The differential equation of the family of curves \(y = Ae^{3x} + Be^{-3x}\), where A and B are arbitrary constants, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{d^2y}{dx^2}=9y\)
There are two arbitrary constants, so they are eliminated by differentiating twice:
\( \displaystyle \frac{dy}{dx} = 3Ae^{3x} - 3Be^{-3x} \)
\( \displaystyle \frac{d^2y}{dx^2} = 9Ae^{3x} + 9Be^{-3x} \)
\( \displaystyle = 9\left(Ae^{3x} + Be^{-3x}\right) = 9y \)