Re: Power series solution to differential equations
- To: mathgroup at smc.vnet.net
- Subject: [mg41569] Re: [mg41533] Power series solution to differential equations
- From: Selwyn Hollis <selwynh at earthlink.net>
- Date: Tue, 27 May 2003 01:47:31 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
At risk of redundancy, here's a much nicer technique for the nonsingular case that's based on "Abbott's method." I can't see how it can be used in the singular case though. T[y_] := D[y,t,t] + Cos[t]*D[y,t] + t*y n = 5; {y[t]+O[t]^(n+2) /. Solve[(T[y[t]]+O[t]^n == 0)/. {y[0]->1,y'[0]->0}, Table[D[y[t], {t,j}], {j,2,n+1}]/.t->0]/. {y[0]->1,y'[0]->0}, y[t]+O[t]^(n+2) /. Solve[(T[y[t]]+O[t]^n == 0)/. {y[0]->0,y'[0]->1}, Table[D[y[t], {t, j}], {j, 2, n + 1}]/.t->0]/. {y[0]->0,y'[0]->1}}//Flatten ----- Selwyn Hollis http://www.math.armstrong.edu/faculty/hollis On Monday, May 26, 2003, at 05:46 AM, Dr. Wolfgang Hintze wrote: > Given a differential equation of the form > > diffeq = a[x] u''[x] + b[x] u'[x] + f[x, u[x]] == 0 > > where ' means d/dx we assume that u[x] has a power series expansion > about x0 of the form (t = x-x0) > > u[t] = Sum[ c[k] t^(k+z) , {k, 0, Infinity }] > > We have to determine z and the coefficients c[k]. > > Question: what is the best way to tackle this problem in Mathematica? > > Any hint is greatly appreciated. > > Wolfgang > > >