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Solving an ODE with eulers formula.

  1. Apr 1, 2012 #1
    If I have [itex] y''+y'+2y=sin(x)+cos(x) [/itex]
    can I just say [itex] y=Ae^{ix} [/itex]
    and then find y' and y'' and then plug them in and solve for A.
    so I get that [itex] A= \frac{1}{1+i} [/itex]
    then i multiply and divided by the complex conjugate.
    then I back substitute in Eulers formula.
    now since I have my original equation has both a real and an imaginary part.
    when I multiply out A times Eulers formula, I will take both real and imaginary parts.
    and I get that y=sin(x) and tested this and it works.
    But is what i did with Eulers formula ok.
  2. jcsd
  3. Apr 1, 2012 #2


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    hi cragar! :smile:
    not really :redface:

    for a particular solution for an RHS of sinx + cosx, you should use Aeix + Be-ix

    (or Ccosx + Dsinx)

    in this case, you're lucky that B = 0 :biggrin:

    (but don't do it again! :wink:)
  4. Apr 1, 2012 #3
    if i substitute what you said. Do I take the real or imaginary part in the end. Or do I just take it all.
  5. Apr 1, 2012 #4


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    you take it all :smile:

    but if you're expecting a real solution anyway, you might as well use Ccosx + Dsinx in the first place! :wink:

    (technically, C and D are both complex, but they'll come out real)
  6. Apr 1, 2012 #5
    thanks for your answers. I like to push the limits of Eulers formula. And see what it can do.
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