First order linear differential equation with $\mu(x,y)$ How can I solve this first order linear differential equation?
$$y'=1-\frac{2}{x+y}$$
I have tried turning it into an inexact differential equation, but I get an integration factor $\mu(x,y)$ and I don't know how to apply it.
 A: Set $v = \frac{2}{x+y},$ then $2\frac{\,dv}{\,dx} = -v^2 \left( 1 + \frac{\,dy}{\,dx}\right)$
The given equation reduces to 
$$
-2\frac{\,dv}{v^2(2-v)} = \,dx.
$$
Integrate. Use partial fractions for the left hand side. Can you take it from here?
A: You can use a substitution to settle this differential equation. Let $s(x)=x+y(x)$. Differentiating this gives us 
\begin{equation*}
\frac{ds}{dx}(x)=1+\frac{dy}{dx}(x)=1+1-\frac{2}{s(x)}.
\end{equation*}
Algebraic manipulation gives
\begin{equation*}
\frac{ds}{dx}(x)=2-\frac{2}{s(x)}=2\left( 1-\frac{1}{s(x)} \right) \\
\Rightarrow \frac{\frac{ds}{dx}(x)}{1-\frac{1}{s(x)}}=2.
\end{equation*}
Integrating both sides with respect to $x$ gives us
\begin{equation*}
\int\frac{\frac{ds}{dx}(x)}{1-\frac{1}{s(x)}}dx=\int 2dx \\
\Rightarrow \ln (s(x)-1)+s(x)=2x+C
\end{equation*}
for a constant $C$. Rearrange for $s(x)$ and substitute back. 
A: If you wollow what Raghav suggested, by a rather simple integration, you obtain "x" as a function of "v".   
The result of this integration leads to x = 1 / v + ArcTanh[1 - v] + C.   
For simplicity, we shall add a boundary condition y(0) = 2 which then reduce to v(0) = 1 so C = -1. So, the solution is just   
x = (1 - v) / v + ArcTanh[1 - v]   
Back to definition, you then have an implicit relation of "y" as a function of "x" that you will need to solve numerically. But you can also consider the solution as a parametric formulation of the problem since (x + y) = 2 / v. Then  
x = (1 - v) / v + ArcTanh[1 - v]
y = (1 + v) / v - ArcTanh[1 - v]     
In my opinion, this is simpler to the direct expression of "y" as a function of "x". For the same boundary conditions, the solution of the differential equation you set (and it is not simple to reach it) is
y = 1 - x + W[Exp(1 + 2 x)]  
where W(z) represents the Lambert function
