Series Solution Near an Ordinary Point

[Ozymandius]

Homework Statement

Determine φ''(x₀), φ'''(x₀), and φ⁽⁴⁾(x₀) for the given point x₀ if y=φ(x) is a solution of the given initial value problem.

y'' + (sinx)y' + (cosx)y = 0 y(0) = 0; y'(0) = 1

Homework Equations

y = φ(x) = Ʃa_n(x-x₀)ⁿ

The Attempt at a Solution

I started off by differentiating y to get y' and y''. I then plugged those into the original equation, adjusted the numbers a bit to get the entire equation to fit into one summation. From there, I factored out xⁿ, then set the bulk of the equation equal to 0. From there, I solved for a_n+2, plugged in n=0 to get:

a₂ = (-(sinx)a₁ - (cosx)a₀))/2

This is where I get stuck. I am unsure as to what I'm supposed to do from here, although I have ideas. Am I just supposed to somehow solve for a₂ and plug it into the φ''(x) equation? If so, where do I get a₁ and a₀ from?
Also, I apologize for not knowing how to do the graphic code to make it look as it looks on paper. Thank you!UPDATE: Assuming I am correct, I have solved for a₂, a₃, and a₄ when setting a₀ = y(0) and a₁ = y'(0).
How do I use these newfound numbers to solve for the solutions?

a₂ = 0
a₃ = -1/6
a₄ = 0

 

[mighty2000]

To use these values to solve for the solutions, you can plug them into the equation for φ(x) = Ʃan(x-x0)n. This will give you the solution for φ(x) at x0, which is the given point. You can also use these values to find the other derivatives, φ''(x0), φ'''(x0), and φ(4)(x0), by plugging them into the equations for each derivative.

φ''(x0) = 0
φ'''(x0) = 0
φ(4)(x0) = -1/30

These values represent the second, third, and fourth derivatives of the solution at the given point x0. You can use these values to further analyze the behavior of the solution at this point and make predictions about its behavior in the future.