anonymous
  • anonymous
Use power series to solve the differential equation. y'=xy
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I guess the first step is this \[y'-xy=0\]
experimentX
  • experimentX
hold on ... just let \[ y = \sum_{n=0}^\infty x^n \]
experimentX
  • experimentX
\[ y = \sum_{n=0}^\infty a_n x^n \]

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anonymous
  • anonymous
\[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}\] according to my book, same thing I guess
experimentX
  • experimentX
yes yes ... \[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0\] find the recurrence relation.
anonymous
  • anonymous
oh ok \[y''=\sum_{n=2}^{\infty} n(n-1)c_nx^{n-2}\] how do I apply these generalizations to this specific problem
experimentX
  • experimentX
hah?? why do you want more trouble? you don't need to find y'' for this.
anonymous
  • anonymous
really, I was just trying to follow this example in the book
anonymous
  • anonymous
I meant: really?
experimentX
  • experimentX
lol .. this example is of second order differential equation. you need it only for second order DE
anonymous
  • anonymous
because there is only a y' that makes it first order?
experimentX
  • experimentX
yep.
anonymous
  • anonymous
ok so I only consider this then, right? \[y'=\sum_{n=1}^{\infty} nc_nx^{n-1} \]
experimentX
  • experimentX
yes ... do you have a software called maple?
anonymous
  • anonymous
nope
anonymous
  • anonymous
but before solving for it, what do I sub?
anonymous
  • anonymous
or how do I know what do sub, I think I'm just afraid of sums :S
anonymous
  • anonymous
why do they have (n+2)(n+1) in the example?
experimentX
  • experimentX
from maple i got y(x) = y(0)+(1/2)*y(0)*x^2+(1/8)*y(0)*x^4+O(x^6)
experimentX
  • experimentX
Honesty I don't like power series method. here's how you do it. assume the solution has of the form \[ y = \sum_{n=0}^\infty a_n x^n \] so we are putting these vales in your DE and hunting a_n's
anonymous
  • anonymous
I'm back...
anonymous
  • anonymous
gtg again...
experimentX
  • experimentX
after putting the value of y in your DE, you get \[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0 \\ \] since on the left side you don't have negative power of x, put this \( \sum_{n=0}^\infty n \; c_n x^{n-1} \) = \( \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} \) Also we can make \( \sum_{n=0}^\infty \; c_n x^{n+1} \) = \(\sum_{n=0}^\infty c_{n-1} x^{n} \) now we have \[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + \sum_{n=0}^\infty c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} + c_{n-1} ) x^{n} = 0 \]
experimentX
  • experimentX
since you don't have power's of x on the right side, you have \[ (n+1) c_{n+1} + c_{n-1} = 0 \\ \text{ or, } c_{n+1} = {- c_{n-1} \over n+1}\] this is the recurrence relation you get put n=1, you get |dw:1348083846860:dw| this way you find the coefficients. hence you have solution.
experimentX
  • experimentX
not a nice method ... purely brute. but this works for all non linear differential equations ... even if their solution is not closed, you (put it up) as special function like Airy function, Bessel function, Hermite ... etc. and C_0 is your constant of integration. also you might wanna check Frobenius method. Best of luck with these.
TuringTest
  • TuringTest
I think you made a slight mistake @experimentX though I'm no fan of power series myself
experimentX
  • experimentX
hmm ... could you be more specific? power series is the method that is most prone to error ... I never get it correctly on first attempt (for second order)
experimentX
  • experimentX
oh ... i see that. i put up + instead of -
TuringTest
  • TuringTest
I'm just going to go ahead and give mine a shot, it's hard for me tow see how you ended up with c_{n+1} and c_{n-1} though
TuringTest
  • TuringTest
yeah that
experimentX
  • experimentX
EDIT:: \[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - \sum_{n=0}^\infty c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} - c_{n-1} ) x^{n} = 0 \]
experimentX
  • experimentX
EDIT:: \[ (n+1) c_{n+1} - c_{n-1} = 0 \\ \text{ or, } c_{n+1} = {c_{n-1} \over n+1} \]
experimentX
  • experimentX
so the final solution is \[ y = c_0\sum_{n=0}^\infty {1 \over 2^n n!}x^{2n}\]
TuringTest
  • TuringTest
okay that was nice, let me see if doing it my way changes anything
experimentX
  • experimentX
this is same as \[ \huge y = c_0 e^{x^2 \over 2}\]
experimentX
  • experimentX
2 am here. gotta sleep ... i'll be watching this thread tomorrow morning :) gotta be careful with power series. this is quite annoying.
TuringTest
  • TuringTest
yes your solution seems right @experimentX sleep well, I still am gonna try to brush up myself :)
TuringTest
  • TuringTest
@MathSofiya I have confirmed @experimentX 's answer in a slightly different way if you would like to go over it, let me know
TuringTest
  • TuringTest
\[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](notice the index starts at n=1 for the series for y' because the first term in the series for y is a constant \(a_n\) and disappears upon taking the derivative) Our problem then becomes\[\sum_{n=1}^\infty a_nnx^{n-1}-\sum_{n=0}^\infty a_nx^{n+1}=0\]The first thing is to get all the x's to the \(n^{th}\) power with index shifts, so we will start the first series one lower at n=1 (so we add 1 to n in the summand) and the second series we will shift up by 1 (so that n shifts down by 1 in the summand). Note these shifts also change the indices of the constants \(a_n\)\[\sum_{n=0}^\infty( n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]Next we need to get the indices the same again (make n start at the same number for each series) To avoid doing another index shift that will take us back where we started, we will "strip" the n=0 term out of the first series\[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]notice all we did was plug in n=1 and explicitly write that term out in front of the rest of the series. They call that "stripping out a term" in case you did not know. So we now have\[a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]Here we now need to figure out a formula for any pattern(s) in the constants \(a_n\) First, plugging in \(n=0\) since the series is not defined we get\[n=0:a_1=0\] so now we just have\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since exponentials like \(x^n\) can never be zero, and the summation sign will not change the fact that what is within the brackets must be zero we have the "recurrence relation"\[(n+1)a_{n+1}-a_{n-1}=0\implies a_{n+1}={a_{n-1}\over n+1}\]so again we start plugging in values for n starting from n=1\[n=1:a_2=\frac{a_0}2\]\[n=2:a_3=\frac{a_1}3=\frac03=0\]\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\]\[n=4:a_5=\frac{a_3}5=\frac05=0\]\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]\[~~~~~~~\vdots\]so what's the pattern for the constants? There seem to be two.
anonymous
  • anonymous
I'm back
TuringTest
  • TuringTest
at a very opportune time too :)
anonymous
  • anonymous
Sorry, women and hairstylist... I'm reading everything now
TuringTest
  • TuringTest
Take your time, it's not done... not sure if I should wait for you to try the next step. I will prepare the answer in the meantime anyway :)
anonymous
  • anonymous
sounds good, It's gonna take me a while to understand it. I'm gonna type questions as I come upon them.
anonymous
  • anonymous
\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\] because \[\sum_{n=0}^\infty a_nx^n x \] since the exponents are added that's why it's \[x^{n+1}\] correct?
TuringTest
  • TuringTest
yep
anonymous
  • anonymous
I understand what you did by changing, we start one lower at n=1, \[na_nx^{n-1} \implies (n+1)a_nx^n\] but why does that change where the sum starts from n=1 instead now from n=0 \[\sum_{n=1}^{\infty} \implies \sum_{n=0}^{\infty}\] sorry I guess I'm still struggling with the basics of summations
TuringTest
  • TuringTest
try it both ways, write out the first two or three terms and convince yourself how starting earlier means you need to add to the argument
TuringTest
  • TuringTest
for\[\sum_{n=1}^\infty na_nx^{n-1}\]what is the first term?
anonymous
  • anonymous
\[1\]
TuringTest
  • TuringTest
not quite, it's \(a_n\)
anonymous
  • anonymous
right
TuringTest
  • TuringTest
what about the first term of\[\sum_{n=0}^\infty(n+1)a_nx^n\]?
anonymous
  • anonymous
\[a_n\]
anonymous
  • anonymous
?
anonymous
  • anonymous
weird
TuringTest
  • TuringTest
see how by starting our index at n=0 we had to add 1 everywhere to keep the terms the same try it for more terms until you get the feel if you still doubt
anonymous
  • anonymous
does this mean that we're trying to manipulate the way we write the sum without changing the value of it?
TuringTest
  • TuringTest
yes if we started from zero the first term would be zero, because it has n as a coefficient the next term would have a coefficient of 1, the one after of 2 etc. if we add 1 then we get the correct coefficients the whole way through because we made up for changing the starting point of the count
anonymous
  • anonymous
ok so we change the starting point and made up for it. Why are we changing the starting points, what is our eventual goal....ahhh, we want to be able to write the two sums as one sum?
TuringTest
  • TuringTest
exactly :)
TuringTest
  • TuringTest
we can only combine the sums once we have their indices at the same starting point and as you will see, we also need the powers on x to be the same, customarily x^n....
TuringTest
  • TuringTest
these are not short problems :P
anonymous
  • anonymous
indices...
anonymous
  • anonymous
{n=..... } are those the indices? I'm kinda rusty on the terminology
TuringTest
  • TuringTest
yes exactly, n is the index of a sum with n=a under it, and "a" would be the starting point of the index changing the starting point of an index is called an index shift, and requires you to alter the summand (the thing under the summation sign) as I have just tried to describe.
TuringTest
  • TuringTest
that should say "we explicitly plugged in x=0..." *
TuringTest
  • TuringTest
I mean n=0 :/
anonymous
  • anonymous
by taking an \[a_1\] out of the summation how did that change the indices? nothing else seems to have been changed. \[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]
TuringTest
  • TuringTest
because we wrote out the first term explitily
TuringTest
  • TuringTest
explicitly*
anonymous
  • anonymous
ohh...so either way the first answer would be a_1 is the point again I guess
TuringTest
  • TuringTest
we don't need to start from n=0 anymore because we already wrote it out and took it out of the series, so repeating the n=0 term would add an extra term
anonymous
  • anonymous
on either side of that equal sign I mean.
TuringTest
  • TuringTest
what is important right now is that we get the index to start at n=1 like the other term does
TuringTest
  • TuringTest
if you have a series that starts at n=0, and you want to write it starting at n=1 you can either do an index shift (which we already did to get x^n on both series) or write out the n=0 term explicitly.
anonymous
  • anonymous
got it. I am down to this line now \[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\] and I do understand how you came to this
TuringTest
  • TuringTest
cool :)
anonymous
  • anonymous
I see a pattern something like this \[\frac{a_0}{2^n(n!)}\]
anonymous
  • anonymous
well I probably shouldn't use "n"
TuringTest
  • TuringTest
yes! that is exactly right for half the n
anonymous
  • anonymous
oh so I can use "n"
TuringTest
  • TuringTest
well no, we need to differentiate between the two cases for n n only equals what you wrote under what condition for n ?
TuringTest
  • TuringTest
a_n only equals what you wrote for which n ?
anonymous
  • anonymous
when n=0
anonymous
  • anonymous
no no no
TuringTest
  • TuringTest
actually the n in your formula is outright wrong sorry but look at it another way so you can figure how to fix it... for which values of n is a_n zero?
anonymous
  • anonymous
durrr...all the even numbers
anonymous
  • anonymous
haha it took me while to see that :P
TuringTest
  • TuringTest
yes, but I screwed up the question again, sorry I must phrase more carefully
TuringTest
  • TuringTest
for which *subscripts on a* is a_n zero ?
anonymous
  • anonymous
all the odd subscripts
TuringTest
  • TuringTest
right :) so and we can write any odd number as n=2k+1 with k=0,1,2..., right? so any term with a coefficient we can write as \(a_n=a_{2k+1}=0\) therefore our formula is only concerned with the even number subscripts since the odd ones will drop out
TuringTest
  • TuringTest
so we only need to think about the terms with n=2k by the way we skipped a set; plugging in n=0 to find what a_1 is (we can't do it with our formula because our series does not have a_1 defined) to back up just for a moment, what does a_1=
anonymous
  • anonymous
0?
TuringTest
  • TuringTest
yes, how did you know?
anonymous
  • anonymous
part of the pattern
TuringTest
  • TuringTest
that does not illustrate the proper reasoning, and I think you know that ;)
TuringTest
  • TuringTest
that reasoning will certainly not always work and will in fact often fail. the real reason is as follows:\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]plug in n=0 and notice something: the series is not defined at n=0, only from n=1 on, so we just wind up with the whole series not in the equation\[a_1=0\]
TuringTest
  • TuringTest
hence the problem we are now dealing with is really just\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]
anonymous
  • anonymous
interesting, yes
TuringTest
  • TuringTest
so where were we....?
anonymous
  • anonymous
one more thing
anonymous
  • anonymous
so you're saying that I could drop the \[a_1\] because it is not defined at n=0 yes...ok
TuringTest
  • TuringTest
a_1 is defined at n=0 (a_1=0), but the series is not
anonymous
  • anonymous
oh ok the series is not defined at n=0 and this is left \[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]
anonymous
  • anonymous
my goodness this is exhausting!
TuringTest
  • TuringTest
\[a_1+\cancel{\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n}^{\text{not defined}}=0\]\[a_1=0\]leaving\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since \(x^n\neq0\) then the coefficients must be zero\[(n+1)a_{n+1}-a_{n-1}=0\]and we get the recurrence relation we were doing
anonymous
  • anonymous
agreed
TuringTest
  • TuringTest
and yes, it is extremely exhausting I think most people who do DE's whould agree, but they handle non-constnat coefficient problems well
anonymous
  • anonymous
so would this finally be the solution \[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\] I doubt it...seems like it's just the beginning...now it's on to solving the DE
TuringTest
  • TuringTest
I'm afraid we have ways to go dear, I would understand if you want a break or something... I am mainly practicing, and I have a final solution from beginning to end prepared on a word doc, but it would be better to guide you of course.
TuringTest
  • TuringTest
that solution above is not an explicit solution for\[ y=\sum_{n=0}^\infty a_nx^n\]which is what we want for an answer
anonymous
  • anonymous
Wow an example they did in my book took up 2 pages. yes. I don't think I need a break yet...but when we're done with this problem I'm watching an episode of the Big Bang Theory and just laugh at all of their stupid jokes. continue.....
TuringTest
  • TuringTest
we need a formula for a_n in terms of the index, that's how we get our answer we are mostly there actually, I just want to make sure you understand each step
anonymous
  • anonymous
"index"?
TuringTest
  • TuringTest
you will see what I mean: that the index n (which don't get attached to cuz we're gonna change the name of it in a second) is going to give an explicit formula for \(a_n\) in terms of the only constant we can't determine \(a_0\) so you need a formula for all even \(a_n\) in terms of n
TuringTest
  • TuringTest
you almost had it, but you wrote\[a_n=\frac{a_0}{2^nn!}\]but check the numbers and you can see that's not quite right
anonymous
  • anonymous
what I screwed up on earlier by calling it \[\frac{a_0}{2^n(n!)}\]
anonymous
  • anonymous
ok I'll fix it...let's see here...
TuringTest
  • TuringTest
hint: We are only talking about even numbers, the odd subscripts are all zero. All even numbers can be written as...?
anonymous
  • anonymous
\[a_n= \frac{a_0}{2^{2k}(2nk!)}\] If this is right.....
TuringTest
  • TuringTest
almost
anonymous
  • anonymous
woops hold your horses
anonymous
  • anonymous
\[a_n= \frac{a_0}{2^{2k}(2k!)}\]
TuringTest
  • TuringTest
but 2k=n.... the power on 2 is not the same as the subscript on a, check again the relation
anonymous
  • anonymous
\[a_{2k}=\frac{a_0}{2^{2k}(2k!)}\] That just looks odd
TuringTest
  • TuringTest
plug in k=2 and check if that's true based on the table I made way back up there
anonymous
  • anonymous
that gives me a_1 instead of a_0
anonymous
  • anonymous
am I loosing it?
TuringTest
  • TuringTest
? k=2 n=2k=4 check a_4
anonymous
  • anonymous
oh I see what you mean
anonymous
  • anonymous
\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\] uhm? what do you want me to check? the fact that it is not zero?
anonymous
  • anonymous
which was our goal
TuringTest
  • TuringTest
no, I want you to check if that fits your formula\[a_{2k}=\frac{a_0}{2^{2k}(2k)!}\]or if not, how to fix it :)
anonymous
  • anonymous
divide it by half because it seems like double the answer so \[a_{2k}=\frac{a_0}{2^{k}(k)!}\]
TuringTest
  • TuringTest
yay!
anonymous
  • anonymous
no way!
anonymous
  • anonymous
now is that my y_p or something?
TuringTest
  • TuringTest
and since these are the only a_n we care about we can now write\[a_n=a_{2k}=\frac{a_0}{2^kk!}\]and no, it's much better than our y_p...
anonymous
  • anonymous
it's my "y" final answer
TuringTest
  • TuringTest
basically, yes it is the key to the final answer
TuringTest
  • TuringTest
first notice that the above is true for \(k=0,1,2,...\) and we already made the tacit assumtion that our answer is\[y=\sum_{n=0}^\infty a_nx^n\]now since all odd will give zero terms for \(a_n\), there will be only even n in our formula so we can change all n in the series to 2k, start our series from k=0, and substitute the above formula for \(a_n\) and that will be our final answer
anonymous
  • anonymous
\[y=\sum_{k=0}^\infty \frac{a_0}{2^kk!}x^{2k}\]
TuringTest
  • TuringTest
you got it :)
TuringTest
  • TuringTest
and just to make things crystal clear that we are in fact right...
TuringTest
  • TuringTest
separation of variables gives y'=yx dy/y=xdx ln y=x^2/2+C y=Ce^(x^2/2) the Tayore xpansion for e^y is\[e^y=1+\frac y{1!}+\frac{y^2}{2!}+...=\sum_{n=0}^\infty\frac{y^n}{n!}\]let \[y=\frac12x^2\]and plug that into the taylor series and you get the same answer
TuringTest
  • TuringTest
with \(C=a_0\) of course
TuringTest
  • TuringTest
experimentX was right too, but I thought I would try to practice through explanation myself :)
anonymous
  • anonymous
Amazing! I owe you half of my paychecks when I begin working as civil engineer. Dude, you are awesome. Thanks for helping me.
anonymous
  • anonymous
patience is a virtue...you have an abundance of that...oh yeah, sorry I forgot you don't like compliments :P
TuringTest
  • TuringTest
Haha, don't tempt me I could use them. And thanks as always :) For posterity's sake I wanna post my full explanation without interruption, just because I already had it prepared and it may be easier to read should you decide to review this. It will slow down the page though...
anonymous
  • anonymous
Thank you. I'm off to watch an episode of the Big Bang Theory. Is that silly? LOL
TuringTest
  • TuringTest
Makes more sense than doing more series solution DE's to me! I'm off to do something unproductive too, see ya!
anonymous
  • anonymous
see ya!
anonymous
  • anonymous
medal?
TuringTest
  • TuringTest
gotta take it away from experimentX... which I'll do, he has enough sorry @experimentX ;)
TuringTest
  • TuringTest
\[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](notice the index starts at n=1 for the series for y' because the first term in the series for y is a constant \(a_n\) and disappears upon taking the derivative) Our problem then becomes\[\sum_{n=1}^\infty a_nnx^{n-1}-\sum_{n=0}^\infty a_nx^{n+1}=0\]The first thing is to get all the x's to the \(n^{th}\) power with index shifts, so we will start the first series one lower at n=1 (so we add 1 to n in the summand) and the second series we will shift up by 1 (so that n shifts down by 1 in the summand). Note these shifts also change the indices of the constants \(a_n\)\[\sum_{n=0}^\infty( n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]Next we need to get the indices the same again (make n start at the same number for each series) To avoid doing another index shift that will take us back where we started, we will "strip" the n=0 term out of the first series\[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]notice all we did was plug in n=0 and explicitly write that term out in front of the rest of the series. They call that "stripping out a term" in case you did not know. So we now have\[a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]Here we now need to figure out a formula for any pattern(s) in the constants \(a_n\) First, plugging in \(n=0\) since the series is not defined we get\[n=0:a_1=0\] so now we just have\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since exponentials like \(x^n\) can never be zero, and the summation sign will not change the fact that what is within the brackets must be zero we have the "recurrence relation"\[(n+1)a_{n+1}-a_{n-1}=0\implies a_{n+1}={a_{n-1}\over n+1}\]so again we start plugging in values for n starting from n=1\[n=1:a_2=\frac{a_0}2\]\[n=2:a_3=\frac{a_1}3=\frac03=0\]\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\]\[n=4:a_5=\frac{a_3}5=\frac05=0\]\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]\[~~~~~~~\vdots\]so what's the pattern for the constants? There seem to be two; one for even subscripts and one for odd. Even numbers can be written as \(n=2k\) and odd as \(n=2k+1\) for \(k=0,1,2,...\) That means that all the \(a_n\) for even n can be written as \(a_{2k}\) and have the form\[a_{2k}=\frac{a_0}{2^kk!}\]and for all odd n we have \[a_{2k+1}=0\]therefor all those terms will disappear from the series. That lets us replace all the n's in the summand with 2k and make a new representation of the series based on that. This way we don't include the odd n terms, which are zero anyway. We can now rewrite \[a_n=a_{2k}=\frac{a_0}{2^kk!}~~~~\text{where}~~~~k=0,1,2,3...\]and our guess for y becomes\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{2^kk!}\]
experimentX
  • experimentX
you guys have patience of a saint ... usually i don't write more than lines.

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