College Math Teaching

October 9, 2013

Fun with complex numbers and particular solutions

I was fooling around with y''+by'+cy = e^{at}cos(bt) and thought about how to use complex numbers in the case when e^{at}cos(bt) is not a solution to the related homogenous equation. It then hit me: it is really quite simple.

First notes the following: e^{rt}cos(st) = \frac{1}{2}(e^{(r + si)t} + e^{(r - si)t}) and e^{rt}sin(st) = \frac{1}{2i}(e^{(r + si)t} - e^{(r - si)t}).

Then it is a routine exercise to see the following: given that z = r+si, \bar{z} = r-si are NOT solutions to p(m)= m^2 + bm + c = 0 p(m) is the characteristic equation of the differential equation. Then: attempt y_p = Ae^{zt} + Be^{\bar{z}t} Put into the differential equation to see y''_p + by'_p + cy_p = A(z^2+bz+c)e^{zt} + B(\bar{z}^2 + b\bar{z} + c)e^{\bar{z}t} .

Then: if the forcing function is e^{rt}cos(st) , a particular solution is y_p = Ae^{zt} + Be^{\bar{z}t} where A = \frac{1}{2(p(z))}, B = \frac{1}{2(p(\bar{z}))}. If the forcing function is e^{rt}cos(st) , a particular solution is y_p = Ae^{zt} - Be^{\bar{z}t} where A = \frac{1}{2i(p(z))}, B = \frac{1}{2i(p(\bar{z}))}.

That isn’t profound, but it does lead to the charming exercise: if z, \bar{z} are NOT roots to the quadratic with real coefficients p(x), then \frac{1}{p(z)} + \frac{1}{p(\bar{z})} is real as is \frac{i}{p(z)} - \frac{i}{p(\bar{z})} .

Let’s check this out: \frac{1}{p(z)} + \frac{1}{p(\bar{z})} = \frac{p(\bar{z})+p(z)}{p(z)p(\bar{z})}. Now look at the numerator and the denominator separately. The denominator: p(z)p(\bar{z})= (z^2 + bz +c)(\bar{z}^2 + b\bar{z} + c) = (z^2 \bar(z)^2) + b(b (z \bar{z}) + (z(z \bar{z}) +\bar{z}(z \bar{z})) + c (\bar{z} + z) + (z^2 + \bar{z}^2)) + (c^2) Now note that every term inside a parenthesis is real.

The numerator: z^2 + \bar{z}^2 + b (z + \bar{z}) + 2c is clearly real.

What about \frac{i}{p(z)} - \frac{i}{p(\bar{z})} ? We need to only check the numerator: i (z^2 - \bar{z}^2 + b(z - \bar{z}) + c-c) is indeed real.

Yeah, this is elementary but this might appear as an exercise for my next complex variables class.

Advertisements

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: