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? 

We can also try it this way,
$$(xy^2 + x)dx + (yx^2 + y)dy=0$$
$$xdx +ydy +xy^2dx+yx^2dy$$
$$\frac{1}{2}(2xdx+2ydy) + \frac{1}{2}(2xy^2dx+2yx^2dy)=0$$
$$\frac{1}{2}d(x^2+y^2) + \frac{1}{2}d(x^2y^2) =0$$
$$x^2+y^2+ x^2y^2 +c=0$$
:)
