# 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

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

So solving the t equation gives

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

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

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

So the solutions are

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

If

$k^2-c^2\lambda^2 < 0$
• 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

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?

Your notation is hard to read. You have not written your independent variables. When you say the wave equation do you mean with no damping term?
The addition of other terms and derivatives will change the form of the solution.
• May 23rd 2011, 11:06 PM
Miika89
Yes, with no damping term. You answered my question with your final sentence thank you.