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Question about linear second order ODE solution coefficients

  1. Feb 16, 2012 #1
    Hey all,

    there is something that has always bugged me in linear second order ODEs. We say that the general solution is:

    [tex]y=C_1e^{r_1x}+C_2e^{r_2x}[/tex]

    where r_1 and r_2 are the solutions of the characteristic polynomial.

    The cases where r1, r2 are real are pretty straightforward. If they are complex, however:

    [tex]r_1=a+ib , r_2=a-ib[/tex]

    after algebraic manipulation we come up with the following:

    [tex]y=e^{ax}[(C_1+C_2)cos(bx)+i(C_1-C_2)sin(bx)]=e^{ax}[c_1 cos(bx)+c_2 sin(bx)][/tex]

    We then say, that this is the general solution of the ODE, and if c_1, c_2 are real, the solutions are also real.

    This would imply that the C_1 and C_2 coefficients are considered to be complex. However, I have not seen such an assumption in any book or description so far. In fact, they are sometimes not defined at all, simply referred to as "arbitrary constants". Does this assumption cover the possibility that they may be complex numbers? Am I missing something obvious?
     
    Last edited: Feb 16, 2012
  2. jcsd
  3. Feb 16, 2012 #2
    Generally you look for real valued solutions. There is another way to come up with the general solution. It is straightforward to prove that if y=u+iv is a complex valued solution of this type of equation then both u and v are real valued solutions and they are linearly independent (try it). The two complex solutions immediately yield the same two real solutions and the general solution is a linear combination of u and v.
     
  4. Feb 16, 2012 #3
    Certainly, a similar thing happens with harmonic conjugates. However, unless I am mistaken, even though the solution is real, it is (or rather, it can be) the superposition of complex functions. Therefore, wouldn't it make sense to explicitly state that the C_1 and C_2 coefficients can be complex?
     
  5. Feb 16, 2012 #4
    I agree. From a strictly mathematical point of view, a complex solution is a solution. But if we are solving a classical physical boundary value problem then the solutions are real. I think that author's are sometimes sloppy in not properly motivating or explaining exactly what solutions we are interested in.

    Sorry. To answer the original question, they can be complex as you assume.
     
  6. Feb 16, 2012 #5
    This clears things up! Thank you!
     
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