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Second-Order PDE Help

  1. Aug 14, 2011 #1
    1. Integrate (by calculus): u''(x) = -4u(x), 0 < x < pi






    2. The attempt at a solution
    I'm not really sure where to start on this one is my problem. I can see that it won't be a e^2x problem because of the negative, which leads me to believe that it will deal with the positive/negative relationship involved when you differentiate cos.
    The answer is u(x) = c1sin(2x) + c2cos(2x) which makes sense since u'(x) = 2cos(2x) - 2sin(2x) and u''(x) = -4sin(2x) - 4cos(2x) = -4(sin(2x) + cos(2x)) = -4u(x)
    But, how do I go about showing my work? How am I supposed to know it's c1sin(2x) + c2cos(2x) in the first place?
     
  2. jcsd
  3. Aug 14, 2011 #2

    rock.freak667

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    Rewrite your equation as

    u''+ 4u = 0


    and since this is a 2nd order ODE with constant coefficients, you can use a trial solution of u = erx to get auxiliary equation which will be for the form

    ar2+br+c = 0 where you can solve for r.

    For more info: here
     
  4. Aug 14, 2011 #3

    dynamicsolo

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    Have you discussed in your course yet how to interpret complex roots from the characteristic equation to a second-order DE? You will find the "Euler Identity" of use:

    e^(ikx) = cos(kx) + i sin(kx) .
     
  5. Aug 14, 2011 #4

    vela

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    You could say you made an educated guess as to what the solution is. Guessing the answer is a legitimate way to solve a differential equation. It's a second-order equation so you need two independent solutions. As you noted, the negative sign suggests the solutions will be sines and cosines. Then it's just a matter of figuring out what you need to do to get the factor of 4 out front.
     
  6. Aug 15, 2011 #5
    Thank you guys for all your help!
    I haven't actually started this class yet, I'm just preparing for it since I know it's going to be quite a *****. It's a graduate level PDE class, and I didn't learn much in my ODE class because of a terrible teacher (even though I got an A).
    I was wondering if guessing the answer is sufficient enough, assuming I prove why it works.
    When you say "you need two independent solutions" do you mean having c1 and c2?
    Would just c1sin(2x) work? After all, if u = sin(2x), then u' = 2cos(2x) and u'' = -4sin(2x), which works.
    Or does that not work since there need to be 2 constant c's because it's second order?

    I haven't learned that (or anything yet), but how would I go about using that as an answer?
    Would I just use e^(i2x)?

    Interesting, I'll look into that more thank you! That makes sense, I need to study it more though.
     
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