<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] Note that LHS integral should give
$$ \int \frac{2+v}{(v+4)(1-v)}dv = -\frac35 \ln(1-v) - \frac25 \ln(4+v) $$
After fixing the sign error, you can transform your answer to match the one given.
$$ 3\ln\left(1-\frac{y-3}{x+1}\right) + 2\ln\left(4+\frac{y-3}{x+1}\right) + 5\ln(x+1) = -5c $$
$$ 3\ln \left(\frac{x-y+4}{x+1} \right) + 2\ln\left(\frac{4x+y+1}{x+1}\right) + 5\ln(x+1) = -5c $$
Using the $\log$ properties, this simplifies to
$$ 3\ln (x-y+4) + 2\ln(4x+y+1) = -5c $$
Taking the exponential gives
$$ (x-y+4)^3(4x+y+1)^2 = e^{-5c} $$
and the given initial point gives $e^{-5c}=16$
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