> No, AFAIK, nothing other than explicit substitution with .subs().
Hello,
there are a few weird results. I'd like to solve this homogenous edo :
$tx'=x+\sqrt{x^2+y^2}$.
using x=tu
sage: t=var('t')
sage: x(t) = function('x',t)
sage: id(t)=t
sage: u=function('u',t)
sage: d=diff(u*id,t)
apparently, you can't substitute inside diff
sage: DE=(t*d==x+sqrt(t**2+x**2)).subs_expr(x==u*id)
sage: desolve(DE,[u,t])
arcsinh(u(t)) == c + integrate(abs(t)/t^2, t)
sage: assume(t>0)
sage: desolve(DE,[u,t])
arcsinh(u(t)) == c + integrate(1/t, t)
sage: desolve(DE,[u,t],[1,0])
oups, first problem :
Division by 0
#0: ic1(soln=asinh(u) = 'integrate(abs(t)/t^2,t)+%c,xc=t = 0,yc=u = 1)
(ode2.mac line 297)
-- an error. To debug this try: debugmode(true);
sage: sol=desolve(DE,[u,t]).simplify_exp()
arcsinh(u(t)) == c + log(t)
OK
sage: desolve(DE,[u,t],[0,1]).simplify_exp()
but the problem reamins the same. Welle, let's make the substitution
by hand :
sage: solp=sol.subs(c=0)
sage: solp
arcsinh(u(t)) == log(t)
sage: Sol=solve(solp,u)[0]
sage: Sol
u(t) == sinh(log(t))
sage: x(t)=t*Sol.rhs()
sage: x(t)
t*sinh(log(t))
here comes our hyp2exp problem...
sage: sh(a)=(exp(a)-exp(-a))/2
sage: x(t).substitute_function(sinh,sh).simplify_full()
t*sinh(log(t))
argh
sage: x
t |--> t*sinh(log(t))
x is what we expect. Is it t ?
sage: x(a).substitute_function(sinh,sh).simplify_full()
a*sinh(log(a))
no...and now, weirder and weirder...
sage: a*sinh(log(a)).substitute_function(sinh,sh).simplify_full()
1/2*a*log(a) - 1/2
sage: t*sinh(log(t)).substitute_function(sinh,sh).simplify_full()
1/2*t^2 - 1/2
this 1/2*a*log(a) - 1/2 is a bit unexpected...
well, I'm discovering sage but these results look strange.
cheers,
Guillaume
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