DSolve Issues

*To*: mathgroup at smc.vnet.net*Subject*: [mg88335] DSolve Issues*From*: donkorgi12 <ringtailinblacklw02 at gmail.com>*Date*: Fri, 2 May 2008 03:42:33 -0400 (EDT)

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.

**Follow-Ups**:**Re: DSolve Issues***From:*"W_Craig Carter" <ccarter@mit.edu>