Closed idontgetoutmuch closed 8 years ago
The initial value is X[t[0]], not X[0].
We get the same solution as GNU-Octave:
Xt0 = 1;
t = [1 2 3]';
r = lsode( @(X,t) X, Xt0, t);
[t r]
ans =
1.0000 1.0000
2.0000 2.7183
3.0000 7.3891
My bad - thanks for the explanation
No problem, thanks for testing!
gives
I.e. we get the solution at time 0 and the solution at times 1 and 2 but not at time 3.