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

one

The correct answer is Option (2) → one ##

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

$\Rightarrow \frac{dy}{y + 1} = \frac{dx}{x - 1} \quad \text{[applying variable separable method]}$

On integrating both sides, we get

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

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

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

$C(y + 1) = (x - 1)$

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

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

So, the required solution is $x - 1 = 0$.

Hence, only one solution exist.