Re: DSolve Issues

*To*: mathgroup at smc.vnet.net*Subject*: [mg88389] Re: [mg88335] DSolve Issues*From*: "W_Craig Carter" <ccarter at mit.edu>*Date*: Sat, 3 May 2008 06:19:44 -0400 (EDT)*References*: <200805020742.DAA05399@smc.vnet.net>

Hello, I can't see your code, so I have to guess what you wish to do. There is a difference between 0. and 0 And for good and useful reasons For your case: Rationalize[ 0.104479, .00001] (*is 7/67*) DSolve[Phi''[r] + 2*r^(-1) Phi'[r] + (7/67) Phi[r] == 0 , Phi[r], r] gives a useful solution Even more useful is DSolve[Phi''[r] + 2*r^(-1) Phi'[r] + ArbitraryConstant Phi[r] == 0 , Phi[r], r] On Fri, May 2, 2008 at 3:42 AM, donkorgi12 <ringtailinblacklw02 at gmail.com> 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 > > Yet, Mathematica treats those "zeros" as something else. > Try this little experiment, it is very instructive for your zero conceptualization (observe the "." in the output) matsym = {{1, 1}, {10^24, 10^24}} Det[matsym] matnum = {{1., 1}, {10^24, 10^24}} Det[matnum] matnumappx = {{1., 1}, {10^24 + 1, 10^24}} Det[matnumappx] matappxsym = {{1, 1}, {10^24 + 1, 10^24}} Det[matappxsym] -- W. Craig Carter

**References**:**DSolve Issues***From:*donkorgi12 <ringtailinblacklw02@gmail.com>