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 $(x-y+4)^3(4x+y+1)^2 = 16$.

I made a summary with the most important steps:
STEP $1$
$$\frac{dY}{dX}=\frac{4(X-1)-Y-3+7}{2(X-1)+Y+3-1}\ \rightarrow\ Y'=\frac{4X-Y}{2X+Y}\rightarrow\ Y'=\frac{4-\frac YX}{2+\frac YX}$$
$$Z=\frac YX\qquad \frac{dY}{dX}=Z+X\frac{dZ}{dX}$$
$$Z+XZ'=\frac{4-Z}{2+Z}\qquad\rightarrow\qquad XZ'=-\frac{Z^2+3Z-4}{Z+2}$$
STEP $2$
$$\ln|CX|=-\int\frac{Z+2}{(Z+4)(Z-1)}dZ$$
Solve the integral by Partial Fractions  then
$$\ln|CX|=-\frac25\ln|Z+4|-\frac35\ln|Z-1|\\\ln|CX(Z+4)^{\frac25}(Z-1)^{\frac35}|=0\\CX(Z+4)^{\frac25}(Z-1)^{\frac35}=1\\(Y+4X)^2(Y-X)^3=C$$
STEP $3$
Substitute Y and X
GENERAL SOLUTION 
$$(4x+y+1)^2(y-x-4)^3=C\\(4x+y+1)^2(x-y+4)^3=C$$
Check the answer
STEP 4
Particular Solution with $y(0)=3$
$$C=-16$$ 
$$(4x+y+1)^2(y-x-4)^3=-16\\(4x+y+1)^2(-(x-y+4))^3=-16$$
$$(4x+y+1)^2(-(x-y+4))^3=-16\qquad \rightarrow \qquad (4x+y+1)^2(x-y+4)^3=16$$
