Question

If $y={ae}^{mx}+{be}^{-mx},\frac{d^{2}y}{d{x}^2}-m^{2}y$ is equal to

• A. $m^2({ae}^{mx}-{be}^{-mx})$
• B. $1$
• C. $0$
• D. $m^2({ae}^{mx}+{be}^{-mx})$

Answer 1

Answer: (C)

$0$

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

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

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

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

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

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

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