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 $(x-y+4)^3(4x+y+1)^2 = 16$.
 A: 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 $e^{-5c}=16$
A: 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.
A: 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 $y(0)=3$
$$C=-16$$ 
$$(4x+y+1)^2(y-x-4)^3=-16\\(4x+y+1)^2(-(x-y+4))^3=-16$$
$$(4x+y+1)^2(-(x-y+4))^3=-16\qquad \rightarrow \qquad (4x+y+1)^2(x-y+4)^3=16$$
