Re: is this solvable?
- To: mathgroup at smc.vnet.net
- Subject: [mg56656] Re: is this solvable?
- From: Peter Pein <petsie at arcor.de>
- Date: Mon, 2 May 2005 01:32:38 -0400 (EDT)
- References: <d4sov7$ale$1@smc.vnet.net> <d51n43$cru$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
dennis wrote:
> The only function I know with its derivative proportional to itself is
> the exponential. Therefore, I think the solution is trivial if p1 and
> p2 are constants.
>
> a[t] = a0*Exp[-p1*t] and b[t] = b0*Exp[-p2*t]
>
> I know this isn't using Mathematica to get the solution, but the
> problem seems trivial if what you want is the solution and not a method
> in Mathematica.
>
> Regards,
> Dennis
>
> ames_kin at yahoo.com wrote:
>
>>a'[t] + b'[t]== -p1 a[t] - p2 b[t]
>>
>>where {a[0]== a0, b[0]== b0}
>>
>>is this solvable in Mathematica? If so, how will I go about doing so?
>>
>>let's assume a[0]== a0, and b[0]==b0
>>
>>if symbolic solution isn't possible, then intial conditions of
>>a[0]== 1, and b[0]==0.5 couild be used...(or any other numbers for
>
> that
>
>>matter)
>>
>>thanks in advance.
>
>
Hi Dennis,
it is indeed very simple to find _one_ possible solution. On the other
hand we can add (nearly) every possible equation. For example:
Simplify[
DSolve[{eq, a[0] == Pi} /. {b -> (Cos[a[#1]] - a[#1] & ),
p1 | p2 -> 1}, a[t], t]]
will give us {{a[t] -> ArcCos[-Exp[-t]]}}
--
Peter Pein
Berlin