Solve the ODE: $(x^{2}-2xy)y'+y^{2}-2xy=0$ solve via exact equation technique Solve the ODE: $$(x^{2}-2xy)y'+y^{2}-2xy=0$$
My try:
$\begin{aligned}{c}
(x^{2}-2xy)y'+y^{2}-2xy=0 \\
\implies (x^{2}-2xy)\frac{\mathrm{d}y}{\mathrm{d}x}+y^{2}-2xy=0\\
\implies (x^{2}-2xy)\mathrm{d}y+(y^{2}-2xy)\mathrm{d}x=0\\
\implies (x^{2}-2xy)e^{-x}\mathrm{d}y+(y^{2}-2xy)e^{-x}\mathrm{d}x=0\\
\\
\end{aligned}$
but the equation doesn't turn exact. I found no relevant integration factor.
Edit
Solve with an exact equation method.
 A: $$(x^{2}-2xy)dy+(y^{2}-2xy)dx=0$$
Divide by $x^2y^2$
$$\left(\dfrac 1 {y^{2}}-\dfrac 2 {xy}\right)dy+\left(\dfrac 1 {x^{2}}-\dfrac 2 {xy}\right)dx=0$$
$$\dfrac 1 {y^{2}}dy-\dfrac 2 {xy}d(x+y)+\dfrac 1 {x^{2}}dx=0$$
$$-d(\dfrac 1 {y})-\dfrac 2 {xy}d(x+y)-d(\dfrac 1 x)=0$$
$$d \left(\dfrac {x+y} {xy}\right)+\dfrac 2 {xy}d(x+y)=0$$
Now  the integrating factor is obvious.
$$\mu (x,y) =\dfrac {xy} {x+y}$$
Multiply by $\mu$ and integrate.
A: $$y'=\frac{2xy-y^2}{x^2-2xy}=\frac{2\frac{y}{x} -\left(\frac{y}{x}\right)^2}{1-2\frac{y}{x}}$$
$y=ux\rightarrow y'=u+u'x$
hence
$$u'x+u =\frac{2u-u^2}{1-2u}$$
and $$u'x =\frac{u^2 +2u -1}{1-2u}$$
and we obtain
$$\frac{1-2u}{u^2 +2u +1} du =\frac{dx}{x}$$
which is easy to solve.
A: We are given
$$(x^2-2xy)\frac{dy}{dx}+y^2-2xy=0$$
$$\underbrace{(y^2-2xy)}_Mdx+\underbrace{(x^2-2xy)}_Ndy=0$$
by which $M_y = 2y-2x$ is not equal to $N_x=2x-2y$. So, the differential equation is not exact. To make it exact, observe that the differential equation is homogenuous and therefore the integrating factor can be found by
$$\mu(x)=\frac{1}{xM+yN}=-\frac{1}{xy(x+y)}$$
which gives
$$\underbrace{\left(-\frac{y}{x(x+y)}+\frac{2}{x+y}\right)}_{{M}^{*}}dx+\underbrace{\left(-\frac{x}{y(x+y)}+\frac{2}{x+y}\right)}_{N^{*}}dy=0$$
The new equation is exact since
$${M^{*}}_y=-\frac{3}{(x+y)^2}={N^{*}}_x$$
A summary of general techniques to make non-exact different equations exact is shown here.
A: $$y'=\frac{2xy-y^2}{x^2-2xy}$$
This is a homogeneous equation. So substitute $y=tx$
$$x\frac{dt}{dx}+t=\frac{2tx^2-t^2x^2}{x^2-2tx^2}$$
$$x\frac{dt}{dx}+t=\frac{2t-t^2}{1-2t}$$
$$x\frac{dt}{dx}=\frac{t+t^2}{1-2t}$$
$$\frac{dx}{x}=\frac{1-2t}{t+t^2}dt$$
$$\int\frac{dx}{x}=-\int\frac{2t+1-2}{t+t^2}dt$$
Hint [Let $t+t^2=u \Rightarrow (2t+1)dt=du$]
I'll let you integrate
