Question

If $y=e^{mx}+e^{-mx}$, then prove that $\frac{d^{2}y}{d{x}^2}=m^{2}y$

Answer 1

Given, $y=e^{mx}=e^{-mx}$ .... (i)
Differentiating w.r. to $x$, weget
$\frac{dy}{dx}=e^{mn}\cdot m+e^{-mx}(-m)$
Again differentiating, we get
$\frac{d^{2}y}{d{x}^2}=e^{mx}\cdot m\cdot m+e^{-mx}(-m)(-m)$
$=m^2[e^{mx}+e^{-mx}]$
$\frac{d^{2}y}{d{x}^2}=m^{2}y$. ....[using (i)]