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Thread: 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]

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

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

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

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

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

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

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

    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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