I have a function which looks like
- 9 t
f (t) = %e
6 (9 c + 43 c + 10 c ) 2 (6 c - 58 c + 50 c )
3 2 1 3 2 1
(- ------------------------ - ------------------------) sin(3 t)
5 5
(----------------------------------------------------------------
6
- 15 t
(9 c + 43 c + 10 c ) cos(3 t) (9 c + 28 c + 10 c ) %e
3 2 1 3 2 1
+ -------------------------------) - -------------------------------
15 15
It's an example solution obtained by desolve(...) for system of linear ODEs
with constant coefficients:
(%o1) desolve(['diff(f[1](t),t,1)-3*f[3](t)-15*f[2](t)+12*f[1](t) = 0,
'diff(f[2](t),t,1)+3*f[3](t)+12*f[2](t)+12*f[1](t) = 0,
'diff(f[3](t),t,1)+9*f[3](t)+26*f[2](t)-34*f[1](t) = 0],
[f[1](t),f[2](t),f[3](t)])$
(%o2) at(%,makelist(f[i](0)=c[i],i,1,3));
How it could be simplified to
f(t) = exp(a_1 t) (sin(b_1 t) P_1 + cos(b_1 t) Q_1) +
exp(a_2 t) (sin(b_2 t) P_2 + cos(b_2 t) Q_2) + ...
Where P_i and Q_i are polynomials over c_i and t (with expanded coefficients)?
I'm not so strong in Maxima simplification and patterns so I don't know a
straightforward way to do it.
--
Alexey Beshenov <al at beshenov.ru>
http://beshenov.ru/
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: application/pgp-signature
Size: 189 bytes
Desc: not available
Url : http://www.math.utexas.edu/pipermail/maxima/attachments/20080325/36e69051/attachment.pgp