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Math Help - Second Order Linear Differential Equations

  1. #1
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    Exclamation Second Order Linear Differential Equations

    i)An electrical circuit consists of an inductor of inductance L in series with a capacitor of capacitance C. An alternating voltage of Eocos(wt) is applied to the circuit at time t=0 with the initial charge q in the circuit being zero and the initial current i (=dq/dt) being zero. Using Kirchoff’s laws the charge in the circuit satisfies the equation[/FONT]

    <br />
                     L \frac{d^2q}{dt^2}  +  \frac{q}{C} = Eocos(wt)<br />
    Show that the charge in the circuit at any time t is given by the equation
     <br />
q = \frac{Eo}{L(w^2 - n^2)}[cos(nt) - cos(wt)]        <br />
    where
    <br />
n^2 = \frac{1}{CL}<br />
    I can get most of this. I dont get the cos(nt) bit and my denominator is
    <br />
n^2 - w^2<br />
    ii)
    What is the current in the circuit at any time t ?.
    What is the charge in the circuit when w = n ?
    Last edited by ben.mahoney@tesco.net; February 15th 2009 at 09:34 AM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    i)An electrical circuit consists of an inductor of inductance L in series with a capacitor of capacitance C. An alternating voltage of Eocos(wt) is applied to the circuit at time t=0 with the initial charge q in the circuit being zero and the initial current i (=dq/dt) being zero. Using Kirchoff’s laws the charge in the circuit satisfies the equation[/FONT]

    <br />
                     L \frac{d^2q}{dt^2}  +  \frac{q}{C} = Eocos(wt)<br />
    Show that the charge in the circuit at any time t is given by the equation
     <br />
q = \frac{Eo}{L(w^2 - n^2)}[cos(nt) - cos(wt)]        <br />
    where
    <br />
n^2 = \frac{1}{CL}<br />
    I can get most of this. I dont get the cos(nt) bit and my denominator is
    <br />
n^2 - w^2<br />

    Since you say you "get most of this", I presume you recognized that the characteristic equation was Lr^2+ \frac{1}{C}= 0 so that r= \pm i\sqrt{\frac{1}{LC}} and the solution to the associated homogeneous equation is q(t)= Acos(\frac{1}{\sqrt{LC}}t)+ Bsin(\frac{1}{\sqrt{LC}}t).

    You are TOLD in the problem that n^2= \frac{1}{LC} so that is q(t)= A cos(nt)+ B sin(nt). That "n" was introduced just to make the writing easier.

    The right hand side of the equation is E_0cos(\omega t) and, because there are no odd derivatives, we try a solution of the form qA(t)= Acos(\omega t). Then q"(t)= -A\omega^2 cos(\omega t). Putting those into the equation, it becomes
    -AL\omega^2 cos(\omega t)+ \frac{A}{C}cos(\omega t)= E_0 cos(\omega t)
    In order for that to be true for all t, we must have
    -AL\omega^2+ \frac{A}{C}= -A(L\omega^2- \frac{1}{C})= E_0
    A= -\frac{E_0}{L\omega^2- \frac{1}{C}}
    That \frac{1}{C} term can be written as \frac{L}{LC}= Ln^2 because we have define n^2= \frac{1}{LC} so A= -\frac{E_0}{L\omega^2- Ln^2}

    So the "particular solution" is Acos(\omega t)= -\frac{E_0}{L\omega^2- Ln^2} cos(\omega t) and the general solution is
    q(t)= A cos(nt)+ Bsin(nt)-\frac{E_0}{L\omega^2- Ln^2} cos(\omega t)

    Our initial conditions are q(0)= 0, q'(0)= 0.
    q(0)= A-\frac{E_0}{L\omega^2- Ln^2} cos(\omega t)= 0
    so
    A= \frac{E_0}{L\omega^2- Ln^2} cos(\omega t)
    and
    q'(0)= nB= 0
    That is
    \frac{E_0}{L\omega^2- Ln^2} cos(\omega t)(cos(nt)- cos(\omega t)

    ii)
    What is the current in the circuit at any time t ?.
    What is the charge in the circuit when w = n ?
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