The particular solution of the differential equation \(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\) when \(\mathrm{x}=\frac{2}{3}\) and \(\mathrm{y}=\frac{1}{3}\) is

\(
\frac{d y}{d x}=\frac{x+y+1}{x+y-1}
\)
Put \(x+y=v \Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x}\)
\(\therefore \frac{d v}{d x}-1=\frac{v+1}{v-1} \Rightarrow \frac{d v}{d x}=\frac{v+1}{v-1}+1=\frac{2 v}{v-1} \)
\( \therefore \int \frac{(v-1) d v}{2 v}=\int d x \)
\( \therefore \int \frac{1}{2} d V-\int \frac{1}{2 v} d v=\int d x=\frac{v}{2}-\frac{1}{2} \log |v|=\) \(x+c \)
\( \therefore \frac{x+y}{2}-\frac{1}{2} \log |x+y|=x+c\)
We have \(\mathrm{x}=\frac{2}{3}, \mathrm{y}=\frac{1}{3}\)
\(
\begin{aligned}
& \therefore \frac{1}{2}-\frac{1}{2} \log |1|=\frac{2}{3}+c \Rightarrow c=\frac{1}{2}-\frac{2}{3}=\frac{-1}{6} \\
& \therefore \frac{x+y}{2}-\frac{1}{2} \log |x+y|=x-\frac{1}{6}
\end{aligned}
\)
\( \therefore \frac{x+y}{2}-\frac{1}{2} \log |x+y|=x-\frac{1}{6} \)
\( \therefore(x+y)-\log |x+y|=2 x-\frac{2}{6} \Rightarrow\) \(y-x+\frac{1}{3}=\log |x+y|\)