Trying to solve differential equation $y'=\frac{3x-y+4}{x+y}$ ÊDIT: Found the second mistake (failed by calculating $u_{1,2}$)!
I might have made a mistake, however, I am not able to detect it. Here we go:

$$y'=\frac{3x-y+4}{x+y}, \quad y(1)=1$$

1.) set $x = X+a$ and $y=Y+b$, so $$\frac{dY}{dX} = \frac{3X-Y+(3a-b+4)}{X+Y+(a+b)}$$ 
2.) Choose $a,b$ with $a=-1$ and $b=1$, then the differential equation is:
$$\frac{dY}{dX} = \frac{3X-Y}{X+Y} = \frac{3-\frac{Y}{X}}{1+\frac{Y}{X}}$$
3.) Now substitution: $Y = uX$, so $Y' = u+ Xu'$ and we have
$$\frac{dY}{dX} = \frac{3-u}{1+u} = u+Xu'$$
4.) Solve this, ending with:
$$\frac{1+u}{-u^2-2u+3} du = \frac{dX}{X}$$
5.) Solving this by integration, ending with:
$$-\frac{1}{2} \ln (-u^2-2u+3) = \ln(X) + \ln(C)$$
$$-u^2-2u+3 = \exp(-2(\ln(CX))) = e^{\ln((CX)^{-2})} = \frac{1}{(CX)^2}$$
$$-u^2-2u+3-\frac{1}{(CX)^2} = 0$$ 
6.) So I get $$u_{1,2} = -1 \pm \sqrt{16-\frac{4}{(CX)^2}}$$
Because of the substitution and by having chosen $Y=y-1$ and $X=x+1$ it follows that
$$y-1 = -x-1 \pm (x+1)\cdot \sqrt{16-\frac{4}{(C(x+1))^2}}$$
Finally:
$$y(x) = -x \pm \sqrt{16(x+1)^2-\frac{4}{C^2}}$$
7.) With the initial values I get
$$C=\frac{1}{\sqrt{15}}$$
8.) However, my result is not true for the differential equation I started with, since left hand side and right hand side are not equal.
Thanks for any advice/hint :) 

Your mistake started from 6 (I'm writing $c=C^2$ here)
$$ u^2 + 2u = 3 - \frac{1}{cX^2} $$
$$ (u+1)^2 = 4 - \frac{1}{cX^2} $$
$$ u= -1 \pm \sqrt{4-\frac{1}{cX^2}} $$
or 
$$ \frac{y-1}{x+1} =-1 \pm \sqrt{4-\frac{1}{c(x+1)^2}} $$
Using the condition $x = 1, y = 1$ we get
$$ -1 \pm \sqrt{4-\frac{1}{4c}} = 0 $$
Only the plus sign satisfies
$$ c = \frac{1}{12} $$
Final solution
$$ y(x) = 1 + (x+1)\left(-1 + \sqrt{4-\frac{12}{(x+1)^2}} \right) $$
Or
$$ y(x)= -x + 2\sqrt{x^2+2x-2}$$
