Series Solution Coefficients for y'' - (sinx)y = cosx with Initial Conditions

[JaysFan31]

Homework Statement

Find the indicated coefficients of the power series solution about x = 0 of the differential equation:
y'' - (sinx)y = cosx, y(0) = -5, y'(0) = 3.
y = _ + _x + _x^2 + _x^3 + _x^4 + O(x^5)

Homework Equations

The Attempt at a Solution

This is going to be a tad confusing in typing it, but I hope it can be read.

I have the summation(anx^n)
This equals y.
y' = summation(nanx^n-1)
y'' = summation (n(n-1)anx^n-2)

Just to make it easier, we end up with
x^n[(n+2)(n+1)an+2-ansinx] = cosx
Thus, an+2 = (cosx + ansinx) / ((n+2)(n+1))

I know obviously that the first two terms (x^0 and x^1) are -5 and 3 respectively. I also know that the x^2 term is 0.5 by plugging in 0 for x. However, this doesn't work for the rest of them. I've done a lot of these types of problems, but this is the first one with sin(x) or cos(x), which puts an "x" in the an+2 equation (which I wrote above). What does x equal in this case? Can anyone just show me how to find the remaining coefficients because I'm pretty sure my equation is correct.

 

[HallsofIvy]

Expand sin(x) and cos(x) in Taylor's series.

 

[JaysFan31]

Yes, but I still don't have x do I?

 

[JaysFan31]

I got it. Thanks for the help.