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Proving the Form of a Homogenous LC Circuit?

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a circuit with input source x(t), which contains also an inductor and a capacitor in series which I have found to be related to the output voltage y(t) (across the capacitor) like so: LC*d2y/dt2 + y(t) = x(t). I have also found its roots through the quadratic equation to be complex, and equal to +/- the imaginary unit j.

    The question asks me to show its homogenous solution has the form K1ej*a*t + K2ej*b*t, where a and b are the roots I found above.

    2. Relevant equations

    Unsure. See above for my progress so far.

    3. The attempt at a solution

    My solution has led me here, having gotten the above equation and its roots. I just do not know how to show that I can express the solution to the differential equation in the requested form. Any help would be greatly appreciated!
     
  2. jcsd
  3. May 21, 2013 #2

    tiny-tim

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    Hi Rome_Leader! :smile:

    If you have a homogeneous quadratic equation in y(t),

    write D = d/dt, and convert the equation into an ordinary quadratic equation in D …

    (D - a)(D - b)(y) = 0​

    then put (D - b)(y) = z, so (D - a)(z) = 0 :wink:
     
  4. May 21, 2013 #3
    I follow, but that does not appear to express it in the form K1*e^(j*a*t) + K2*e^(j*b*t). I don't think it makes it any simpler either, at least not to my eyes.
     
  5. May 21, 2013 #4

    tiny-tim

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    yes it does, try it :smile:
     
  6. May 21, 2013 #5
    Expanding, I get d^2y/dt^2 - (dy/dt)*b - (dy/dt)*a + ab.

    Your z substitution is a little confusing, but then I get (dy/dt-a)*(dy/dt-b) = 0 again?
     
  7. May 21, 2013 #6

    tiny-tim

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    solve (D - a)z(t) = 0 first …

    z = … ? :wink:

    (and then solve the non-homogeneous equation (D - b)y(t) = z(t))
     
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