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Question about complex solutions to DiffEqns

  1. Jan 22, 2005 #1
    I am seeking information on the following problem:

    The problem statement tells that a function [tex]z(t) = x(t) + j y(t)[/tex] (where x(t) and y(t) are real functions of t) satisfies the following Inhomogenous differential equation (That i can see corresponds to a damped, driven harmonic oscillator)

    [tex]z'' + 2bz' + \omega^2 z = Fe^{j\omega_ot}[/tex]

    All derivatives are with respect to time, and j = root -1.

    And the question states to find the DIFF EQN that is satisfied by [tex]x(t)[/tex] that is the real part of [tex]z(t)[/tex].

    At first, I was thinking, since x(t) corresponds to the real part of the solution to the DE above, that the differential equation must not have a characteristic equation with complex solutions ( that is [b^2 - 4w^2] CANNOT be less than zero, and therefore b^2 = 4w^2 AT LEAST. But I am given no value of any of the parameters in the DE. I also thought about excluding the right side forcing function also because that has complex numbers in it, so i assumed that we cannot include that in the answer--- however I am not sure. I am not being asked to solve the DE above, but rather find one for x(t), but I tried solving the DE anyway to see what the COMPLEX part and the REAL part would be and perhaps reverse engineer the problem back to get a DE that would be the answer... the way to do that would be to either use a particular solution, laplace transform or variation of parameters... but I dont want to waste my time doing that since we are "technically" not supposed to know how to solve Inhomog DEs just yet.

    Any help would be appreciated, thanks a lot.
     
  2. jcsd
  3. Jan 22, 2005 #2
    Hmmm--- one more thought--- could Euler's Identity [tex]e^{j \omega t} = cos (\omega t) + j sin(\omega t)[/tex] help for this problem? I see it could maybe be used to change that forcing function on the right hand side....
     
  4. Jan 23, 2005 #3

    Hurkyl

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    That is incorrect -- complex-linear combinations of your two complex solutions can be purely real.

    For example, try solving the ODE [itex]z'' - z = 0[/itex].

    Then the method you mentioned will give you these two independent solutions: [itex]e^{jt}[/itex] and [itex]e^{-jt}[/itex]. However, as you well know, the solution space is also spanned by these two independent solutions: [itex]\cos t[/itex] and [itex]\sin t[/itex].

    Exercise: you've already shown how to write the exponential function as a linear combination of the two trig functions. Now, show how to write the trig functions as linear combinations of the exponential functions.


    Anyways, Euler's identity will help -- your goal is to simlpy take the real part of the equation, and that identity tells you the real part of the RHS.
     
  5. Jan 23, 2005 #4
    Wait a minute--- What about just having the right hand side be [tex]Fcos(\omega_o t)[/tex]? The solution to the DE then would be only give a real solution --- wouldnt that correspond to x(t)?
     
  6. Jan 23, 2005 #5

    HallsofIvy

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    That's pretty much what Hurkyl just said!
     
  7. Jan 23, 2005 #6

    dextercioby

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    How about pluggin' in the eq.the complex solution and then identifying the real & the imaginary parts of each side of the equation...???

    Daniel.
     
  8. Jan 23, 2005 #7
    thanks guys i think i have it, and btw i cannot plug in the complex solution because he didnt give me it! He just said it was in x + ji form ---- but your method would have been great had i known them!!
     
  9. Jan 23, 2005 #8

    dextercioby

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    Well
    [tex] z(t)=:x(t)+jy(t) [/tex]

    is the "complex solution" i was referring to...

    Daniel...


    P.S.YW.
     
  10. Jan 23, 2005 #9

    Hurkyl

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    x(t) + j y(t) is the complex solution. :tongue2:
     
  11. Jan 23, 2005 #10
    thanks ill try that.
     
    Last edited: Jan 23, 2005
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