The number of Solution s of $\frac{{dy}}{{dx}} = \frac{{y + 1}}{{x - 1}}$, when $y(1) = 2$ is

Given that, $\frac{{dy}}{{dx}} = \frac{{y + 1}}{{x - 1}}$

$\Rightarrow$ $\frac{{dy}}{{y + 1}} = \frac{{dx}}{{x - 1}}$

On integrating both sides,

we get
$\log (y + 1) = \log (x - 1) - \log C$
$C(y + 1) = (x - 1)$

$\Rightarrow$ $C = \frac{{x - 1}}{{y + 1}}$

When $x = 1$ and $y = 2$, then $C = 0$

So, the required solution of the differential equation is $x - 1 = 0$.
Hence, only one Solution exist.