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Math Help - PDE - Wave eqn with Neumann

  1. #1
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    Unhappy 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.

    thanks for reading.
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  2. #2
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    Quote Originally Posted by primenumbers View Post
    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.

    thanks for reading.
    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

     <br />
\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

     <br />
\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
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    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

     <br />
\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

     <br />
\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.
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  4. #4
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    Quote Originally Posted by primenumbers View Post
    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.
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