       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 == 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==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.
>
>
>

```

• Prev by Date: Re: list of dates
• Next by Date: Re: DSolve Issues
• Previous by thread: Re: DSolve Issues
• Next by thread: Re: DSolve Issues