If \(y = 3e^{2x} + 2e^{3x}\), then \(\displaystyle\frac{d^2y}{dx^2}+6y\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(5\displaystyle\frac{dy}{dx}\)
Differentiating twice, using \(\displaystyle\frac{d}{dx}e^{mx} = me^{mx}\):
\( \displaystyle \frac{dy}{dx} = 6e^{2x} + 6e^{3x} \)
\( \displaystyle \frac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x} \)
Adding 6y:
\( \displaystyle \frac{d^2y}{dx^2} + 6y = (12e^{2x} + 18e^{3x}) + (18e^{2x} + 12e^{3x}) \)
\(= 30e^{2x} + 30e^{3x}\)
\( \displaystyle = 5\left(6e^{2x} + 6e^{3x}\right) = 5\frac{dy}{dx} \)
Shortcut: For a single term \(e^{mx}\), \(y'' + 6y = (m^2 + 6)e^{mx}\) and \(5y' = 5me^{mx}\). Since \(m^2 + 6 = 5m\) for both m = 2 and m = 3, the relation \(y'' + 6y = 5y'\) holds for each term, and so for their sum.
NCERT (2026-27 edition):
Second Order Derivative
