Second Order Linear Differential Equations

**ben.mahoney@tesco.net** · February 15th 2009, 09:11 AM

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]

$ 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
$ q = \frac{Eo}{L(w^2 - n^2)}[cos(nt) - cos(wt)] $
where
$ n^2 = \frac{1}{CL} $
I can get most of this. I dont get the cos(nt) bit and my denominator is
$ 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 ?

**HallsofIvy** · February 15th 2009, 10:15 AM

Originally Posted by ben.mahoney@tesco.net

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]

$ 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
$ q = \frac{Eo}{L(w^2 - n^2)}[cos(nt) - cos(wt)] $
where
$ n^2 = \frac{1}{CL} $
I can get most of this. I dont get the cos(nt) bit and my denominator is
$ n^2 - w^2 $

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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