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Question on showing general formula of solution

  1. Aug 10, 2016 #1
    1. The problem statement, all variables and given/known data
    show that the general solution of the differential equation d^2/dt^2 + 2 *alpha * dr/dt + omega^2 * r = 0,

    where alpha and w are constant and R is a function of time "t" is R = e^(-alpha * t) * [ C1*sin( sqrt(omega^2 - alpha^2) * t) + C2*cos( sqrt(omega^2 - alpha^2) * t)

    IF alpha^2 - omega^2 < 0.


    I was struggling over how to solve this problem and I would be very grateful if someone have me a badly needed hint.
    2. Relevant equations

    None come to mind as of now



    3. The attempt at a solution

    I said let R = f(t) = e^(d * t).

    I then had the expression when I plugged it into the differential equation

    d^2 * e^(d * t) + 2 *alpha * d * e^(d * t) + omega^2 * e^(d * t) = 0

    I canceled the e^(d * t) term and I got the characteristic equation:
    d^2 + 2*alpha*d + omega^2 = 0

    I then applied the quadratic formula to solve for d:
    d = [ -2 *alpha +/- sqrt( (2*alpha)^2 - 4*omega^2 ) ] / 2.

    I thus get d = -alpha +/- sqrt (alpha^2 - omega^2 )

    Now I have a huge problem. If alpha^2 - omega^2 < 0, then my d is non-real and I can not proceed from here.
    I tried substituting in the functions sin(d*t) and cos(d*t) instead of e^(d*t) into the differential equation and solving, but then I become stuck as I can not eliminate the plugged in functions at all from the general expression.

    Could anyone please be kind enough to help?

    Thanks, and have a great day!
     
  2. jcsd
  3. Aug 10, 2016 #2

    Ray Vickson

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    Homework Helper

    The absolutely 100% standard method of showing something is a solution is to substitute it into the equation to see if it "works". you have an explicit formula for r(t), so nothing prevents you from taking the derivatives, etc.
     
  4. Aug 11, 2016 #3

    pasmith

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    Homework Helper

    Have you seen the identity
    [tex]e^{ix} = \cos x + i \sin x[/tex] before? It's very important to the theory of second-order linear constant coefficient ODEs. Now for real [itex]a[/itex] and [itex]b[/itex] the quantities [itex]e^{a + ib}[/itex] and [itex]e^{a - ib}[/itex] are complex conjugates, so it follows that the linearly independent combinations [tex]
    \frac12(e^{a + ib} + e^{a - ib})[/tex] and [tex]\frac1{2i}(e^{a + ib} - e^{a - ib})[/tex] are real, and you can easily establish that they are equal to [itex]e^{a}\cos b[/itex] and [itex]e^a \sin b[/itex] respectively.
     
  5. Aug 11, 2016 #4
    Hi. No, I have not seen this identity before. Thanks for drawing it to my attention.

    I understand how one might be able to simplify the expression with the given identity, but I do not understand why one would after finding two complex conjugate expressions do the linear combinations
    which you say are equal to the final results
    .

    Could you please explain why you feel the need to take the linear combinations by adding/subtracting the two conjugate pairs and then dividing by 2 or 2i respectively?
    What is the theory behind your actions? How does taking these linear combinations help?

    Thanks so much for your help, and have a great day!
     
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