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Math Help - Linear diff equation help... please!

  1. #1
    rak
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    Linear diff equation help... please!

    Hello everyone, I have an exam in Linear differential equations and transforms soon and could do with some basic-ish help on how to solve second order diff equations.

    The question states:

    The function u(x,t) satisfies the equation

    (d2U/dU2) = (1/c^2) (d2u/dt2) + (2b/c) (du/dt)

    For 0 < x < l , t > 0 , and c > 0 , 0 < b < pi/l are constants.

    Suppose u(x,t) is subject to homogeneous boundary conditions u(0,t) = u(l,t) = 0 and initial condition u(x,0) = 0 for 0 < x < l.


    Use separation of variables to derive general solution:

    u(x,t) = SUM(n-inf) [ Bn * exp(-bct) * sin(wn)t * sin((n pi x)/l)

    Where Bn is the nth value of the constant B

    And wn = c * sqrt ( (n^2 pi^2)/l^2 - b^2)


    I so far have separated variables to find

    X''T = 1/c^2 XT'' + 2b/c XT'

    So

    X''/X = 1/c^2 T''/T + 2b/c T'/T = -k^2

    And

    X'' + k^2 X = 0 => X(x) = Acoskx + Bsinkx


    And similarly


    T''+2bcT' + c^2*k^2*T = 0


    Could anybody confirm that this is the right approach? And where do I go from here, how do I solve T?


    Thanks very much, its much appreciated, and I apologise for the crude writing of equations I tried my best to make them look clear!
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  2. #2
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    Quote Originally Posted by rak View Post
    Hello everyone, I have an exam in Linear differential equations and transforms soon and could do with some basic-ish help on how to solve second order diff equations.

    The question states:

    The function u(x,t) satisfies the equation

    (d2U/dU2) = (1/c^2) (d2u/dt2) + (2b/c) (du/dt)

    For 0 < x < l , t > 0 , and c > 0 , 0 < b < pi/l are constants.

    Suppose u(x,t) is subject to homogeneous boundary conditions u(0,t) = u(l,t) = 0 and initial condition u(x,0) = 0 for 0 < x < l.


    Use separation of variables to derive general solution:

    u(x,t) = SUM(n-inf) [ Bn * exp(-bct) * sin(wn)t * sin((n pi x)/l)

    Where Bn is the nth value of the constant B

    And wn = c * sqrt ( (n^2 pi^2)/l^2 - b^2)


    I so far have separated variables to find

    X''T = 1/c^2 XT'' + 2b/c XT'

    So

    X''/X = 1/c^2 T''/T + 2b/c T'/T = -k^2

    And

    X'' + k^2 X = 0 => X(x) = Acoskx + Bsinkx


    And similarly


    T''+2bcT' + c^2*k^2*T = 0


    Could anybody confirm that this is the right approach? And where do I go from here, how do I solve T?


    Thanks very much, its much appreciated, and I apologise for the crude writing of equations I tried my best to make them look clear!
    Right so far! The k is found from the BC (it's k=\frac{\pi}{l} )

    The ODE T''+2bcT' + c^2 k^2 T = 0 is constant coefficient so use the usual method, seek a solution of the form T = e^{mt}. See the tutorial at the top of this forum for a refresher. A note, since 0 < b < \frac{\pi}{l},\;m will be complex.
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