Re: DSolve Issues

*To*: mathgroup at smc.vnet.net*Subject*: [mg88378] Re: DSolve Issues*From*: "David Park" <djmpark at comcast.net>*Date*: Sat, 3 May 2008 06:17:45 -0400 (EDT)*References*: <fveguc$5ed$1@smc.vnet.net>

Clear[Phi] {Phi''[r] + 2*r^(-1) Phi'[r] + 0.104479*Phi[r] == 0 , Phi[0] == K} DSolve[%, Phi, r] Phi[r_] = First@(Phi[r] /. % // Chop) Limit[Phi[r], r -> 0] -- David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ "donkorgi12" <ringtailinblacklw02 at gmail.com> wrote in message news:fveguc$5ed$1 at smc.vnet.net... >I am solving the following Differential Equation > > Phi''[r]+2*r^(-1)Phi'[r]+0.104479*Phi[r]==0 and Phi[0]==K (some > constant) ; kinda has a cos/sin solution > > 2.71828^(-0.323232 \[ImaginaryI] r) ((0.+ > 0. \[ImaginaryI]) + (0.+ 0. \[ImaginaryI]) 2.71828^( > 0.646465 \[ImaginaryI] r) + (0.+ 1.54687 \[ImaginaryI]) K - (0.+ > 1.54687 \[ImaginaryI]) 2.71828^(0.646465 \[ImaginaryI] r) K) > > all divided by r. > > My problem is that Mathematica is not treating those "zeros".... as > well zeros. Thus, the solution cannot really be used. In fact, if I > manually reproduce the solution and remove those "zeros", then the > solution is fine. > > I have another similar ODE Anyone have any ideas. > > Phi''[r]+2*r^(-1)Phi'[r]- 287.31*Phi[r]==0 and Phi[3R]==0(some > constant) ; kinda has a cosh/sinh solution > > As you might have notice I need to match these two solutions and their > derivatives at some point. The ratio of which gives me what I desire > the value of R. > > Yet, Mathematica treats those "zeros" as something else. > > >