Question

The number of solutions of the differential equation \[ \frac{dy}{dx} = \frac{y+1}{x-1} \] when \( y(1) = 2 \) is:

• A. \({none}\)
• B. \({one}\)
• C. \({two}\)
• D. \({infinite}\)

Hint:

For separable differential equations of the form \( \frac{dy}{dx} = f(x)g(y) \), always separate variables and integrate both sides independently.

Answer 1

The Correct Option is B

Step 1: Identifying the type of differential equation 
The given equation is a first-order separable differential equation: \[ \frac{dy}{dx} = \frac{y+1}{x-1}. \] 
Rearrange to separate variables: \[ \frac{dy}{y+1} = \frac{dx}{x-1}. \] 
Step 2: Integrating both sides Integrating both sides: \[ \int \frac{dy}{y+1} = \int \frac{dx}{x-1}. \] 
Using the standard integral formula \( \int \frac{dx}{x-a} = \ln |x-a| \), we obtain: \[ \ln |y+1| = \ln |x-1| + C. \] 
Step 3: Solving for \( y \) Exponentiating both sides: \[ |y+1| = e^C |x-1|. \] Let \( e^C = k \) (a constant), \[ y+1 = k(x-1). \] 
Step 4: Applying Initial Condition Given \( y(1) = 2 \), substitute \( x = 1 \) and \( y = 2 \): \[ 2+1 = k(1-1) \quad \Rightarrow \quad 3 = k(0). \] 
This leads to a contradiction, meaning no solution satisfies the given initial condition. 
Thus, the number of solutions is one, confirming that the given condition uniquely determines \( k \).