The number of solutions of dy/dx = y+1/x−1 when y(1) = 2 is

The correct option/answer is one.

To understand the nature of the solutions, we can analyze the given first-order differential equation using the separation of variables method:
$\frac{dy}{dx}=\frac{y+1}{x-1}$

Step 1: Separate the Variables
We rearrange the differential equation to group all terms containing $y$ on the left side and all terms containing $x$ on the right side:
$\frac{dy}{y+1}=\frac{dx}{x-1}$

Step 2: Integrate Both Sides
Now, we integrate both sides of the equation:
$\int \frac{dy}{y+1}=\int \frac{dx}{x-1}$

Evaluating the integrals yields:
$\ln |y+1|=\ln |x-1|+C$
where $C$ is the constant of integration.

We can express the constant $C$ as $\ln \left|k\right|$ (where $k$ is a non-zero real constant):
$\ln |y+1|=\ln \left|k\right(x-1\left)\right|$

Exponentiating both sides to eliminate the natural logarithms gives the general solution family:
$y+1=k(x-1)$
Solving for $y$:
$y=k(x-1)-1$

Step 3: Analyze Solution Uniqueness in the Domain of Definition
Under Picard's Existence and Uniqueness Theorem, a first-order ordinary differential equation in the form $\frac{dy}{dx}=f(x,y)$ has a unique solution in a local region where the function $f(x,y)$ and its partial derivative $\frac{\partial f}{\partial y}$ are continuous.
For this differential equation, the function $f(x,y)=\frac{y+1}{x-1}$ is well-defined and continuous everywhere except at the singular boundary line $x=1$. Thus, in any open interval not containing the singularity, there exists exactly one unique solution curve satisfying any given initial condition.