anonymous
  • anonymous
Use power series to solve the differential equation \[y''+xy'+y=0\] \[y=\sum_{n=0}^{\infty} a_n x^n\] \[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\] \[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}\] \[xy'=\sum_{n=1}^{\infty}a_nnx^n\] \[\sum_{n=2}^{\infty} n(n-1)a_nx^{-2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
yep, now what's first?
anonymous
  • 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
  • TuringTest
no we don't strip out the terms yet, we just do an index shift

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More answers

TuringTest
  • TuringTest
the first order of business is to get all the x's to be to the nth power through using index shifts
TuringTest
  • 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
  • anonymous
you mean on \[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\]
TuringTest
  • TuringTest
yes, we need to start from a different n to get x^n through an index shift
anonymous
  • 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
  • anonymous
nooooo
anonymous
  • anonymous
hold on
anonymous
  • anonymous
\[\sum_{n=0}^\infty n(n+1)a_nx^n\]
TuringTest
  • TuringTest
nice :)
anonymous
  • anonymous
still not right
anonymous
  • anonymous
because that would give us zero
TuringTest
  • TuringTest
no but closer
TuringTest
  • TuringTest
you didn't add 2 to every n
anonymous
  • anonymous
\[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]
TuringTest
  • TuringTest
and I do mean \(every\) n (subscripts included)
TuringTest
  • TuringTest
again better, but...^
anonymous
  • anonymous
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n \]
anonymous
  • anonymous
can I add them?
TuringTest
  • TuringTest
now that, I do believe, is correct :)
anonymous
  • anonymous
or multiply
anonymous
  • anonymous
or just leave as is?
TuringTest
  • TuringTest
add what? oh that? I wouldn't yet
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
yeah leave as is asd now rewrite what you have
TuringTest
  • TuringTest
and*
TuringTest
  • TuringTest
the whole expression please I'm lazy
anonymous
  • 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
  • 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
  • anonymous
the middle
TuringTest
  • TuringTest
so what do we do to fix it?
TuringTest
  • TuringTest
you have to be a bit clever here...
anonymous
  • anonymous
\[ xa_n +\sum_{n=0}^\infty a_n(n+1)x^n \]
TuringTest
  • TuringTest
did you try to do an index shift?
TuringTest
  • TuringTest
it looks like you combined the idea of the index shift with stripping out the first term
anonymous
  • anonymous
oh my...let's try this again
TuringTest
  • TuringTest
it is a bit tricky so let me give you a hint...
TuringTest
  • 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
  • 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
  • anonymous
just have the addition of an a_n up front like in the last example?
TuringTest
  • TuringTest
no, answer my question above and you will see why...
anonymous
  • anonymous
a 1
anonymous
  • anonymous
times a_n
TuringTest
  • TuringTest
no, be more careful
TuringTest
  • TuringTest
what would be the first term of\[\sum_{n=0}^\infty a_n nx^n\]
anonymous
  • anonymous
zero
TuringTest
  • TuringTest
right, so...
TuringTest
  • 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
  • TuringTest
so we can go ahead and just change the starting point of that series without further manipulation
TuringTest
  • TuringTest
always watch for that, it comes in handy
anonymous
  • 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
  • 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
  • anonymous
ok
anonymous
  • 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
  • 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
  • TuringTest
good, so now make it all into one summation...
anonymous
  • anonymous
\[\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0\]
TuringTest
  • TuringTest
nice, so what will you set to zero to find the recurrence relation?
TuringTest
  • TuringTest
(I am going to type minimally now as I am eating dinner at the same time :)
anonymous
  • anonymous
food comes first in my book... :P Ok let's see here...
TuringTest
  • TuringTest
I can eat and respond with "yes" or "no" pretty well, no worries :)
anonymous
  • 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
  • anonymous
si?
TuringTest
  • TuringTest
si, pero si quieras podrias factor la \(-a_n\)
anonymous
  • anonymous
ok \[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]
TuringTest
  • 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
  • TuringTest
go ahead and list out the first 5 or so and let's see if we see a relation
anonymous
  • 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
  • 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
  • anonymous
\[n=0:a_2=-a_0\frac{-1}{2}\]
anonymous
  • anonymous
\[n=0:a_2=a_0\frac{1}{2}\]
anonymous
  • anonymous
whatcha eating?
TuringTest
  • TuringTest
torta de pollo=chicken sandwich on a french sort of bread
TuringTest
  • TuringTest
con chipotle
anonymous
  • anonymous
nice
anonymous
  • anonymous
oh yeah...the pattern
TuringTest
  • 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
  • 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
  • TuringTest
each one*
anonymous
  • anonymous
in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]
anonymous
  • anonymous
take you're time...i'm gonna watch a few more mins of my episode
TuringTest
  • 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
  • TuringTest
okay I got a fairly simple answer, let me see if I can confirm it...
TuringTest
  • TuringTest
stupid wolfram is of no help :/
anonymous
  • 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
  • anonymous
how did I do?
TuringTest
  • 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
  • TuringTest
let me try on paper again...
TuringTest
  • TuringTest
here I think I can demonstrate the benefit of not multiplying out terms and writing things over-explicitly to find patterns...
anonymous
  • anonymous
ok
TuringTest
  • 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
  • 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
  • TuringTest
so try to do the same thing as last time name all even numbers and make a new series with index k
TuringTest
  • TuringTest
then represent the coefficients and plug that in for \(a_n\)
anonymous
  • anonymous
n=2k \[a_n=a_{2k}=\frac{a_0}{k!}\]
TuringTest
  • TuringTest
is it over k! ?
anonymous
  • anonymous
hhm...you know I thought about 2k!, but wouldn't that me too much? \[a_n=a_{2k}=\frac{a_0}{2k!}\]
anonymous
  • anonymous
*be
TuringTest
  • 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
  • anonymous
\[a_n=a_{2k}=\frac{a_0}{(2k)!}\]
TuringTest
  • TuringTest
yeah, the sub it into the original guess for y
TuringTest
  • TuringTest
then*
anonymous
  • anonymous
\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\]
TuringTest
  • 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
  • anonymous
I forgot, what did you intend to study? Electrical Engineering or Mathematics?
TuringTest
  • TuringTest
it definitely has the e^(-x^2) think in it though just by glancing, so I am even more confident now
anonymous
  • anonymous
why the e^?
TuringTest
  • TuringTest
Electrical engineering, though I admit I vacillate at times We'll seee how I feel after a year in uni
TuringTest
  • 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
  • anonymous
don't have the brain cells left to do so, but I trust your reasoning
TuringTest
  • TuringTest
haha, well thanks. I hope I'm right. Nos vemos!
anonymous
  • anonymous
see ya!
anonymous
  • anonymous
my spanish sucks as you might be able to tell
TuringTest
  • TuringTest
Nos vemos ~ we'll see each other
anonymous
  • anonymous
I was right!
TuringTest
  • TuringTest
sweet :)
anonymous
  • anonymous
we've made a mistake
anonymous
  • 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
  • TuringTest
let's see...
TuringTest
  • TuringTest
and why should 24 be 12 exactly?
anonymous
  • anonymous
because we need 12 4*3 is 12
anonymous
  • anonymous
I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one
TuringTest
  • TuringTest
you dropped a 2...
anonymous
  • anonymous
where?
anonymous
  • anonymous
\[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]
anonymous
  • anonymous
oh I see
TuringTest
  • 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
  • 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
  • anonymous
I'll finish y''=y tomorrow...I got it except for the parts stated above
TuringTest
  • 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
  • anonymous
ok
TuringTest
  • 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
  • anonymous
g'night
anonymous
  • anonymous
|dw:1348252850986:dw|
anonymous
  • anonymous
|dw:1348252918564:dw|
anonymous
  • anonymous
|dw:1348253019736:dw|
anonymous
  • anonymous
|dw:1348253097421:dw|
anonymous
  • anonymous
|dw:1348253165705:dw|
anonymous
  • 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
  • TuringTest
yep, exactly
TuringTest
  • TuringTest
oh no, you dropped an x
TuringTest
  • TuringTest
at n=1 the exponent on x is 1
TuringTest
  • 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
  • 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
  • anonymous
much better
TuringTest
  • TuringTest
thought so, probably should have done that a while ago, hehe :P
anonymous
  • anonymous
Makes perfect sense now, no doubt in my mind anymore. Haha, the brain exercise was good though
anonymous
  • anonymous
cool thanks !
TuringTest
  • TuringTest
Awesome, always a pleasure :D
anonymous
  • 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
  • TuringTest
hm... I guess you may be right
anonymous
  • anonymous
yuuss!!!!
anonymous
  • anonymous
^^^^my way of saying: yes!
anonymous
  • anonymous
so it's an alternating series? for 2k+1?
anonymous
  • anonymous
I mean 2k
TuringTest
  • TuringTest
sorry I was double-checking this whole time yeah it alternates, so you know the only change we're going to make?
anonymous
  • anonymous
let's see if I remember, can I cheat?
TuringTest
  • 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
  • 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
  • TuringTest
good idea brb
anonymous
  • anonymous
k
anonymous
  • 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
  • TuringTest
what can we do to -1 to make it even?
TuringTest
  • TuringTest
sorry, positive*
anonymous
  • anonymous
\[(-1)^n\]
TuringTest
  • TuringTest
yes
anonymous
  • anonymous
AWESOME!
TuringTest
  • TuringTest
very awesome!!!!!!!!!!!!
anonymous
  • anonymous
hey I got 240 for a_6
TuringTest
  • TuringTest
you love multiplying the numbers out, I just call it 6!
anonymous
  • anonymous
no that's 720
TuringTest
  • 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
  • anonymous
sounds good... sorry don't mean to be nitpicking :S
TuringTest
  • TuringTest
oh no, I;m glad you pointed it out, but.... must... eat
anonymous
  • anonymous
dude go for it LOL I'll see if I can figure this pattern out
anonymous
  • 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
  • anonymous
\[-\frac{a_n}{n+2}\]
anonymous
  • anonymous
\[a_{n+2}=-\frac{a_n}{n+2}\]
TuringTest
  • 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
  • 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
  • anonymous
yuuus!!!
TuringTest
  • 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
  • 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
  • TuringTest
well you're catching all my mistakes, so it seems to be working :)
TuringTest
  • TuringTest
the pattern seems harder to write concisely for me now... though it seems obvious
anonymous
  • 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
  • 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
  • anonymous
yeah it's quite strange, but the formula is correct, right? could we have still made an error there?
TuringTest
  • 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
  • anonymous
sure I'll start from the beginning and post it in a new post...this one is getting quite long
TuringTest
  • 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
  • anonymous
oh ok, I'll see ya trmw then
TuringTest
  • TuringTest
hasta luego
anonymous
  • anonymous
hasta luego mi amigo!

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