Series Solution for O.D.E (Frobenius method?)

[JacobHempel]

Homework Statement

(For Physics 306: Theoretical methods of Physics... Text book: Mathematical Tools for Physics (really good!)...
Assumption: 300 level class = considerably junior level class)

-- Find a series solution about x=0 for y''+ysec(x) = 0, at least to a few terms.

(Ans: a$_{0}$[1-(1/2)x$^{2}$+0x$^{4}$+1/720x$^{6}$+...]+a$_{1}$[x-(1/6)x$^{3}$-(1/60)x$^{5}$+...]

Homework Equations

Frobenius Solution: Supposing there is a singularity point at x=x0, a possible solution is given by the power series, that is

Ʃ(from k=0 to inf) a$_{k}$x$^{k+s}$

The Attempt at a Solution

There is a singularity point not at x=0, but at x= npi/2. The problem specifies, though, to try finding a power series solution about x=0.

Let x(t) = Ʃ a$_{k}$x$^{k+s}$

Using this, plug it into the equation to obtain

Ʃ a$_{k}$x$^{k+s-2}$ (k+s)(k+s-1) +Ʃ a$_{k}$x$^{k+s}$ = 0

after that I matched the terms up by pushing the first summation up by two indicese. To do this, I wrote out two terms, then shifted the k values by +2.

a$_{0}$(s)(s-1)x$^{s-2}$+a$_{1}$(1+s)(s)x$^{s-1}$+Ʃ a$_{k+2}$x$^{k+s}$ (k+s+2)(k+s+1) +Ʃ a$_{k}$x$^{k+s}$ = 0

This is where I stopped. I'm unsure as to what to do next. You match up the terms in a regular Forbenius method, but this time it has a cos(x) in the second term, which is what makes this problem strange to me.

I think the next step is to find s, then just write out a few terms of both the series.

Also sorry about the mess regarding the equations. I got rid of all the sigma indices, so when you see a sigma, it is always from k=0 to k=∞.

 

[voko]

Expand the secant function into a power series.

 

[vela]

Since there's no singularity at x=0, you can use a plain old power series and avoid trying to find s from an indicial equation.