# Can anybody help me with taylor series/differential equations?

1. May 23, 2008

### jeebs

This is for revision purposes (not homework so im not trying to cheat my way out of it!) and its too late in the week to see my lecturer about this. I dont 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 dont really have a clue what im meant to do here. im 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 thats proven that the x^2 term does not appear in the final answer. other than that i am stuck.

Last edited: May 23, 2008
2. May 23, 2008

### 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