Question

Help Finding the first four nonzero terms through power series expansion.
y''-(cosx)y'-y=0 ; y(pi/2)=1 y'(pi/2)=1

Answer 1

\[y''-(cosx)y'-y=0\]

Answer 2

\[y(\frac{\pi}{2})=1,y'(\frac{\pi}{2})=1\]

Answer 3

which crazy guy give this problem to you?

Answer 4

my crazy differential equations prof

Answer 5

good luck ... this is going to be terribly awful. first look into Fuchs-Frobenius theorem and hunt for singularities.

Answer 6

Okay I have the solution, but am having a hard time writing it out

http://www.math.ualberta.ca/~runde/files/sol201-8.pdf
Scroll down to number 6

Answer 7

then use power series for y(x) ...and expland cos(x) and then compare the coefficients of 'x'

Answer 8

only 5 pages ??

Answer 9

No it's several different problems, the question is number 6, it's worked out already

Answer 10

An explanation would be beautiful.

Answer 11

okay okay ... let's see what can I do.

Answer 12

http://en.wikipedia.org/wiki/Fuchs's_theorem

Answer 13

http://www.math.ualberta.ca/~runde/files/sol201-8.pdf pleaes look at number 6

Answer 14

yes yes ... I am looking

Answer 15

it's 5000 degrees in my country and the heat is getting to me im sorry.

Answer 16

lol ... it's pretty hot here, but 00:19 in the morning!!
so fine :) ... u go to alberta?? i thought it was in the north??

Answer 17

I'm exaggerating, but please do explain if possible.

Answer 18

haha i know ... by Fuchs theorem , at pi/2 p(x) is analytic.

Answer 19

so it has a power series expansion of the form
http://upload.wikimedia.org/math/6/b/3/6b34399a973da121c349586088ad1760.png

Answer 20

where a is pi/2

Answer 21

yes

Answer 22

that 's' thing is useless there ... it's only useful when you have singularities in p(x) or q(x). so set s=0

Answer 23

keep going

Answer 24

differentiate this thing once and twice ....