Re: Re: Power series solution to differential equations
- To: mathgroup at smc.vnet.net
- Subject: [mg41600] Re: [mg41569] Re: [mg41533] Power series solution to differential equations
- From: Bobby Treat <drmajorbob+MathGroup3528 at mailblocks.com>
- Date: Wed, 28 May 2003 04:57:31 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Here's a simpler version of that:
n = 7;
T[y_] := D[y, t, t] + Cos[t]*D[y, t] + t*y
diffeq = {T[y[t]] + O[t]^n == 0};
initial = {{y[0] == 1, y'[0] == 0}, {y[0] == 0, y'[0] == 1}};
derivatives = Table[D[y[t], {t, j}], {j, 0, n + 1}] /. t -> 0;
result = y[t] + O[t]^(n + 2);
(result /. Solve[Join[diffeq, #], derivatives]) & /@
initial // Flatten
Bobby
-----Original Message-----
From: Selwyn Hollis <selwynh at earthlink.net>
To: mathgroup at smc.vnet.net
Subject: [mg41600] [mg41569] Re: [mg41533] Power series solution to differential
equations
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
>
>
>