$(2x-5y+3)dx=(2x+4y-6)dy$ This is not a new question. Sorry to revive it, but I cannot find another way because it seems that the original question disappeared (It's not me who posted the original question).
Someone (I don't remember who) asked about solving the ODE :
$$(2x-5y+3)dx=(2x+4y-6)dy$$
Yesterday, I gave an answer to this question, but there was a mistake in it. At this time I had not enough time to make the correction and to rewrite it. So I deleted my answer.
Today, as I wanted to post my corrected answer, I cannot find where the original question as gone. 
Since I suppose that the unusual method used in my answer would interest some readers, I post the problem as a new question and the answer will be immediately posted in the answer section.
This procedure was suggested by Max in order to keeps it off the unanswered queue.
 A: 
The solution is expressed on the form of an implicit equation. If one want the explicit solution $y(x)$ one have to solve the cubic equation for $y$. Of course, it is possible, but tiresome.
Note that the method used above is somehow related to what is sometimes done in the "method of characteristics" for solving PDE.
In addition : CHECKING THE RESULT.
The total derivative of $(x-4y+3)(2x+y-3)^2=C$ is :
$\left( (2x+3y-3)^2+4(x-4y+3)(2x+y-3)\right)dx+\left( -4(2x+y-3)^2+2(x-4y+3)(2x+y-3)\right)dy=0$
After simplification : $3(2x+y-3)\left( (2x-5y+3)dx-(2x+4y-6)dy\right)=0$
$$(2x-5y+3)dx=(2x+4y-6)dy$$
We recover exactly the original ODE. Thus the result $(x-4y+3)(2x+y-3)^2=C$ is correct.
A: $$(2x−5y+3)dx=(2x+4y−6)dy$$
$$(2x−5y+3)dx+(6-2x-4y)dy=0$$
$$\frac{b_2}{b_1}=\frac{4}{5}$$
$$\frac{a_2}{a_1}=\frac{-2}{2}=-1$$
Notice that $$\frac{a_2}{a_1} \neq \frac{b_2}{b_1}$$
Suppose that $(h,k)$ is a solution then we can write them as
$$2h-5k+3=0 $$
$$2h+4k-6=0$$
Solving $k=-1$,$h=1$
Then with transformation
$$x=X+h$$
$$y=Y+k$$
Our result will be
$$x=X+1$$
$$y=Y+1$$
$$dx=dX$$
$$dy=dY$$
$$(2(X+1)-5(Y+1)+3)dX+(6-2(X+1)-4(Y+1))dY=0$$
$$(2X-5Y)dX+(-2X-4Y)dY=0$$
$$\frac{dY}{dX}=\frac{2X-5Y}{2X+4Y}$$
Notice that this is a homogeneous linear equation then, with transformation $Y=vX$ $\frac{dY}{dX}=v+X\frac{dv}{dX}$
$$v+X\frac{dv}{dX}=\frac{2-5v}{2+4v}$$
$$X\frac{dv}{dX}=\frac{2-5v-2v-4v^2}{2+4v}$$
$$X\frac{dv}{dX}+\frac{4v^2+7v-2}{2+4v}=0$$
$$\frac{dX}{X}+\frac{2+4v}{4v^2+7v-2}dv=0$$
By method of partial fractions we have
$$\frac{2+4v}{(4v^2+7v-2)}=\frac{A}{v+2}+\frac{B}{4v-1}$$
Solving we have
$$\int \frac{6}{9(v+2)}+\frac{4}{3(4v-1)}dv= -\int \frac{dX}{X}$$
$$\ln|(v+2)^2(4v-1)|=-\ln|X^3|+c$$
$$ln\frac{(2x+y-3)^2(4y-x-3)(x-1^3)}{(x-1)^3}=c$$
$$(2x+y-3)^2(4y-x-3)=C$$
Saw this. Couldn't resist answering.
