Question

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

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

Answer 1

Answer: (C)

$y=ae^{mx}+be^{-mx}\quad\ldots\left(i\right)$
Differentiating w.r.t. $x$, we get
$\frac{dy}{dx}=mae^{mx}-mbe^{-mx}=m\left(ae^{mx}-be^{-mx}\right)$
Again differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{dx^{2}}=m\left(ame^{mx}+bme^{-mx}\right)$
$=m^{2}\left(ae^{mx}+be^{-mx}\right)=m^{2}y$ [using $(i)$]
$\Rightarrow \frac{d^{2}y}{dx^{2}}-m^{2}y=0$