logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

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

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

Can you answer this question?
 
 

1 Answer

0 votes
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 $C$
answered May 21, 2013 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...