y = ae^mx + be^-mx satisfies which of the following differential equa…

The correct answer is Option (3) → $\frac{d^2y}{dx^2} - m^2y = 0$ ##

Given that, $y = ae^{mx} + be^{-mx}$

To get differential equation we have to eliminate $a$ and $b$.

On differentiating both sides w.r.t. $x$, we get

$\frac{dy}{dx} = mae^{mx} - bme^{-mx}$

Again, differentiating both sides w.r.t. $x$, we get

$\frac{d^2y}{dx^2} = m^2ae^{mx} + bm^2e^{-mx}$

$\Rightarrow \frac{d^2y}{dx^2} = m^2(ae^{mx} + be^{-mx})$

$\Rightarrow \frac{d^2y}{dx^2} = m^2y \quad [∵y = ae^{mx} + be^{-mx}]$

$\Rightarrow \frac{d^2y}{dx^2} - m^2y = 0$