Take a look at this differential equation:
$$ (x^3 + xy^2-y)dx + (y^3+yx^2+x)dy = 0$$
For equation to be exact, we need to prove,
$$ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} $$
but here,
$ M = x^3+xy^2-y $ and $\frac {\partial M}{\partial y} =2xy-1$
and,
$ N = y^3+yx^2+x $ and $\frac {\partial N}{\partial x} =2xy+1$
Clearly, $\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}$
But now, if we rearrange the differential equation like this,
$$ xdx + ydy + \frac{xdy-ydx}{x^2+y^2} = 0$$
then we get,
$\frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$ = $\frac {y^2-x^2}{(x^2+y^2)^2}$
Now I know this differential equation IS exact, as i have taken this from an example in a book, but to solve this exact differential equation, they have rearranged it first to the form I have mentioned above. Why can't we prove $\frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$ in its initial form?
Best Answer
Your original differential equation $$ (x^3+xy^2-y)\,dx+(y^3+yx^2+x)\,dy=0 $$ is not exact. What one might do is to multiply it by a function (called integrating factor) $\mu(x,y)$ to make $$ \mu(x,y)(x^3+xy^2-y)\,dx+\mu(x,y)(y^3+yx^2+x)\,dy=0 $$ exact.
In general, if one starts with $$ P(x,y)\,dx+Q(x,y)\,dy=0, $$ then the differential equation $$ \mu(x,y)\bigl(P(x,y)\,dx+Q(x,y)\,dy\bigr)=0 $$ is exact if (check this) $\mu$ satisfies $$ P\mu'_y-Q\mu'_x+\mu\bigl(P'_y-Q'_x\bigr)=0.\tag{*} $$ You do not need to find the general solution to this partial differential equation, it suffices to find a particular solution. In your case, one solution to $(^*)$ is given by $$ \mu(x,y)=\frac{1}{x^2+y^2}. $$ This is why your differential equation becomes exact after multiplying with that $\mu$.