Q. $y = ae^{mx} + be^{- mx}$ 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) $\frac{d^{2}y}{dx^{2}} - m^{2}y = 0$
Explanation
(d)
$y = ae^{mx} + be^{- mx}$
On differentiating $w.r.t.x$, we get
$\frac{dy}{dx} = mae^{mx} - mbe^{- mx}$
Again, on differentiating, we get
$\frac{d^{2}y}{dx^{2}} = m^{2}ae^{mx} + m^{2}be^{- mx}$
$= m^{2}\left( ae^{mx} + be^{- mx} \right) = m^{2}y$
$\Rightarrow \ \frac{d^{2}y}{dx^{2}} - m^{2}y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.