<s>[INST] Show that the substitution $x=X-1$ and $y=Y+3$ turns $\frac{dy}{dx}=...$ into a homogenous equation. $$\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$$ 
After substitution of the given variable I get $$\frac{dY}{dX} = \frac{4-\frac{Y}{X}}{2+\frac{Y}{X}}$$ which seems to give a homogenous equation. (There is no answer given for this part of the question)
But then the next question says find the particular solution with $x=0$ and $y=3$ giving your answer in the form $f(x,y)=c$. Then after this point I get an answer that is completely different to the answer given in the book.
Let $Y=vX$. Then $\frac{dY}{dX} = X\frac{dv}{dX}+v$
Replace, then $$X\frac{dv}{dX}+v=\frac{4-v}{2+v}$$
Making the Integrals
$$\int \frac{1}{\frac{4-v}{2+v}-v}dv=\int\frac{1}{X}dX$$
$$\frac{3\ln(1-v)}{5}-\frac{2\ln(4+v)}{5}-\ln(X)=c$$
Replacing everything back to $y$ and $x$.
$$\frac{3\ln(1-\frac{y-3}{x+1})}{5}-\frac{2\ln(4+\frac{y-3}{x+1})}{5}-\ln(x+1)=c$$
According to this $c=-\frac{2\ln(4)}{5}$
Even before the replacement stages the answer is already completely different to that of the answer given.
What did I do wrong?
Just for reference, answer given is $(x-y+4)^3(4x+y+1)^2 = 16$.
 [/INST] Solving for $u,v$
$$
4x-y+7=v\\
2x+y-1=u
$$
and substituting into
$$
\frac{dy}{dx} = \frac{4x-y+7}{2x+y-1}\rightarrow \frac{dv}{du} = \frac{4u-v}{2u+v}\;\;\;\mbox{which is homogeneous}
$$
Now making $z = v = \lambda u$ and correspondingly $dz = \lambda du+ud\lambda$ we have the new DE
$$
\frac{dv}{du} = \frac{dz}{du} = \lambda + u \frac{d\lambda}{du} = \frac{4-\lambda}{2+\lambda}
$$
or
$$
u\frac{d\lambda}{du} = \frac{4-\lambda}{2+\lambda}-\lambda
$$
This is a separable DE so
$$
\frac{du}{u} = \frac{d\lambda}{\frac{4-\lambda}{2+\lambda}-\lambda}
$$
and solving
$$
\log u = -\frac{3}{5} \log (1-\lambda )-\frac{2}{5} \log (\lambda +4)+C_0
$$
or
$$
u = \frac{C_1}{(1-\lambda)^{\frac 35}(\lambda+4)^{\frac 25}}
$$
or
$$
u^5 = \frac{C_2}{(1-\lambda)^{3}(\lambda+4)^{2}}
$$
etc.
</s>