How Do You Solve a Taylor Series Problem with a Differential Equation?

[jeebs]

This is for revision purposes (not homework so I am not trying to cheat my way out of it!) and its too late in the week to see my lecturer about this. I don't have much of an attempt at the solution because i haven't got a clue where to start. It looks like just a short one though. Here goes...

I've got this problem, show that the taylor series for the function y=y(x) satisfying the differential equation

d^2y / dx^2 - (x^3)dy/dx = tan^-1(x)

with initial conditions y(0) = y'(0) = 1, begins

y = 1 + x + (x^3)/6 + (x^5)/30 +...

where

tan^-1(x) = x - (x^3)/3 + (x^5)/5 -...i don't really have a clue what I am meant to do here. I am pretty new to Taylor series. could somebody give me some instructions about how to tackle this problem?

thanks.all I've got so far is:

the n=0 term is 1, and the coefficient of x is just 1, and also

d^2y / dx^2 - (x^3)dy/dx = x - (x^3)/3 + (x^5)/5 -...

d^2y / dx^2 - 0(1) = 0 therefore d^2y / dx^2 = 0 when x = 0 (y''(0) = 0)

so i that's proven that the x^2 term does not appear in the final answer. other than that i am stuck.

 

[scottie_000]

So you have $$\frac{d^2y}{dx^2} - x^3 \frac{dy}{dx} = x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots$$

and you want a power series solution of the form $$y= \sum_{n=0}^{\infty} a_n x^n$$ where you need to find the coefficients $$a_n$$ from n=0 to 5

So differentiate the sum, twice, then plug in what you get and compare coefficients on both sides