# First Order Inhomogeneous ODE

1. Jul 10, 2014

### 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?

2. Jul 10, 2014

### the_wolfman

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 $\omega L$.

3. Jul 11, 2014

### Waxterzz

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