Q.The number of solutions of $\frac{dy}{dx}=\frac{y+1}{x-1}$ when $y(1)=2$ is:
(A) none
(B) one
(C) two
(D) infinite

The key idea is that the differential equation is separable, but the initial condition lies on the vertical line $x=1$ where the right-hand side is undefined. This makes it an Initial Value Problem that fails the existence condition at the given point.

Step 1: Separate variables (for $x \neq 1$):

Step 2: Integrate both sides:

Step 3: Exponentiate to get the general solution:

Let $k = \pm e^C$, so $y = k(x-1) - 1$, valid for $x \neq 1$. …

Why this is an Initial Value Problem — and why it’s tricky

When you see $\frac{dy}{dx} = \frac{y+1}{x-1}$ together with $y(1)=2$, you’re looking at a first-order ordinary differential equation with an initial condition. The natural instinct is to separate variables and integrate. But before doing any algebra, pause and check the domain of the right-hand side.

The function $f(x,y) = \frac{y+1}{x-1}$ is undefined when $x=1$. The initial condition specifies $x=1$. That’s a red flag: the equation itself isn’t even defined at the point where you’re supposed to start.

The Existence and Uniqueness Theorem (Picard–Lindelöf) requires $f(x,y)$ to be continuous in a neighbourhood of $(x_0, y_0)$. Here, $f$ has a vertical asymptote at $x=1$ — no neighbourhood exists. So the theorem gives no guarantee. But could there still be a solution that somehow “passes through” that point? Let’s check carefully.

Step-by-step reasoning

1. Separate variables (formally)

   Write $\frac{dy}{y+1} = \frac{dx}{x-1}$. This step assumes $y \neq -1$ and $x \neq 1$. The initial condition $y(1)=2$ gives $y \neq -1$, so that’s fine — but $x=1$ is exactly the point we’re about to integrate through.

2. Integrate both sides

3. Exponentiate

Let $K = e^C > 0$. Then $y+1 = \pm K (x-1)$, or more compactly:

where $A$ is any non-zero real constant (the sign is absorbed into $A$). This is the general solution — a family of straight lines through the point $(1, -1)$.

4. Apply the initial condition $y(1)=2$ Substitute $x=1$, $y=2$:

This gives $3 = 0$, which is impossible. No constant $A$ can satisfy this.

5. Interpret the result …

Method: Counting Solutions of an IVP at a Singular Point

Not every initial value problem has a solution. When the initial point makes the right-hand side undefined, solve the general equation and test whether any solution curve can pass through that point.

Steps

Step 1: Separate and find the general solution.

For $\frac{dy}{dx}=\frac{y+1}{x-1}$, separate: $\frac{dy}{y+1}=\frac{dx}{x-1}$, giving $\log|y+1|=\log|x-1|+C$, i.e. $y+1=k(x-1)$.

Step 2: Substitute the initial condition.

Put the given point into the general solution and check consistency.

Step 3: Interpret the outcome. …

Common Mistakes

Mistake 1: Assuming a separable equation always has exactly one solution through a given point.

Why it's wrong: existence can fail where the right-hand side is undefined; here $x=1$ makes $\frac{y+1}{x-1}$ undefined, so no solution curve passes through $(1,2)$. Correct approach: check whether the initial point lies on a singular line before counting solutions. …