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