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Second Order Differential Equation Help

  1. Oct 3, 2008 #1
    A 2-[tex]\mu F[/tex] capacitor is charged to 20 V and then connected across a 6-[tex]\mu H[/tex] inductor forming an LC circuit.
    (a) Find the initial charge on the capacitor

    (b) At the time of connection, the initial current is zero. Assuming no resistance, find the amplitude, frequency and phase of the current. Plot the graph of the current versus time.

    2. Relevant equations

    I was able to get part (a) no problem. It's simply:
    [tex]Q = E_{c}C[/tex] where [tex]E_{c}[/tex] = Voltage drop

    Relevant equation for part (b):

    LdI/dt + Q/C + RI = E
    I = dQ/dt
    LI" +RI' + I/C = dE/dt

    L is the inductance.

    3. The attempt at a solution

    The second order differential equation's solution should be of the form

    y = acost(wt) + bsin(wt).

    you can divide through by L to get:

    I" + RI'/L + I/(LC) = dE/dt(1/L)....since R = 0

    I" + I/ (LC) = dE/dt(1/L)

    I set the right side equal to zero and was able to get a for the solution equation, but couldn't get b. I am not sure how to proceed now. I realize I can't just use 20V for E because that's just for the capacitor, not the whole circuit. Thanks for your help.
  2. jcsd
  3. Oct 3, 2008 #2


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    Shouldn't your differential equation be [itex] \ddot{I}(t)+\frac{1}{LC} I(t)=0[/itex]? And so [itex]I(t)=Asin(\omega t)+Bcos(\omega t)[/itex] where [itex]\omega=\sqrt{\frac{1}{LC}}[/itex]? Is this what you are getting for [itex]I(t)[/itex]? If so then what does I(0)=0 tell you? What does that make Q(t)? What then does [itex]Q(0)=40\mu C[/itex] tell you?
    Last edited: Oct 3, 2008
  4. Oct 3, 2008 #3
    Well, I thought so too, but I am pretty sure I got a value of a = 0 and b = 0. That means essentially that I(t) = 0, which isn't right. Also, I did get a value for b that when used to find the Amplitude A....differs from the answer in the back of the book.
  5. Oct 3, 2008 #4


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    I(0)=0 means that B=0 right? What is Q(t) then?
  6. Oct 3, 2008 #5
    Well if we let dE/dt = 0 then
    I = acost(wt) + bsin(wt).
    I(0) = acost(0) + bsin(0) = 0
    Therfore a = 0

    w in this is the frequency which would be equal to.... [tex]\sqrt{1/(LC)}[/tex]
  7. Oct 3, 2008 #6


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    Okay, if you write that way then you have [itex]I(t)=bsin(\omega t)[/itex]. What is [itex]Q(t)[/itex] then? Remember, [itex]I(t)=\frac{dQ(t)}{dt}[/itex]
  8. Oct 3, 2008 #7
    Then [tex] Q(t) = \int I dt = \frac{-bcos(wt)}{w}[/tex]

    And then [tex] Q(0) = \frac{-bcos(0)}{w} = 4.0 x 10^{-5} [/tex]
    [tex] b = (1/\sqrt{LC})*4 x 10 ^{-5} [/tex]

    But, if a = 0 then the amplitude [tex] A^{2} = \sqrt{a^{2} + b^{2}} can simplify to A^{2} = b^{2} or A = b...and A is supposed to be 4.0 x 10^{-5}...and b = 11.55 roughly...should I not set the constant in the integration to 0?
  9. Oct 3, 2008 #8


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    You can't set the integration constant to zero. The actual relationship should be:

    [tex]\int_{Q(0)}^{Q(t)} dq= \int_{0}^{t} I(t')dt'[/tex]

    To find [itex]b[/itex] you will need at least one more piece of information. Remember, you also know that the voltage across the capacitor is 20V at t=0.
  10. Oct 3, 2008 #9
    Thanks for your help...at the moment I am pretty tired, so I will come back tomorrow and look at it. If I have questions tomorrow I will post them. Again thanks so far.

    Edit: Well, turns out I can't leave it alone for the night. I did the integral and provided everything is correct with:

    I(t) = bsin(wt) with [tex]w =\sqrt{1 / LC}[/tex]

    I got:

    [tex]Q(t) - Q(0) = -b\sqrt{LC}cos(\sqrt{1 / LC}t) + b\sqrt{LC}[/tex]
    [tex]Q(t) - 4 x 10^{-5} = b[\sqrt{ LC} - \sqrt{LC}cos(\sqrt{1 / LC}t)][/tex]

    So provided I did the above correctly I need to find Q(t). It's possible I am too tired to think clearly and am going in circles...which seems to be what I have done...
    Last edited: Oct 3, 2008
  11. Oct 3, 2008 #10


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    I don't think your approach there would help. As said the initial voltage of the capacitor is 20V, so what is the initial voltage drop across the inductor at that time? Relate the expression you have for the potential drop across the inductor for that initial condition.
  12. Oct 4, 2008 #11
    Well if we just have an LC circuit then it should be -20V. The system will only use enough energy to move the current around it...and no more otherwise the charge could build up indefinitely.
  13. Oct 4, 2008 #12
    Could someone do this problem with the numbers in the first post and tell me what you get for the amplitude? I believe this book has been wrong before.
  14. Oct 4, 2008 #13


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    I get b=11.547 A

    Also, I guess I steered you in the wrong direction; calculating Q(t) is a pointless step here.Just use the fact that [itex]V_L(t)=-L\frac{dI}{dt}[/itex] and that you have [itex]V_L(0)[/itex].
  15. Oct 4, 2008 #14
    That's an answer I have gotten...so perhaps they accidentally put down Q(0) in the back for the amplitude. No worries, you were trying to help me.

    Edit, I just noticed that if you use...


    And plug in all the values that you get the 11.547 for A
    Last edited: Oct 4, 2008
  16. Oct 6, 2008 #15
    I wouldn't have messed with Q at all.
    I would solve v''+(1/LC)v=0 for v(t) with v(0)=20 v'(0)=0
    and then i(t) = -cv'

    C --- + i --> $
    --- V $ L
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