On Wed, Nov 21, 2012 at 5:54 PM, MatthiasK <[email protected]> wrote:
> Hi,
> I am a bit confused about the state of solving ODE's in sympy. It seems
> that I cannot directly solve initial value problems, right?
>
Yes, but it should be simple to implement. Just no one has done it yet.
>
> Is there a good way to solve the resulting equation system in simple cases?
> e.g. my code right now:
> t = sym.Symbol('t')
> i = sym.Function('i')
> prob = Lsym.Derivative( i(t), t, 2 ) + sym.Derivative(i(t), t) + i(t)
> sol = sym.dsolve(prob, i(t))
>
> now I'd like to do something like:
> equations = [sol.subs(t,0) == var(1), sym.diff(sol,t).subs(t,0) == var(2)]
> solve(equations)
>
> sym.diff(sol,t) does not return a fully substituted result however.
> Instead I receive:
> Subs(Derivative(i(_t) == (_t*C2 + C1)*exp(-_t), _t), (_t,), (0,))
>
> Can I still solve with this somehow or do I need to rethink my approach?
>
Eq() does not automatically distribute most operations across it,
unfortunately (see https://code.google.com/p/sympy/issues/detail?id=1931and
https://code.google.com/p/sympy/issues/detail?id=1932). Your best bet is
to just work with the solution as a regular expression, i.e., work with
sol.lhs - sol.rhs (solve() assumes that just plain expressions are equal to
0). Then what you are trying will work. Or, since this particular
solution is already solved for i(t), you can just work with the right-hand
side.
Aaron Meurer
> Best Regards,
> Matthias
>
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