Using the taylor series, how do you solve y'' + xy = 0, initial conditions y(0) = 1, y'(0) = -3.?

Answer 1

The function is y= y(x)
Now expand this function around x=0
You can neglect higher order terms:
Then this expansion becomes:

y(x) = y(0)+ y'(0)[x-0] + y''(0) [x-0]^2 / 2! + y'''(0) [x-0]^3 / 3! + other terms...
So neglecting 3rd order term and putting the values given, we get
y= 1 -3x+ (-xy)x^2 /2

or y= 1-3x-x^3 *y/2
or y(1+x^3/2) = 1-3x
or
y= (1-3x) / (1+ x^3 /2)

Answer 2

Expand xy according to Taylor's series and solve.

Answer 3

let y= sum( x^n)