Solving a First Order Inhomogeneous ODE for a Regular LR Circuit with AC Voltage

[Waxterzz]

For a regular LR circuit (L and R in series) and with a AC voltage:

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?

 

[the_wolfman]

Your solutions looks correct.

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

\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}

The trick is to figure out what \theta is in terms of R,\omega, L. To do this think of a right triangle with sides R and \omega L.

 

[Waxterzz]

Your solutions looks correct.

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

\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}

The trick is to figure out what \theta is in terms of R,\omega, L. To do this think of a right triangle with sides R and \omega L.

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 going to 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. :)