Question

$y={ae}^{mx}+{be}^{-mx}$ satisfies which of the following differential equations:

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

Answer 1

Answer: (C)

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

Solution:

We have, $y=a{e}^{mx}+b{e}^{-mx}$

On differentiating, we get

$\frac{dy}{dx}=am{e}^{mx}-bm{e}^{-mx}$

Again differentiating, we get

$\frac{d^{2}y}{d{x}^2}=a{m}^2{e}^{mx}+b{m}^2{e}^{-mx}$

or, $\frac{d^{2}y}{d{x}^2}=m^2(a{e}^{mx}+b{e}^{-mx})$

or, $\frac{d^{2}y}{d{x}^2}=m^{2}y$ (Given, $y=a{e}^{mx}+b{e}^{-mx}$)

or, $\frac{d^{2}y}{d{x}^2}-m^{2}y=0$

Hence, C is the correct option.