The general solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y+1}{x+y-1}\) is

\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y+1}{x+y-1}...(i)\)
Put \(x+y=\mathrm{v}\)...(ii)
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{dv}}{\mathrm{~d} x}-1...(iii)\)
Substituting (ii) and (iii) in (i), we get
\(\begin{aligned}
& \frac{d v}{d x}-1=\frac{v+1}{v-1} \\
& \Rightarrow \frac{d v}{d x}=\frac{2 v}{v-1} \Rightarrow \frac{v-1}{2 v} d v=d x
\end{aligned}\)
Integrating on both sides, we get
\(\begin{aligned} & \frac{\mathrm{v}}{2}-\frac{1}{2} \log \mathrm{v}=x+\mathrm{c}_1 \\ & \Rightarrow \mathrm{v}-\log \mathrm{v}=2 x+2 \mathrm{c}_1 \\ & \Rightarrow x+y-\log (x+y)=2 x+2 \mathrm{c}_1 \\ & \Rightarrow y=x+\log (x+y)+\mathrm{c}, \text { where } \mathrm{c}=2 \mathrm{c}_1\end{aligned}\)