What kind of ODE is this and how to solve it? $$ (2x-4y+6)dx+(x+4y-3)dy=0 $$
I converted it to this form $ dy/dx = (2x-4y+6)/(x+4y-3) $ in the hope I can use $ z=y/x $ substitution for homogeneous ODEs but because of constants 6 and 3 I can't.
 A: Do first a change of variables:
$$ u = x + 1 \qquad v = y-1 $$
Then $\mathrm{d}u = \mathrm{d}x$ and $\mathrm{d}v = \mathrm{d}y$. Now 
$$ 2u - 4v = 2x + 2 - 4y + 4 = 2x - 4y + 6$$
and
$$ u + 4v = x + 1 + 4y - 4 = x + 4y - 3$$
So in the new variables you are solving
$$ (2u-4v) \mathrm{d}u + (u+4v) \mathrm{d}v = 0 $$
or (away from the line $u+4v = 0$)
$$ \frac{2u-4v}{u+4v} \mathrm{d}u + \mathrm{d}v = 0 $$
Then you can do the substitution: write $v = uz$ you get $\mathrm{d}v = z\mathrm{d}u + u \mathrm{d}z$:
$$ \left( \frac{2-4z}{1+4z} + z\right) \mathrm{d}u + u \mathrm{d}z = 0 $$
A: This is an ODE that can be solved in the following way. Write
$$ (2x-4y+6)dx+(x+4y-3)dy=0 $$
as
$$\frac{-2x+4y-6}{x+4y-3} =\frac{dy}{dx}$$
What we need now is to write the RHS as 
$$\frac{ax+by}{cx+dy} $$
so that it becomes homogeneous.
So we need to find $k$ and $h$ such that
$$-2(x_1+h)+4(y_1+k)-6=-2x_1 +4y_1$$
$$ (x_1+h)+4(y_1+k)-3 =x_1 +4y_1$$
This is, we need that
$$\eqalign{
  &  - 2h + 4k - 6 = 0  \cr 
  & h + 4k - 3 = 0 \cr} $$
This gives,
$$\eqalign{
  & h =  - 1  \cr 
  & k = 1 \cr} $$
So we have
$$\frac{{ - 2{x_1} + 4{y_1}}}{{{x_1} + 4{y_1}}} = \frac{{d{y_1}}}{{d{x_1}}}$$
But this ODE is homogeneous (as we wanted), so we can put
$$\frac{{{y_1}}}{{{x_1}}} = v$$
and get
$$\frac{{ - 2 + 4v}}{{1 + 4v}} = v'{x_1} + v$$
This results in a separable ODE,
$$\frac{{d{x_1}}}{{{x_1}}} =  - \frac{{4v + 1}}{{4{v^2} - 3v + 2}}dv$$
I guess you can take it from here.
NOTE: In general, the equation
$$\frac{{ax + by + c}}{{dx + ey + f}} = \frac{{dy}}{{dx}}$$
can be made homogeneous by solving
$$\eqalign{
  & ah + bk + c = 0  \cr 
  & dh + ek + f = 0 \cr} $$
and substituting $X = x+h$, $Y=y+k$.
In turn, the homogeneous ODE can be made separable by putting $\dfrac{Y}{X} = v$
