z'(x^2+1)=-2xz-x$. Hence you have $z'+\frac{2x}{x^2+1}z=-\frac{x}{x^2+1}$. This gives you $z=\frac{k(x)}{x^2+1}$. Substituting, you have $k'(x)=-x$, which implies $k(x)=-\frac{x^2}{2}+C$. Hence $z=-\frac{x^2}{2(x^2+1)}+\frac{C}{x^2+1}$. Finally, $y^2=z$.
From here you get $2y^2(x^2+1)+x^2=C$, which is the answer you got.
