Use power series to solve the differential equation
\[y''+xy'+y=0\]
\[y=\sum_{n=0}^{\infty} a_n x^n\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\]
\[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}\]
\[xy'=\sum_{n=1}^{\infty}a_nnx^n\]
\[\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

- anonymous

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- schrodinger

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- TuringTest

yep, now what's first?

- anonymous

changing the sums so that they all begin with n=2
changing the last sum from n=0 to n=1 would be
\[a_1+\sum_{n=1}^\infty a_nx^n\]

- TuringTest

no we don't strip out the terms yet, we just do an index shift

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## More answers

- TuringTest

the first order of business is to get all the x's to be to the nth power through using index shifts

- TuringTest

the only series that has x to some power other than n is the first one
how can we change the index of the first series to make the exponent on x be n?

- anonymous

you mean on
\[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\]

- TuringTest

yes, we need to start from a different n to get x^n through an index shift

- anonymous

so I'm shifting the starting point up by 2 so I have to subtract 2 from {n=2-2}
\[-\sum_{n=0}^\infty n(n-1)a_nx^n\] negative to make up for the negative?

- anonymous

nooooo

- anonymous

hold on

- anonymous

\[\sum_{n=0}^\infty n(n+1)a_nx^n\]

- TuringTest

nice :)

- anonymous

still not right

- anonymous

because that would give us zero

- TuringTest

no but closer

- TuringTest

you didn't add 2 to every n

- anonymous

\[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]

- TuringTest

and I do mean \(every\) n (subscripts included)

- TuringTest

again better, but...^

- anonymous

\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n
\]

- anonymous

can I add them?

- TuringTest

now that, I do believe, is correct :)

- anonymous

or multiply

- anonymous

or just leave as is?

- TuringTest

add what? oh that? I wouldn't yet

- anonymous

ok

- TuringTest

yeah leave as is asd now rewrite what you have

- TuringTest

and*

- TuringTest

the whole expression please I'm lazy

- anonymous

\[\sum_{n=0}^{\infty} (n+2)(n-1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

- TuringTest

\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]okay next order of business is what you tried to do earlier; get all indices to start at the same n value
the only problem child here is which term?

- anonymous

the middle

- TuringTest

so what do we do to fix it?

- TuringTest

you have to be a bit clever here...

- anonymous

\[ xa_n +\sum_{n=0}^\infty a_n(n+1)x^n \]

- TuringTest

did you try to do an index shift?

- TuringTest

it looks like you combined the idea of the index shift with stripping out the first term

- anonymous

oh my...let's try this again

- TuringTest

it is a bit tricky so let me give you a hint...

- TuringTest

you cannot do another index shift. If you do you will change the exponent on x, which means we are screwed again, so we need to do something else to get this to start from n=0

- TuringTest

we can't really strip out a term because we don't have this series defined for n<1, but ask yourself, if it was defined at n=0, what would the first term be?

- anonymous

just have the addition of an a_n up front like in the last example?

- TuringTest

no, answer my question above and you will see why...

- anonymous

a 1

- anonymous

times a_n

- TuringTest

no, be more careful

- TuringTest

what would be the first term of\[\sum_{n=0}^\infty a_n nx^n\]

- anonymous

zero

- TuringTest

right, so...

- TuringTest

\[0+\sum_{n=1}^\infty a_n nx^n=\sum_{n=0}^\infty a_n nx^n\]so we actually didn't need to change anything; the first term we "stripped out" was a zero anyway! It makes no difference.

- TuringTest

so we can go ahead and just change the starting point of that series without further manipulation

- TuringTest

always watch for that, it comes in handy

- anonymous

oh ok uhm.....I'll revisit the last example later, because the I'm not convinced that we needed to do a index shift (or was it stripping of a term)

- TuringTest

first to get the x^n on all terms we did an index shift (where we add or subtract from each n in the summand)
next to get all the starting points for each series the same, we strip out terms/or do the trick we just did
they are different tricks, and the order in which you do them is imortant

- anonymous

ok

- anonymous

\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]

- TuringTest

it is important for you to try to understand why the index shift requires you to add/subtract each n in the summand while stripping out a term does not

- TuringTest

good, so now make it all into one summation...

- anonymous

\[\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0\]

- TuringTest

nice, so what will you set to zero to find the recurrence relation?

- TuringTest

(I am going to type minimally now as I am eating dinner at the same time :)

- anonymous

food comes first in my book... :P
Ok let's see here...

- TuringTest

I can eat and respond with "yes" or "no" pretty well, no worries :)

- anonymous

\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}\]

- anonymous

si?

- TuringTest

si, pero si quieras podrias factor la \(-a_n\)

- anonymous

ok
\[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]

- TuringTest

and now comes the tedious part that you can do while I eat
plug in n=0,1,2,3
and check what results from that recurrence relation until you notice a pattern like last time
so just cranks it out; plug in n=0,n=1,n=2
and if a_2 can be written in terms of a_0 or something like last time, do it. That is how we find the pattern.

- TuringTest

go ahead and list out the first 5 or so and let's see if we see a relation

- anonymous

\[n=1:a_3=0\]
\[n=2:a_4=-a_2\frac{1}{12}\]
\[n=3:a_5=-a_3\frac{2}{20}\]
\[n=4:a_6=-a_4\frac{3}{30}\]
\[n=5:a_7=-a_5\frac{4}{42}\]

- TuringTest

what about n=0 ?
we need to start from n=0 since our series starts from n=0 this time
(last time it started from n=1)

- anonymous

\[n=0:a_2=-a_0\frac{-1}{2}\]

- anonymous

\[n=0:a_2=a_0\frac{1}{2}\]

- anonymous

whatcha eating?

- TuringTest

torta de pollo=chicken sandwich on a french sort of bread

- TuringTest

con chipotle

- anonymous

nice

- anonymous

oh yeah...the pattern

- TuringTest

so you did well in your subs I think, but now you must try to always sub all the way back as far as you can..

- TuringTest

yeah, the pattern
write each on as some finite number of unknowns, in this case everything in terms of \(a_0\) and \(a_3\)

- TuringTest

each one*

- anonymous

in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]

- anonymous

take you're time...i'm gonna watch a few more mins of my episode

- TuringTest

it's not different, it's the same as last time
odd subscript gives zero
even has a pattern reducible to a_0

- TuringTest

okay I got a fairly simple answer, let me see if I can confirm it...

- TuringTest

stupid wolfram is of no help :/

- anonymous

\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\]
\[n=1:a_3=0\]
\[n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}\]
\[n=3:a_5=0\]
\[n=4:a_6=-a_4\frac{3}{30}=\frac{a_0}{5(3!)}\]
\[n=5:a_7=0\]

- anonymous

how did I do?

- TuringTest

almost, I actually realized that I made a mistake because I think I see a mistake from you
so you did as well as me I'd say :D

- TuringTest

let me try on paper again...

- TuringTest

here I think I can demonstrate the benefit of not multiplying out terms and writing things over-explicitly to find patterns...

- anonymous

ok

- TuringTest

\[n=0:a_2=\frac{a_0}{2\cdot 1}\]\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}\]now reevaluate the pattern

- anonymous

\[n=0:a_2=\frac{a_0}{2\cdot 1}=\frac{a_0}{2!}\]
\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\]

- TuringTest

so try to do the same thing as last time
name all even numbers and make a new series with index k

- TuringTest

then represent the coefficients and plug that in for \(a_n\)

- anonymous

n=2k
\[a_n=a_{2k}=\frac{a_0}{k!}\]

- TuringTest

is it over k! ?

- anonymous

hhm...you know I thought about 2k!, but wouldn't that me too much?
\[a_n=a_{2k}=\frac{a_0}{2k!}\]

- anonymous

*be

- TuringTest

well that is the answer I get... so unless you have a way of confirming it apart from this I say that is correct
btw you should put the (2k)! with parentheses to indicate that it is not 2(k!)

- anonymous

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

- TuringTest

yeah, the sub it into the original guess for y

- TuringTest

then*

- anonymous

\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\]

- TuringTest

...and that's what I got :D
I sure think it's right as it comes out quite nicely, as opposed to wolf's answer
http://www.wolframalpha.com/input/?i=y''%2Bxy'%2By%3D0&t=crmtb01

- anonymous

I forgot, what did you intend to study? Electrical Engineering or Mathematics?

- TuringTest

it definitely has the e^(-x^2) think in it though just by glancing, so I am even more confident now

- anonymous

why the e^?

- TuringTest

Electrical engineering, though I admit I vacillate at times
We'll seee how I feel after a year in uni

- TuringTest

oh because the tayor expansion of e^(ax^2) always has x^(2k)/(2k)! in it
(try the expansion yourself and see why, it's not that hard if you remember Taylor series)

- anonymous

don't have the brain cells left to do so, but I trust your reasoning

- TuringTest

haha, well thanks. I hope I'm right. Nos vemos!

- anonymous

see ya!

- anonymous

my spanish sucks as you might be able to tell

- TuringTest

Nos vemos ~ we'll see each other

- anonymous

I was right!

- TuringTest

sweet :)

- anonymous

we've made a mistake

- anonymous

\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\]
is 24 not 12...so the pattern is probably incorrect too

- TuringTest

let's see...

- TuringTest

and why should 24 be 12 exactly?

- anonymous

because we need 12 4*3 is 12

- anonymous

I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one

- TuringTest

you dropped a 2...

- anonymous

where?

- anonymous

\[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]

- anonymous

oh I see

- TuringTest

you wrote\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\]\[n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}\]but\[a_2=\frac{a_0}2\]and\[\frac1{12}=\frac1{2(3!)}\]so
1)\[a_4=a_2\frac1{2(3!)}=\frac{a_0}2\cdot\frac1{2(3!)}=\frac{a_0}{4!}\]and that minus sign you put that minus sign does not belong

- anonymous

I did not pick up on the\[ a_2=\frac{a_0}{2}\] idea...i just randomly subbed a_0 for a_2
:/

- anonymous

I'll finish y''=y tomorrow...I got it except for the parts stated above

- TuringTest

once you find a base formula in terms of \(a_0\) or whatever the lowest you can express is, milk it to find the pattern

- anonymous

ok

- TuringTest

and like I say, sometime the writing out each term, even if it's just multiplying by one, helps make the pattern obvious
g'night then!

- anonymous

g'night

- anonymous

|dw:1348252850986:dw|

- anonymous

|dw:1348252918564:dw|

- anonymous

|dw:1348253019736:dw|

- anonymous

|dw:1348253097421:dw|

- anonymous

|dw:1348253165705:dw|

- anonymous

@TuringTest would this statement about striping a term be true (generally speaking)
\[0+a_1+\sum_{n=2}^\infty a_n nx^n\]

- TuringTest

yep, exactly

- TuringTest

oh no, you dropped an x

- TuringTest

at n=1 the exponent on x is 1

- TuringTest

\[\sum_{n=0}^\infty a_nnx^n=a_0(0)x^0+\sum_{n=1}^\infty a_nnx^n=0+a_1(1)x^1+\sum_{n=2}^\infty a_nx^n\]\[=a_1x+\sum_{n=2}^\infty a_nx^n\]

- TuringTest

easier example, stripping out terms for the sum of the first n integers\[\sum_{i=1}^n i=1+\sum_{i=2}^n=1+2+\sum_{i=3}^n i=1+2+3+\sum_{i=4}^n i\]

- anonymous

much better

- TuringTest

thought so, probably should have done that a while ago, hehe :P

- anonymous

Makes perfect sense now, no doubt in my mind anymore.
Haha, the brain exercise was good though

- anonymous

cool thanks !

- TuringTest

Awesome, always a pleasure :D

- anonymous

@TuringTest
shouldn't \[a_4 \] have a minus sign? I know you said it shouldn't but
\[a_2=\frac{a_0}{2}\]
and now we have
\[a_4=-a_2\frac{1}{12}=-\frac{a_0}{24}\]

- TuringTest

hm... I guess you may be right

- anonymous

yuuss!!!!

- anonymous

^^^^my way of saying: yes!

- anonymous

so it's an alternating series? for 2k+1?

- anonymous

I mean 2k

- TuringTest

sorry I was double-checking this whole time
yeah it alternates, so you know the only change we're going to make?

- anonymous

let's see if I remember, can I cheat?

- TuringTest

cheating would be having me tell you, any way you wanna figure out how to make a series change sign every other term is up to you :)

- anonymous

cheating would be using a the google search engine, but let's see if I can figure it out on my own without google.com

- TuringTest

good idea
brb

- anonymous

k

- anonymous

argh!
\[\sum_{n=0}^{\infty}x^n=0,x,x^2,etc.\]
\[\sum_{n=0}^{\infty}(-1)x^n=0,-x,-x^2\]

- TuringTest

what can we do to -1 to make it even?

- TuringTest

sorry, positive*

- anonymous

\[(-1)^n\]

- TuringTest

yes

- anonymous

AWESOME!

- TuringTest

very awesome!!!!!!!!!!!!

- anonymous

hey I got 240 for a_6

- TuringTest

you love multiplying the numbers out, I just call it 6!

- anonymous

no that's 720

- TuringTest

haha whatever I'm not even checking
if you are having a problem with the constants again give me a min and I'll brb again to help

- anonymous

sounds good... sorry don't mean to be nitpicking :S

- TuringTest

oh no, I;m glad you pointed it out, but.... must... eat

- anonymous

dude go for it LOL
I'll see if I can figure this pattern out

- anonymous

I think this is incorrect
\[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]
because
\[a_{n+2}=\frac{-a_nn-1}{(n+2)(n+1)}\neq-a_n\frac{n-1}{(n+2)(n+1)}\]
It should be
\[-a_n\frac{(n+1)}{(n+2)(n+1)}\]

- anonymous

\[-\frac{a_n}{n+2}\]

- anonymous

\[a_{n+2}=-\frac{a_n}{n+2}\]

- TuringTest

\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}\]is what you wrote, let's see if that is correct...

- TuringTest

\[(n+2)(n+1)a_{n+2}+a_n(n+1)=0\implies a_{n+2}=-a_n\frac{n+1}{(n+2)(n+1)}\]aw crap you're right again

- anonymous

yuuus!!!

- TuringTest

well, being right that we were wrong is always a little bittersweet, haha
gotta do the pattern over, but at least it's only\[a_{n+2}=-\frac{a_n}{n+2}\]which should be easier I would think

- anonymous

Yeah, let's see if I can figure the pattern out...
(btw I did yoga today...feeling more alert, or it could be plain coincidence =)

- TuringTest

well you're catching all my mistakes, so it seems to be working :)

- TuringTest

the pattern seems harder to write concisely for me now... though it seems obvious

- anonymous

\[a_2=-\frac{a_0}{2}=-\frac{a_0}{2!}\]
\[a_3=-\frac{a_1}{3}\]
\[a_4=-\frac{a_2}{4}=\frac{a_0}{3!}\]
\[a_5=-\frac{a_3}{5}\]
\[a_6=-\frac{a_4}{6}=-\frac{a_0}{36}\]
what's withe the 36?

- TuringTest

it's like the odds get cut out of the factorials and I'm trying to figure out how to write that
same for the evens

- anonymous

yeah it's quite strange, but the formula is correct, right? could we have still made an error there?

- TuringTest

a mistake in the recursion relation? we could have... to be safe we ought to take it from the top, but I can't do that, I have to leave...

- anonymous

sure I'll start from the beginning and post it in a new post...this one is getting quite long

- TuringTest

good idea
I am thinking\[a_{2k}=(-1)^k\frac{a_0}{2^kk!}\]and do not yet have an expression for the odds

- anonymous

oh ok,
I'll see ya trmw then

- TuringTest

hasta luego

- anonymous

hasta luego mi amigo!

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