If \(y=500e^{7x}+600e^{-7x}\) and \(\displaystyle\frac{d^2y}{dx^2}=ky\), then the value of k is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 49
Differentiating twice:
\( \displaystyle \frac{dy}{dx} = 3500e^{7x} - 4200e^{-7x} \)
\( \displaystyle \frac{d^2y}{dx^2} = 24500e^{7x} + 29400e^{-7x} \)
\( \displaystyle = 49\left(500e^{7x} + 600e^{-7x}\right) \)
\(= 49y\)
Comparing with \(\displaystyle\frac{d^2y}{dx^2} = ky\) gives k = 49.
NCERT (2026-27 edition):
Second Order Derivative
