If \(y = Ae^{2x} + Be^{-3x}\), where A and B are constants, then \(\displaystyle\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(6y\)
Differentiating twice, using \(\frac{d}{dx}e^{kx} = k\,e^{kx}\):
\( \displaystyle \frac{dy}{dx} = 2Ae^{2x} - 3Be^{-3x} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = 4Ae^{2x} + 9Be^{-3x} \)
Adding the two:
\( \displaystyle \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} = 6Ae^{2x} + 6Be^{-3x} \)
\( = 6\left(Ae^{2x} + Be^{-3x}\right) \)
\( = 6y \)
Both exponentials get the same factor 6, since \(4 + 2 = 6\) and \(9 - 3 = 6\).