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The general solution of $\frac{d y}{d x}=\frac{x+y+1}{y-x+1}$ is
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Verified Answer
The correct answer is:
$2 x y+(x+1)^2-(y+1)^2=C$
$\begin{aligned}
& \text { We have, } \frac{d y}{d x}=\frac{x+y+1}{y-x+1} \\
& \Rightarrow \quad \begin{aligned}
y d y-x d y+d y & =x d x+y d x+d x \\
(y+1) d y & =(x+1) d x+y d x+x d y \\
(y+1) d y & =(x+1) d x+d(x y)
\end{aligned}
\end{aligned}$
On integrating, $\frac{(y+1)^2}{2}=\frac{(x+1)^2}{2}+x y+C$
$\begin{aligned}
& \Rightarrow \quad(x+1)^2-(y+1)^2+2 x y=-2 C \\
& \Rightarrow \quad 2 x y+(x+1)^2-(y+1)^2=C
\end{aligned}$
& \text { We have, } \frac{d y}{d x}=\frac{x+y+1}{y-x+1} \\
& \Rightarrow \quad \begin{aligned}
y d y-x d y+d y & =x d x+y d x+d x \\
(y+1) d y & =(x+1) d x+y d x+x d y \\
(y+1) d y & =(x+1) d x+d(x y)
\end{aligned}
\end{aligned}$
On integrating, $\frac{(y+1)^2}{2}=\frac{(x+1)^2}{2}+x y+C$
$\begin{aligned}
& \Rightarrow \quad(x+1)^2-(y+1)^2+2 x y=-2 C \\
& \Rightarrow \quad 2 x y+(x+1)^2-(y+1)^2=C
\end{aligned}$
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