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Second Order Homogeneous Differential Equation.

  1. Dec 1, 2008 #1
    I would very much appreciate if anyone can help me with this problem; I have approached it from many different angles to no avail.

    The position x(t) of a particle moving along the x-axis is governed by the differential
    equation:

    [tex]x'' + kx' + (n^2)x = 0[/tex] , and initially [tex]x(0) = a[/tex], [tex]x'(0) = u[/tex].

    Show that:

    [tex]
    \int_{0}^{Infinity} x^{2}dt = \frac{1}{2kn^2}((u + ka)^2 + n^2a^2)
    [/tex]

    Show that, as a function of k, this is a minimum when:

    [tex]k^2 = n^2 + \frac{u^2}{a^2}[/tex]
     
  2. jcsd
  3. Dec 1, 2008 #2

    gabbagabbahey

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    You have a pretty straight forward constant coefficient 2nd order ODE. The first step is to solve it for x(t)...Have you at least been able to do that? What have you tried?
     
  4. Dec 1, 2008 #3
    [tex]x'' + kx' + (n^2)x = 0[/tex], let [tex]x = ce^{Yx}[/tex]

    So: [tex]x' = Yce^{Yx}[/tex]
    And: [tex]x'' = (Y^2)ce^{Yx}[/tex]

    So eventually: [tex]Y^2 + kY + n^2 = 0[/tex]

    I believe the next step would be to factorise this, and then depending on whether there are two roots or a repeated root it would be of the form [tex]y = Ae^{(Y1)x} + Be^{(Y2)x}[/tex] or [tex]y = (A + Bx)e^{Yx}[/tex] respectively. However, I am unsure what to do with the quadratic I obtained which contains constants k and n. Any ideas? Again, thanks for taking the time to help me.

    Oh, and sorry it takes me so long to reply; I'm still getting the hang of this LaTeX thing...
     
  5. Dec 1, 2008 #4

    gabbagabbahey

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    Just use the quadratic equation; you should get two distinct roots.
     
  6. Dec 1, 2008 #5
    The general solution I obtain:

    [tex]y = Ae^{(1/2)(-b + \sqrt[2]{k^2-4n^2})x} + Be^{(1/2)(-b - \sqrt[2]{k^2-4n^2})x}[/tex]

    I was able to get this far when doing it by myself, it is here that I hit the wall. I fail to see what anything I have done so far has to do with proving the Integral at the beginning involving x^2.
     
  7. Dec 1, 2008 #6

    gabbagabbahey

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    First, you should be obtaining x(t) not y(x); and second -b=-k

    [tex]\Rightarrow x(t)=Ae^{\frac{-k + \sqrt{k^2-4n^2}}{2}t} + Be^{\frac{-k - \sqrt{k^2-4n^2}}{2}t}[/tex]

    Follow?

    From here, you apply your initial conditions to determine A and B. Once you've done that, you can compute the integral [tex]\int_0^{\infty} x^2 dt[/tex] and show (after a fair bit of work) that you get what you're supposed to get.
     
  8. Dec 1, 2008 #7
    Maybe I just don't get differential equations - I don't know. All I manage to get for my constants A and B is the one in terms of the other and a or u.

    x(0) yields: [tex]A+B=a[/tex]
    x'(0) yields: [tex](1/2)[k(-A-B) + \sqrt[2]{k^2-4n^2}(A-B)] = u[/tex]

    As a result of this I can only get the constants in terms of two other constants; I cannot solve like I usually would if I had actual integers instead of letters.
     
  9. Dec 1, 2008 #8

    gabbagabbahey

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    You'll need to solve for A and B in terms of 'u' and 'a' only...you have two equations, and two variables which you wish to solve for (A and B). So...?
     
  10. Dec 1, 2008 #9
    I'm at a total loss for what to do next. I think and think but nothing occurs to me. I don't just have two variables, no matter what I do I will always get A and B in terms of two other things; I can never solve for them since there are two letters a and u and not numbers.
     
  11. Dec 1, 2008 #10

    gabbagabbahey

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    You should find that A=f(a,u) and B=g(a,u) where f and g are some functions. Your answers will be in terms of a and u.

    For example, if I had the equations [tex]2A-B=u[/tex] and [tex]A+3B=7a-10u[/tex], I would solve the first one for B to get [itex]B=2A-u[/itex] and then substitute it into the second to get:

    [tex]A+3(2A-u)=7a-10u \quad \implies A=a-u[/tex] and [tex]B=2A-u=2a-3u[/tex]

    Do you follow?
     
  12. Dec 1, 2008 #11
    Actually yes... Surprisingly. Just looking at it now I can understand what you meant when you said a fair amount of work! Thank you.
     
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