
cos(x)y'' + xy'  2y = 0
Find the first three nonzero terms in each of two linearly independent power series solutions centred at of the equation
What do you expect the radius of convergence of the solutions to be?
I tried to right in its taylor series form but I got stuck somehow... is a regular singular point that I've checked...please help thanks!

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I think zero is an ordinary point, not singular. I can derive a series solution with two arbitrary constants but can't figure out how to extract two linearly independent solutions from it. Maybe someone here can help. Writing it as:
where:
and if we assume a solution of the form
and substitute all the sums into the DE and shift the indexes I get:
collecting terms then:
and if I choose then for we have:
and I can let and be arbitrary and it's not hard to code this in Mathematica and show at least graphically, a good fit between this series solution and a numerical solution for some IVP such as:
The plot below superimposes a numerical solution to this IVP against the series solution using the first 20 terms. The agreement is not bad for my money.
Code:
pseries = Series[x/Cos[x], {x, 0, 25}];
Subscript[b, n_] := SeriesCoefficient[
pseries, n]
qseries = Series[2/Cos[x], {x, 0, 25}];
Subscript[b, n_] := SeriesCoefficient[
pseries, n]
Subscript[d, n_] := SeriesCoefficient[
qseries, n]
Subscript[a, 0] = 1;
Subscript[a, 1] = 0;
mycoef = Table[Subscript[a, n] =
(1/(n*(n  1)))*
(Sum[k*Subscript[a, k]*Subscript[
b, n  k  1], {k, 1,
n  1}]  Sum[Subscript[a, k]*
Subscript[d, n  k  2],
{k, 0, n  2}]), {n, 2, 20}];
myfun[x_] = Sum[Subscript[a, n]*x^n,
{n, 0, 20}]
series = Plot[myfun[x], {x, 0, 2},
PlotStyle > Red]
sol = NDSolve[
{Cos[x]*Derivative[2][y][x] +
x*Derivative[1][y][x]  2*y[x] ==
0, y[0] == 1, Derivative[1][y][0] ==
0}, y, {x, 0, 2}]
numeric = Plot[Evaluate[y[x] /. sol],
{x, 0, 2}, PlotStyle > Blue]
Show[{series, numeric}]

Look,this is what I'm gonna' do with this then: I'm going to let and . Then:
Likewise, I'll let and . Then:
and these are obviously linearlyindependent so then I can (probably) write the solution as:
. . . there's your five. Now, note he asked what do you "think" the radius of convergence is and not what "IS" the radius of convergence. Then that part is easy. I think the radius of convergence is since and and are singular there even though my plot goes past this point. And if you wanted to be more rigorous, then I'd recommend "Intermediate Differential Equations" by Rainville where I got most of this and where it is shown that this type of problem expressed as the series above has a radius of convergence at least as large as where this expression is derived in the text.
If that's not it then I bet it's close . . . B is ok with me. :)