<s>[INST] Reduce $\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$ to a homogenous equation by substituting $x=X-1$ and $y=Y+3$ 
By substituting $x=X-1$ and $y=Y+3$ reduce the differential equation
  $$
\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}
$$
  to a homogeneous equation and hence find the general solution in terms of $x$ and $y$.

All I have so far is:
$$
\frac{dy}{dx}=\frac{4X-Y}{2X+Y}
$$
Then I thought about doing something like:
$$
X=x+1\\
dX=dx
$$
But I wasn't sure whether that was correct or useful here. Can somebody help point me in the right direction? The answer listed is $(y-x-4)^3(y+4x+1)^2=A$ where $|A|=e^c$. 
 [/INST] Notice that $$\frac{dY}{dX}=\frac{4X-Y}{2X+Y}= \frac{4 -\frac{Y}{X}}{2 + \frac{Y}{X}}.$$
So, let $V= Y/X$, so $Y=VX\implies dY/dX = V + X(dV/dX)$. With these, we substitute: 
$$V + X\frac{dV}{dX} = \frac{4 -V}{2 +V}\implies \frac{2+V}{4-3V-V^2}dV = \frac{dX}{X}.$$
This can be broken up using partial fractions. After we will exploit the properties of logarithms to get an answer in the form of $$(1-V)^3(V+4)^2X^5=e^C.$$
Since $V=Y/X$, we substitute that back in, simplify, and then substitute $X=x+1$, and $Y=y-3$, and simplify again. It's more of an algebra exercise!
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