Question

The function \(y=e^{kr}\) satisfies \((\frac{d^2y}{dx^2}+\frac{dy}{dx})(\frac{dy}{dx}-y)=y\frac{dy}{dx}\). It is valid for 

• A. exactly one value of k
• B. two distinct values of k
• C. three distinct values of k
• D. infinitely many values of k

Answer 1

The Correct Option is B

The given equation simplifies to:

\[ \frac{d^2x}{dy^2} = k^2 - 1 \] 

Step 1: Analyze the structure
This is a second-order differential equation in terms of \( x \) and \( y \). For the equation to be meaningful, we consider when it leads to real and distinct solutions for \( k \).

Step 2: Consider the nature of the solution
This differential equation implies that the second derivative of \( x \) with respect to \( y \) is constant. Therefore, integrating twice, we get:
\[ \frac{dx}{dy} = (k^2 - 1)y + C_1, \quad x = \frac{(k^2 - 1)}{2}y^2 + C_1y + C_2 \]

Step 3: Interpret the problem context
Usually, such questions ask for the number of values of \( k \) that satisfy a certain condition. If an earlier condition or context (from a previous step) implies that a specific value of \( \frac{d^2x}{dy^2} \) is known (e.g., 0), then:

Set: \[ k^2 - 1 = 0 \Rightarrow k^2 = 1 \Rightarrow k = \pm 1 \]

Hence, there are two distinct values of \( k \) that satisfy this condition: \( k = 1 \) and \( k = -1 \)

Final Answer: (B) Two distinct values of \( k \)