Re: I have an operator, How do I DSolve it?

*To*: mathgroup at smc.vnet.net*Subject*: [mg96213] Re: I have an operator, How do I DSolve it?*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Mon, 9 Feb 2009 05:36:35 -0500 (EST)*Organization*: The Open University, Milton Keynes, UK*References*: <gmndhs$s04$1@smc.vnet.net>

In article <gmndhs$s04$1 at smc.vnet.net>, Aaron Fude <aaronfude at gmail.com> wrote: > I have defined a differential operator, but what's the syntax for > using it in DSolve? > > I have the Bessel operator > > B[m_, f_] := (D[f[#], {#, 2}] + > 1/# D[f[#], {#, 1}] + (1 - m^2/#^2) f[#]) & > > It's the Bessel operator for now, but I will soon replace it with a > different one. This is for testing puposes only. > > Now, how do I use it in DSolve? DSolve[B[m, S][r] == 0, S[r], r] > So far I use > > DSolve[S''[r] + 1/r S'[r] + (1 - m^2/r^2) S[r] == 0, S[r], r] > > but that's not a good solution because I have to different expressions > for the same thing. I want to used B only. In[1]:= B[m_, f_] := (D[f[#], {#, 2}] + 1/# D[f[#], {#, 1}] + (1 - m^2/#^2) f[#]) & In[2]:= DSolve[B[m, S][r] == 0, S[r], r] Out[2]= {{S[r] -> BesselJ[m, r] C[1] + BesselY[m, r] C[2]}} In[3]:= DSolve[S''[r] + 1/r S'[r] + (1 - m^2/r^2) S[r] == 0, S[r], r] Out[3]= {{S[r] -> BesselJ[m, r] C[1] + BesselY[m, r] C[2]}} In[4]:= %% === % Out[4]= True Regards, --Jean-Marc