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Help to solve 4th order ode

  1. Dec 26, 2009 #1

    I try to solve this PDE as below:


    With boundary conditions of x=0.y=y''=0 and x=-h,y=y'=0.
    k and h are constant.
    The solution look like from the Fourier transforms.
    There is one term in the solution which is :

    beta_n is the positive roots of the equation of beta_n=tan(beta_n). n is from 1 to infinity.

    I have no idea how this term come from. If someone can help me, I really appreciate it!!

  2. jcsd
  3. Dec 26, 2009 #2
    First of all your ODE has the following form

    z''-kz+1=0, where z=y''

    The general solution to z is obtained by the wery standard method which lead to ODE

    y''= exp(k^(1/2)*x)*C1+exp(-k^(1/2)*x)*C2+1/k

    and after double integration

    y = C1/k*exp(k^(1/2)*x)+C2/k*exp(-k^(1/2)*x)+1/2/k*x^2+C3*x+C4

    what is the general solution to your ODE. Substitution of your boundary conditions gives you 4 algebraic equations for unknowns C1,C2,C3 and C4.
  4. Dec 26, 2009 #3


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    Almost the same thing: the characteristic equation for the differential equation y""- ky"+ 1= 0, which is the same as y""- ky"= -1, is [itex]r^4- kr^2= r^2(r^2- k)= r^2(r- \sqrt{k})(r+ \sqrt{k})[/itex] which has 0 as a double root and roots [itex]\sqrt{k}[/itex] and [itex]-\sqrt{k}[/itex].

    The general solution to the associated homogeneous equation is [itex]z= C_1x+ C_2+ C_3e^{x\sqrt{k}}+ C_4e^{-x\sqrt{k}}[/itex]. Since the right hand side is a constant, normally we would try a constant solution but since a constant and x already satisfy the homogeneous solution, we try [itex]z= Ax^2[/itex]. Then [itex]z'= 2Ax[/itex], [itex]z"= 2A[/itex] and [itex]z"'= z""= 0[/itex]. The equation becomes 0+ k(2A)= -1 so A= -1/(2k). The general soution to that equation is
    [tex]z(x)= C_1x+ C_2+ C_3e^{x\sqrt{k}}+ C_4e^{-x\sqrt{k}}-\frac{1}{2k}x^2[/tex]
  5. Dec 27, 2009 #4

    Thank you for response. If I want to use Fourier transform to solve this one, how can I find the boundary conditons for y''(-h)? I need to use inverse Laplace transform to convert the y function. Therefore, it will become much easier to use the Fourier transforms.

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