Question

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

• A. $m^2(a{e}^{mx}-b{e}^{-mx})$
• B. $1$
• C. $0$
• D. none of these

Answer 1

Answer: (C)

$y=a{e}^{mx}+b{e}^{-mx}$
$\therefore \frac{dy}{dx}=am{e}^{mx}-mb{e}^{-mx}$
Again $\frac{d^{2}y}{d{x}^2}=a{m}^2{e}^{mx}+m^{2}b{e}^{-mx}$
$=m^2(a{e}^{mx}+b{e}^{-mx})=m^{2}y$
or $\frac{d^{2}y}{d{x}^2}-m^{2}y=0$