       Re: Re: Problem with a 1st order IV ODE

• To: mathgroup at smc.vnet.net
• Subject: [mg102646] Re: [mg102581] Re: [mg102564] Problem with a 1st order IV ODE
• From: Leonid Shifrin <lshifr at gmail.com>
• Date: Mon, 17 Aug 2009 04:31:35 -0400 (EDT)
• References: <200908140959.FAA01407@smc.vnet.net>

```@ Daniel

Daniel, thanks, you are right

@ Virgil

Virgil, sorry, I started with the wrong problem.

Regards,
Leonid

On Sun, Aug 16, 2009 at 7:50 AM, <danl at wolfram.com> wrote:

> > Hi Virgil,
> >
> > while I don't have an authorative explanation of the behavior of NDSolve
> > in
> > I do have an alternative approach to suggest.
> >
> > First of all, I have to disappoint you: the solutions you (or, more
> > precisley, DSolve) obtained
> > look nice, but don't seem to work for me:
> >
> > In =
> > Clear[R, k];
> > sol = DSolve[{h'[t] == 1/(h[t] (2 R - h[t])) - k,
> >      h == 0} /. {R -> 10, k -> 0.01}, h[t], t] // FullSimplify
> >
> > Out =
> > {{h[t]->-0.005 t-0.005 Sqrt[t (4000.+t)]},{h[t]->-0.005 t+0.005 Sqrt[t
> > (4000.+t)]}}
> >
> > In =
> >
> > fn1 = FullSimplify[D[#, t] - 1/(# (2 R - #) - k) /. {R -> 10., k ->
> 0.01}]
> > &[(h[ t] /. sol)[]]
> >
> > Out = -0.005+(-10.-0.005 t)/Sqrt[t (4000.+t)]+1/(0.01+0.1 Sqrt[t
> > (4000.+t)]+t (0.2+0.00005 t+0.00005 Sqrt[t (4000.+t)]))
> > [...]
>
> Your grouping is amiss. Try with:
>
> fn1 = FullSimplify[
>    D[#, t] - 1/(# (2 R - #)) + k /. {R -> 10, k -> 1/100}] &[(h[
>      t] /. sol)[]]
>
> Daniel
>
>

```

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