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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.
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