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

Answer: (B)

two distinct values of k

Answer 2

This simplifies to: $k^2 - 1 = (\frac{d^2x}{dy^2})$
Now, let's analyze the resulting expression $k^2 - 1 = (\frac{d^2x}{dy^2})$ :
Valid Domain for k: The equation $k^2 - 1 = (\frac{d^2x}{dy^2})$ is valid for all real values of k. There are no restrictions on k based on this equation.
Valid Domain for $(\frac{d^2x}{dy^2})$: Since $(\frac{d^2x}{dy^2})$ represents the second derivative of x with respect to y, this term is valid as long as x is a function that can be differentiated twice with respect to y. There are no specific restrictions on the domain of x based on this equation.
Therefore, the differential equation is valid for all real values of k and for functions x that are twice differentiable with respect to y.
The correct option is (B): two distinct values of k.