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I guess the first step is this
\[y'-xy=0\]

hold on ... just let
\[ y = \sum_{n=0}^\infty x^n \]

\[ y = \sum_{n=0}^\infty a_n x^n \]

\[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}\]
according to my book, same thing I guess

hah?? why do you want more trouble? you don't need to find y'' for this.

really, I was just trying to follow this example in the book

I meant: really?

lol .. this example is of second order differential equation. you need it only for second order DE

because there is only a y' that makes it first order?

yep.

ok so I only consider this then, right?
\[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}
\]

yes ... do you have a software called maple?

nope

but before solving for it, what do I sub?

or how do I know what do sub, I think I'm just afraid of sums :S

why do they have (n+2)(n+1) in the example?

from maple i got
y(x) = y(0)+(1/2)*y(0)*x^2+(1/8)*y(0)*x^4+O(x^6)

I'm back...

gtg again...

I think you made a slight mistake @experimentX though I'm no fan of power series myself

oh ... i see that.
i put up + instead of -

yeah that

EDIT::
\[ (n+1) c_{n+1} - c_{n-1} = 0 \\
\text{ or, } c_{n+1} = {c_{n-1} \over n+1} \]

so the final solution is
\[ y = c_0\sum_{n=0}^\infty {1 \over 2^n n!}x^{2n}\]

okay that was nice, let me see if doing it my way changes anything

this is same as
\[ \huge y = c_0 e^{x^2 \over 2}\]

yes your solution seems right @experimentX
sleep well, I still am gonna try to brush up myself :)

I'm back

at a very opportune time too :)

Sorry, women and hairstylist...
I'm reading everything now

yep

for\[\sum_{n=1}^\infty na_nx^{n-1}\]what is the first term?

\[1\]

not quite, it's \(a_n\)

right

what about the first term of\[\sum_{n=0}^\infty(n+1)a_nx^n\]?

\[a_n\]

weird

exactly :)

these are not short problems :P

indices...

{n=..... } are those the indices? I'm kinda rusty on the terminology

that should say "we explicitly plugged in x=0..." *

I mean n=0
:/

because we wrote out the first term explitily

explicitly*

ohh...so either way the first answer would be a_1 is the point again I guess

on either side of that equal sign I mean.

what is important right now is that we get the index to start at n=1 like the other term does

cool :)

I see a pattern
something like this
\[\frac{a_0}{2^n(n!)}\]

well I probably shouldn't use "n"

yes! that is exactly right for half the n

oh so I can use "n"

a_n only equals what you wrote for which n ?

when n=0

no no no

durrr...all the even numbers

haha it took me while to see that :P

yes, but I screwed up the question again, sorry I must phrase more carefully

for which *subscripts on a* is a_n zero ?

all the odd subscripts

0?

yes, how did you know?

part of the pattern

that does not illustrate the proper reasoning, and I think you know that ;)

interesting, yes

so where were we....?

one more thing

so you're saying that I could drop the \[a_1\] because it is not defined at n=0
yes...ok

a_1 is defined at n=0 (a_1=0), but the series is not

my goodness this is exhausting!

agreed

"index"?

what I screwed up on earlier by calling it \[\frac{a_0}{2^n(n!)}\]

ok I'll fix it...let's see here...

\[a_n= \frac{a_0}{2^{2k}(2nk!)}\]
If this is right.....

almost

woops hold your horses

\[a_n= \frac{a_0}{2^{2k}(2k!)}\]

but 2k=n....
the power on 2 is not the same as the subscript on a, check again the relation

\[a_{2k}=\frac{a_0}{2^{2k}(2k!)}\]
That just looks odd

plug in k=2 and check if that's true based on the table I made way back up there

that gives me a_1 instead of a_0

am I loosing it?

?
k=2
n=2k=4
check a_4

oh I see what you mean

which was our goal

divide it by half because it seems like double the answer
so \[a_{2k}=\frac{a_0}{2^{k}(k)!}\]

yay!

no way!

now is that my y_p or something?

it's my "y" final answer

basically, yes
it is the key to the final answer

\[y=\sum_{k=0}^\infty \frac{a_0}{2^kk!}x^{2k}\]

you got it :)

and just to make things crystal clear that we are in fact right...

with \(C=a_0\) of course

experimentX was right too, but I thought I would try to practice through explanation myself :)

Thank you. I'm off to watch an episode of the Big Bang Theory. Is that silly? LOL

see ya!

medal?

gotta take it away from experimentX...
which I'll do, he has enough
sorry @experimentX ;)

you guys have patience of a saint ... usually i don't write more than lines.