# Thread: PDE - Wave eqn with Neumann

1. ## 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.

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 $u_{tt} = c^2 u_{xx}$ upon a separation of variables $u = T(t)X(x)$ we get the separated equation

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

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

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

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

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

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

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

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.