Question

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

• A. two
• B. one
• C. zero
• D. infinite

Hint:

Always check whether the differential equation is defined at the given initial pointIf undefined, no solution exists through that point.

Answer 1

The Correct Option is C

Step 1: Identify the differential equation.
\[ \frac{dy}{dx} = \frac{y+1}{x-1} \]
This is a separable differential equation.

Step 2: Separate variables.
\[ \frac{dy}{y+1} = \frac{dx}{x-1} \]

Step 3: Integrate both sides.
\[ \int \frac{dy}{y+1} = \int \frac{dx}{x-1} \]
\[ \ln|y+1| = \ln|x-1| + c \]

Step 4: Simplify the solution.
\[ \ln|y+1| - \ln|x-1| = c \]
\[ \ln \left|\frac{y+1}{x-1}\right| = c \]
\[ \frac{y+1}{x-1} = k \]

Step 5: Apply initial condition \( y(1) = 2 \).
\[ \frac{2+1}{1-1} = \frac{3}{0} \]
This is undefined.

Step 6: Interpretation.
The differential equation is not defined at \( x = 1 \), since denominator \( x-1 = 0 \).
Hence no solution can pass through the point \( (1,2) \).

Step 7: Final conclusion.
\[ \boxed{\text{Number of solutions} = 0} \]