Integrating Factor - Exact Equation problem. I have stumbled with a problem I can't seem to solve. 
$$(x^2 - y ^2)dx - 5xy dy = 0$$
We know that 
$$u(x,y) = \frac{1}{(x M + y N)}$$
if the equation is HDE (Which it is..I believe).
Excuse my notation, Im not very good at the edit part.
$$M = \frac{1}{(x(x^2 - y^2)},\sim N = \frac{1}{y(-5 xy)}$$
Now we know that 
$\frac{dM}{dy} = \frac{dN}{dx}$
Is it me or the result isn't equivalent?
If it is, what is it?
Thank you for your time and help.
 A: $$
(x^2-y^2)\,\mathrm{d}x-5xy\,\mathrm{d}y=0
$$
use $y\,\mathrm{d}y=\frac12\mathrm{d}y^2$
$$
(x^2-y^2)\,\mathrm{d}x-\frac52x\,\mathrm{d}y^2=0
$$
add $y^2\,\mathrm{d}x$
$$
x^2\mathrm{d}x=y^2\,\mathrm{d}x+\frac52x\,\mathrm{d}y^2
$$
multiply by the integrating factor $\frac25x^{-3/5}$
$$
\frac25x^{7/5}\mathrm{d}x=\frac25x^{-3/5}y^2\,\mathrm{d}x+x^{2/5}\,\mathrm{d}y^2
$$
integrate
$$
\frac16x^{12/5}+C=x^{2/5}y^2
$$
multiply by $x^{-2/5}$
$$
\bbox[5px,border:2px solid #C00000]{\frac16x^2+Cx^{-2/5}=y^2}
$$

The differential $y^2\,\mathrm{d}x+\frac52x\,\mathrm{d}y^2$ is reminiscent of $\mathrm{d}(x^\alpha y^2)=\alpha x^{\alpha-1}y^2\,\mathrm{d}x+x^\alpha\,\mathrm{d}y^2$. In fact, if we multiply the first by $\alpha x^{\alpha-1}$, we get the second with $\alpha=\frac25$. This is why we used the integrating factor of $\frac25x^{-3/5}$.
A: i think i can solve the differential equation $\dfrac{dy}{dx} = \dfrac{x^2 - y^2}{5xy}.$
split the differential equation into a system of two 
$$\frac{dy}{dt} = \dfrac{1}{2y}, \ \dfrac{dt}{dx} = \dfrac{2(x^2-y^2)}{5x}.$$ the first has an integral $$  y^2 = t + C\tag 1$$
we will now treat the second as a differential equation for $t$ and using (1) it is $$ \dfrac{dt}{dx} = \dfrac{2(x^2-t-C)}{5x} = -\dfrac{2}{5}\dfrac{t}{x} + \dfrac{2(x^2-C)}{5x} \tag 2$$
the homogeneous solution associated with $(2)$ is $$t =Bx^{-2/5} \tag 3$$ we will use variation of parameters,that is assume a solution of $(2)$ in the form of $(3)$ where $B$ is now a function of $x$ to be decided. differentiating $(3),$ we get 
$\dfrac{dt}{dx} = \dfrac{dB}{dx}x^{-2/5} -\dfrac{2}{5}Bx^{-7/5}$ putting this in $(2)$ we get 
$$\dfrac{dB}{dx}x^{-2/5} -\dfrac{2}{5}Bx^{-7/5} = -\dfrac{2}{5}\dfrac{Bx^{-2/5}}{x} + \dfrac{2(x^2-C)}{5x}  $$ which simplifies to 
$$ \dfrac{dB}{dx} =\dfrac{2}{5}x^{7/5} -\dfrac{2C}{5} x^{-3/5}.$$ on integration gives 
$$B = A + \dfrac{1}{6}x^{12/5} -Cx^{2/5}$$ therefore from $(2)$ 
$$t =  Ax^{-2/5} + \dfrac{1}{6}x^2 -C \tag 4$$
finally by (1) we get the solution 
$$ y^2 = Ax^{-2/5} + \dfrac{1}{6}x^2 \text{ where $A$ is an arbitrary constant.} $$

edit: i forgot that there is an easier way to solve this. this is homogeneous in the sense of euler of degree 2. introduce the change of variable $y = mx$
so $\dfrac{dy}{dx} = m + x\dfrac{dm}{dx}$ so the $m$ satisfies $$ m + x\dfrac{dm}{dx} = \dfrac{1-m^2}{5m}$$ simplfies to $$x\dfrac{dm}{dx} = \dfrac{1-6m^2}{5m} $$
now separate the variables to get $$\dfrac{5mdm}{1-6m^2} = \dfrac{dx}{x}$$ on integration yields $$ 1 - 6m^2 = 1 - 6\dfrac{y^2}{x^2} = 6Cx^{-12/5} $$ 
is the same as before but in much shorter time, $$ y^2 = \dfrac{1}{6}x^2 + Cx^{-2/5}$$
