Homogeneity of a differential equation. Convert $$\frac{dy}{dx} = \frac{-(2x-4y+6)}{x+y-3}$$ into a homogeneous differential equation.
Any help appreciated!  
 A: $$\frac{dy}{dx} = \frac{-(2x-4y+6)}{x+y-3}$$
Cross multiply to get
$$(x+y-3) \, dy +(2x-4y+6) \, dx =0$$
This can be made into a homogeneous differential equation by a linear transformation of both variables:
$$z=x+a, s=y+b,\text { where $ a, b $ are constants.}$$
Can you see why?
A: The given equation is not a homogeneous equation. In order to make it a homogeneous one you need to eliminate the constant terms (Also note that the coefficients of $x$ and $y$ in the parantheses are not proportional). This can be done by means of the change of variables $x=u+h$ and $v=y+k$, where $h$ and $k$ are constants satisfying the equations $$h+k-3=0$$ $$2h-4k+6=0$$. Solving this system of linear equations we get $h=1$ and $k=2$. So our DE becomes $$\frac{dv}{du}=-\frac{2u-4v}{u+v}.$$ Now set $$U=\frac{v}{u}$$ or $v=uU$ which implies $$\frac{dv}{du}=U+u\frac{dU}{du}.$$ Substituting these into the equation we have $$U+u\frac{dU}{du}=-\frac{2-4U}{1+U}.$$ Thus after a few algebra you arrive at a separable equation in $u$ and $U$, which requires only integration to find the solution:$$\frac{1+U}{3U-2-U^2}dU=\frac{du}{u}.$$
