Re: DSolve Issues
- To: mathgroup at smc.vnet.net
- Subject: [mg88372] Re: DSolve Issues
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Sat, 3 May 2008 06:16:40 -0400 (EDT)
- References: <fveguc$5ed$1@smc.vnet.net>
Hi,
DSolve[{Phi''[r] + 2*r^(-1) Phi'[r] + 0.104479*Phi[r] == 0,
Phi[0] == K}, Phi[r], r] // Chop
??
Regards
Jens
donkorgi12 wrote:
> 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.
>
>
>