Question

If$y=a{e}^{mx}+b{e}^{-mx}$, then $\frac{d^{2}y}{d{x}^2}-m^{2}y$ equals to ?

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Finding the value of the expression:

Given function,

$y=a{e}^{mx}+b{e}^{-mx}...\left(1\right)$

Differentiate with respect to $x$,

$\frac{dy}{dx}=am{e}^{mx}–bm{e}^{-mx}\left[\because \frac{d}{dx}{e}^x=e^x\right]$

Differentiate again with respect to $x$.

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=a{m}^2{e}^{mx}+b{m}^2{e}^{-mx} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=m^2\left(a{e}^{mx}+b{e}^{-mx}\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=m^{2}y\left[\mathrm{from}1\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}-m^{2}y=0 \end{gathered}$$

Hence, the correct option is B.