Show that $(y-2x+3)^3=(y-x+1)^2$ is a solution of the differential equation $(2x-4y)dx+(x+y-3)dy=0$ I want to show that 
$$(y-2x+3)^3=(y-x+1)^2$$
is a solution for:
$$(2x-4y)dx+(x+y-3)dy=0$$
what I did so far is:
$$\frac{dy}{dx}=\frac{4y-2x}{x+y-3}$$
any suggestions?
$(y-2x+3)^3=(y-x+1)^2$

differentiating this we have

$3(y-2x+3)^2(dy-2dx)=2(y-x+1)(dy-dx)$

using the original equation and substituting we have

$3(y-x+1)(dy-2dx)=2(y-2x+3)(dy-dx)$

$(y+x-3)dy+(-4y+2x)dx=0$

$(2x-4y)dx+(y+x-3)dy=0$

$\fbox{}$
Now it's time to abuse notation. 

"Cross multiply" to get 

$$(x+y-3)\,dy = (4y-2x)\,dx$$

Then re-arrange to get

$$-(4y-2x)\,dx + (x+y-3)\,dy = 0\implies (2x-4y)\,dx + (x+y-3)\,dy = 0$$