Rodrigo da Silva Guerra <tioguerra at gmail.com> writes:
> /* I was calculating a Laplace transform, like below: */
> ex1:exp(-3*t)*'diff(x(t),t);
> lap_ex1:laplace(ex1,t,s);
>
> /* Then I added an initial condition: */
>
> atvalue(x(t),t=0,0);
>
> /* And I managed to re-evaluate the expression: */
>
> lap_ex1:ev(lap_ex1);
Following you so far, I get:
(%i4) lap_ex1:ev(lap_ex1);
!
(%o4) (s + 3) (laplace(x(t), t, s)! )
!s = s + 3
> /* Now I what I wanted to do is to plug an specific x(t) into that
> expression: */
>
> x(t):=1
>
The easiest way to do this sort of thing is syntactically: instead of
saying to Maxima "please resimplify the expression, knowing also that
x(t)=1", just say "replace all occurrences of x(t) with 1 in here".
(%i5) subst(x(t)=1, lap_ex1);
!
(%o5) (laplace(1, t, s)! ) (s + 3)
!s = s + 3
This has the added benefit of not clobbering x(t) for future work: maybe
you want to look at the laplace transform for other functions too (eg
using the Laplace transform's linearity and expanding a periodic
function as a sum of trig functions or whatever)
Of course, you need to tell the laplace code to get run again (what
looks like a function call in %o5 is actually a noun form, as I guess
you've realised, given the title of the thread).
(%i6) %, nouns;
(%o6) 1
Tada! If you want to read more about this, have a look at the
documentation for the ev function. The line "expr, nouns" is just a
shorthand for "ev(expr, nouns)".
Rupert
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