http://www.artofproblemsolving.com/F...d359c1cf90.gif

Printable View

- Mar 25th 2007, 09:30 PMtukeywilliamsComplex cofficients in differencial equation
- Mar 26th 2007, 05:04 AMThePerfectHacker
Post in correct subforum. This has absolutely nothing to do with algebra!

- Mar 27th 2007, 06:11 AMtukeywilliams
Why did you set c1 = c2 = 1/2 and c1 = 1/2i and c2 = -1/2i? How did you get that to equal e^(\alpha t) \cos t and e^(\alpha t) \sin t?

- Mar 27th 2007, 06:31 AMtopsquark
NOTE: I have corrected the 9:30AM version of this!

You need to split y into its real and complex parts:

I'm going to call alpha = a and beta = b and I^2 = -1. (Don't confuse these with the coefficients of the differential equation!)

y = k1*exp{(a + Ib)t} + k2*exp{(a - Ib)t}

y = k1*exp{at}*exp{Ibt} + k2*exp{at}*exp{-Ibt}

y = exp{at}*[k1*exp{Ibt} + k2*exp{-Ibt}]

y = exp{at}*[k1*cos(bt) + I*k1*sin(bt) + k2*cos(bt) - I*k2*sin(bt)]

Define k1 = c + Id and k2 = e + If where c, d, e, f are real.

Then

y = exp{at}*[c*cos(bt) + I*d*cos(bt) + I*c*sin(bt) - d*sin(bt) + e*cos(bt) + I*f*cos(bt) - I*e*sin(bt) + f*sin(bt)]

y = exp{at}*[c*cos(bt) - d*sin(bt) + e*cos(bt) + f*sin(bt)] + I*exp{at}*[d*cos(bt) + c*sin(bt) + f*cos(bt) - e*sin(bt)]

The complex part of y is

exp{at}*[d*cos(bt) + c*sin(bt) + f*cos(bt) - e*sin(bt)]

= exp{at}*[(d + f)*cos(bt) + (c - e)*sin(bt)]

This must be 0. To be true for all b and all t we must have that

d + f = 0

c - e = 0

Thus f = -d

Thus e = c.

So k2 = e + If = c - Id, which is the complex conjugate of k1 = c + Id.

-Dan - Mar 27th 2007, 09:30 AMtukeywilliams
- Mar 27th 2007, 10:07 AMtopsquark