Re: Integro-differential analog of Love's equation via power series

*To*: mathgroup at smc.vnet.net*Subject*: [mg79846] Re: Integro-differential analog of Love's equation via power series*From*: chuck009 <dmilioto at comcast.com>*Date*: Tue, 7 Aug 2007 01:21:57 -0400 (EDT)

I made a mistake posting my code above: The 1 in the equation I'm coding as g[x_]:=1 and then I use g[xs] in the rhs calculations. I've also used this code to solve: f'[x]=1-1/3x+Integrate[xt f[t],{t,0,1}] with good results, the solution of which is just f[x]=x. a = -1; b = 1; y0 = 1/2; n = 40; g[x_] := 1; xs = N[Table[x, {x, -1, 1, (b - a)/(n - 1)}], 6]; cs = Table[Subscript[c, k], {k, 0, n}]; lhs = cs.Table[(i - 1)*xs^(i - 2), {i, 1, n + 1}]; rhs = g[xs] + ( 1/Pi)*cs.Table[NIntegrate[Evaluate[t^i/((xs - t)^2 + 1)], {t, -1, 1}], {i, 0, n}]; sol = Solve[lhs == rhs /. Subscript[c, 0] -> y0, Drop[cs, 1]] f[x_] = Sum[Subscript[c, i]*x^i, { i, 0, n}] /. First[sol] /. Subscript[c, 0] -> y0 Plot[f[x], {x, -1, 1}] fd[x_] = D[f[x], x] Plot[fd[x], {x, -1, 1}] Plot[1 + (1/Pi)*NIntegrate[f[t]/((x - t)^2 + 1), {t, -1, 1}] - fd[x], {x, -1, 1}, PlotRange -> All];