# Separating variables - Wave Eqn

• May 23rd 2011, 05:15 PM
Miika89
Separating variables - Wave Eqn
[IMG]file:///F:/Users/Miika/AppData/Local/Temp/moz-screenshot-1.png[/IMG]Hi,

In the question below I dont understand where the e^(-kt) comes from? In general when I have been using laplace and fourier transforms the answers are in the form 4/(n^3)*(Pi^3) before the series and not an exponential. Any help on how e^(-kt) is derived would be much appriciated.

http://i114.photobucket.com/albums/n...ika89/wave.jpg
• May 23rd 2011, 06:03 PM
TheEmptySet
When you separate the equations you should get

$\displaystyle X\ddot{T}+2kX\dot{T}=c^2X''T \iff \frac{\ddot{T}+2k\dot{T}}{c^2T}=\frac{X''}{X}=-\lambda^2$

So solving the t equation gives

$\displaystyle \ddot{T}+2k\dot{T}+c^2\lambda^2=0$

This is a 2nd order ODE with constant coefficients this gives

$\displaystyle m^2+2km=-c^2\lambda^2 \iff (m+k)^2=k^2-c^2\lambda^2 \iff m=-k \pm \sqrt{k^2-c^2\lambda^2}$

So the solutions are

$\displaystyle T(t)=e^{-kt}(c_1\cos([\sqrt{k^2-c^2\lambda^2}]t)+c_2\sin([\sqrt{k^2-c^2\lambda^2}]t))$

If

$\displaystyle k^2-c^2\lambda^2 < 0$
• May 23rd 2011, 06:43 PM
Miika89

In my notes it states that the wave eqn uses the form T = Acos(m) + Bsin(m) and not T = Ae^(m) + Be^(-m). Is the above question a combination of both?
• May 23rd 2011, 07:33 PM
TheEmptySet
Quote:

Originally Posted by Miika89