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Fun with DEs
Solve the following, anyone?
<pre>
[1 0 1] [e^t ]
X' = [0 0 2] X + [0 ]
[0 0 1] [e^-t]
</pre>
X and X' are vectors, obviously. I got
<pre>
[t] [1] [0]
e^t(c1 [2] + c2 [0]) + c3 [1]
[1] [0] [0]
</pre>
for the undriven case, which I'm fairly certain of, and
<pre>
[te^t(1+K3) + e^t(K1) + .25e^-t]
[2e^t(K3) + e^-t + (K2 - 2K3) ]
[e^t(K3) - .5e^-t ]
</pre>
for the driven case.
Take your best shot; I guess I'm just looking to make sure I'm right.
Last edited by -N-; 05-09-2005 at 09:40 PM.
Reason: misprint
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