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