Browse Questions

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

$(A)\;\frac{dy}{dx}+my=0 \quad (B)\;\frac{dy}{dx}-my=0\quad(C)\;\frac{d^2y}{dx^2}-m^2y=0 \quad (D)\;\frac{d^2y}{dx^2}+m^2y=0$

Toolbox:
• The general solution of a differential equation is a relation between dependent and independent variable having n arbitary constant.
• The general solution may have more than one form but the arbitary constants must be the same in numbers
Given $y=ae^{mx}+be^{-mx}$
On differentiaiting w.r.t x we get
$\large\frac{dy}{dx}$$=mae^{mx}+(-m)be^{-mx} =mae^{mx}-mbe^{-mx} Again differentiating w.r.t. x we get \large\frac{d^2y}{dx^2}$$=m^2ae^{mx}-m(-m)be^{-mx}$
$=m^2ae^{mx}+m^2be^{-mx}$
$=m^2[ae^{mx}+be^{-mx}]$
But $ae^{mx}+be^{-mx}=y$
$\large\frac{d^2y}{dx^2}$$=m^2y =>\large\frac{d^2y}{dx^2}$$-m^2y=0$
Hence the correct option is $C$