# PDE - Wave eqn with Neumann

• May 12th 2009, 07:19 AM
PDE - Wave eqn with Neumann
Im studying the wave equation with Neumann boundary conditions, and got stuck at the following. can anyone help out?

i have the following equation:
T''(t) = lambda * (c^2)*T(t)
where lamba = - (n*pi*c/l)^2

i want to see how :

T''(t) = lambda * (c^2)*T(t) reduces to...
T(t) = Asin(lambda*t) + Bcos (lamdba*t)

where did the sin and cos come from? is this solved by separation of variables?i tried every way but cant understand.

• May 12th 2009, 08:25 AM
Jester
Quote:

Originally Posted by primenumbers
Im studying the wave equation with Neumann boundary conditions, and got stuck at the following. can anyone help out?

i have the following equation:
T''(t) = lambda * (c^2)*T(t)
where lamba = - (n*pi*c/l)^2

i want to see how :

T''(t) = lambda * (c^2)*T(t) reduces to...
T(t) = Asin(lambda*t) + Bcos (lamdba*t)

where did the sin and cos come from? is this solved by separation of variables?i tried every way but cant understand.

Let back up a bit. From your original PDE $\displaystyle u_{tt} = c^2 u_{xx}$ upon a separation of variables $\displaystyle u = T(t)X(x)$ we get the separated equation

$\displaystyle \frac{T''}{T} = c^2 \frac{X''}{X}$ and as you said if $\displaystyle \frac{X''}{X} = \lambda = - \frac{n^2 \pi^2}{l^2}$ (b/c of the boundary condition). So

$\displaystyle \frac{T''}{T} = -c^2 \omega^2$ where $\displaystyle \omega = \frac{n \pi}{l}$

so $\displaystyle T'' + c^2\omega^2 T = 0$ and the general solution of this ODE is $\displaystyle T = c_1 \sin c \omega t + c_2 \cos c \omega t$
• May 12th 2009, 08:49 AM
Quote:

Originally Posted by danny arrigo
Let back up a bit. From your original PDE $\displaystyle u_{tt} = c^2 u_{xx}$ upon a separation of variables $\displaystyle u = T(t)X(x)$ we get the separated equation

$\displaystyle \frac{T''}{T} = c^2 \frac{X''}{X}$ and as you said if $\displaystyle \frac{X''}{X} = \lambda = - \frac{n^2 \pi^2}{l^2}$ (b/c of the boundary condition). So

$\displaystyle \frac{T''}{T} = -c^2 \omega^2$ where $\displaystyle \omega = \frac{n \pi}{l}$

so $\displaystyle T'' + c^2\omega^2 T = 0$ and the general solution of this ODE is $\displaystyle T = c_1 \sin c \omega t + c_2 \cos c \omega t$

sorry, i dont understand one thing only. why would that be the general solution? that is where i got lost. would i have to know this by heart?

thanks for helping.
• May 12th 2009, 10:02 AM
Jester
Quote:

Originally Posted by primenumbers
sorry, i dont understand one thing only. why would that be the general solution? that is where i got lost. would i have to know this by heart?

thanks for helping.

Go to Chris L T521's differential equation tutorial at the top of this subforum. You'll find out there how to solve this type of ODE.