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First Order Inhomogeneous ODE

  1. Jul 10, 2014 #1
    For a regular LR circuit (L and R in series) and with a AC voltage:

    R8yRjNY.jpg

    I tried to derive the solution myself.

    https://www.dropbox.com/s/jmsu9j0vt91ze8x/LRcircuit.jpg

    So first I solved with undetermined coefficients, plugged them in, and then solved with Cramer's Rule.

    Then I added the term (the solution for the homogeneous case) with the solution I got with undertermined coefficients.

    Then I used initial value: the current i(t) at t=0 is 0.

    Then I got the general solution: an exponential term, a cosine term and a sine term. But my solution is quite different from the book.


    Can anyone help me?
     
  2. jcsd
  3. Jul 10, 2014 #2
    Your solutions looks correct.

    To get it into the above form you have to use the cos difference identity...

    [itex]\cos {\left(\left(\omega t + \phi\right) - \theta\right)}=\cos {\left(\omega t + \phi\right)}\cos {\theta}+\sin {\left(\omega t + \phi\right)} \sin {\theta} [/itex]

    The trick is to figure out what [itex]\theta[/itex] is in terms of [itex]R,\omega, L[/itex]. To do this think of a right triangle with sides [itex]R[/itex] and [itex] \omega L [/itex].
     
  4. Jul 11, 2014 #3
    I made a mistake in my solution bye the way.

    In evaluating the coefficient of C of the exponential term, when I took t = 0 I let the sine term vanishes but that doesn't vanishes sin (wt + q) it become sin (q). I was thinking about sin (wt) becomes zero, yes but not with a phase angle between the brackets. Stupid mistake.

    This the correct one:

    https://www.dropbox.com/s/ne1wo9wknk3s1mw/20140711_121219~2.jpg

    Then I used a numerical example to comparize mine solution with the one for the book.

    No idea how I will get it in that form from the book. :) I'm gonna try later this day. Thanks for your identity.


    Edit: I got it!! Thanks again for your identity. I will post solution to be complete soon. :)
     
    Last edited: Jul 11, 2014
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