How to solve $ (x + 2y - 4)dx - (2x + y - 5)dy = 0$ How to solve the differential equation $(x + 2y - 4)dx - (2x + y - 5)dy = 0$. It's not separable, nor exact nor homogeneous. The solution is $(x - y -1)^3 = C(x + y - 3)$. How can I achieve it?
The other equations similar to this are:
$(1+ x + y)dy - (1- 3x - 3y)dx = 0$
Answer: $3x + y + 2ln(-x - y +1) = k$
$(3x - y + 2)dx + (9x - 3y +1)dy = 0$
Answer: $2x + 6y + C = ln(6x - 2y +1)$
If someone point me out how to solve the first equation I will be likely to solve the others. Thank you very much.
Update: Given Orangutango and Chris help I moved to a solvable d.e. But didn't get the same answer my professor listed. Did I miss some step?
(X + 2Y)dX = (2X+Y)dY
dY/dX = (X + 2Y)/(2X + Y)
Making a substitution to solve the now homogenous:
Y = VX, Y'= V + V'X
V+V'X = (X + 2VX)/(2X + VX)
V+V'X = (1 + 2V)/(2 + V)
V'X = ((1 + 2V)/(2 + V)) - V
V'X = (1 - V^2)/(2 + V) 
(2 + V)dV/(1-V^2) = dX/X
Integrating the left side I got:
int (2 + V)dV/1-V^2 =
2 * int dV / (1-V^2) + int V dV / (1-V^2) =
log | (v + 1) / (v-1) | - 1/2 log | V^2 + 1 | + c1
Integrating the right side:
log X + c2
Then replacing V=Y/X and then X=x-2 and Y=y-1 don't seems to yield the proposed answers. Where did my professor got this? Is the above solution correct? Thanks again.

$$  (x + 2y - 4)dx - (2x + y - 5)dy = 0 $$
Let $X = x + a$ and $Y= y+ b$ and let's see if we can choose suitable values for $a$ and $b$.
$$ (X - a + 2Y - 2b - 4)dX - (2X-2a + Y - b - 5)dY =0$$
It's easy to find what the "suitable" values should be, since we just want them to cancel out the constants! So we solve the following system of equations:
$$\begin{align*}
-4 &= a + 2b\\
-5 &= 2a + b
\end{align*}$$
I'll let you verify that the correct values are $a=-2$ and $b=-1$. This leaves us with
$$(X + 2Y)dX - (2X+Y)dY=0$$
Can you take it from here?
EDIT: Now let $Y=VX$, and hence $dY = VdX + XdV$.
$$\begin{align*}
X(1 + 2V)dX - X(2+V)(VdX+XdV) &=0\\
X(1-V^2)dX - X^2(2+V)dV &=0\\
X(1-V^2)dX &= X^2(2+V)dV\\
\dfrac{1}{X}dX &= \dfrac{2+V}{1-V^2}dV\\
\log(X) + C_1 &= \dfrac{1}{2}(-3 \log(1-V)+\log(1+V))\\
C_2X^2 &= \dfrac{1+V}{(1-V)^3}\\
C_2X^2(1-V)^3 &= 1+V\\
C_2X^2(1-\dfrac{Y}{X})^3 &= 1+\dfrac{Y}{X}\\
C_2\dfrac{(X-Y)^3}{X}&= 1+\dfrac{Y}{X}\\
C_2(X-Y)^3 &= X + Y\\
C_2(x-y-1)^3 &= x + y - 3
\end{align*}$$
So your teacher has the correct solution.
