Question

The differential equation of $y = Ae^{2x} + Be^{-2x} $ is :

• A. $\frac{dy}{dx} - 4y = 0 $
• B. $\frac{d^2 y}{dx^2 } - 4y = 0 $
• C. $\frac{d^2y}{dx^2} = y^2 $
• D. $\frac{d^2y}{dx^2} - y = 0 $

Answer 1

Answer: (B)

Given : $y = Ae^{2x} + Be^{-2x}$
$ \therefore \frac{dy}{dx} = Ae^{2x} . 2 + Be^{-2x} \left(-2\right) $
$\Rightarrow \frac{dy}{dx} = 2 Ae^{2x} - 2 Be^{-2x} $
$\frac{d^{2}y}{dx^{2} } = 2Ae^{2x}.2- 2 Be^{-2x} (- 2)$
$ = 4 Ae^{2x} + 4 Be^{-2x} $
$\Rightarrow \frac{d^{2}y}{dx^{2} } = 4 y \Rightarrow \frac{d^{2}y}{dx^{2}} - 4y = 0 $