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Homework Help: Differential Equations

  1. Dec 8, 2009 #1
    The general solution of the second order DE

    [tex]y'' + a^2y = 0[/tex]

    is

    [tex]y = A cos ax + B sin ax[/tex]

    WTF?

    o_O

    Someone explain why?
     
  2. jcsd
  3. Dec 8, 2009 #2

    Mark44

    Staff: Mentor

    The characteristic equation is r2 + a2 = 0, which has solutions r = +/- ai. This means that the general solution is all linear combinations of {eiax, e-iax} = {cos ax + i sin ax, cos ax - i sin ax). By taking suitable linear combinations of these two complex functions, you can get two real solutions, sin(ax) and cos(ax).
     
  4. Dec 8, 2009 #3
    Why does the linear combinations include complex conjugate?
     
  5. Dec 8, 2009 #4

    Mark44

    Staff: Mentor

    A 1st order differential equation has 1 basic solution, a 2nd order differential equation has 2 basic solutions, a 3rd order differential equation has 3 basic solutions, and so on.

    The basic solutions are all of the form erx, where r is a root of the characteristic equation. If the characteristic equation has complex roots, these roots always come in pairs - the complex conjugates. For the sake of convenience, instead of writing e(a + bi)x and e(a - bi)x, we do a little algebra and write these functions as eaxcos(bx) and eaxsin(bx) so that we don't have to mess with imaginary numbers at all.
     
  6. Dec 8, 2009 #5
    Yes I understand that, I am proficient with complex numbers.
     
  7. Dec 8, 2009 #6

    Mark44

    Staff: Mentor

    So did I answer your question?
     
  8. Dec 9, 2009 #7

    crd

    User Avatar

    because the solutions form a 2-dim vector space over the field of complex numbers, so by properties of vector spaces all scalar multiples are elements of said solution space, which in this case would include linear combinations of complex conjugates
     
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