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MathSofiya
Use power series to solve the differential equation. y'=xy
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
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.
oh ok \[y''=\sum_{n=2}^{\infty} n(n-1)c_nx^{n-2}\] how do I apply these generalizations to this specific problem
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
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?
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?
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)
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
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 \[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + \sum_{n=0}^\infty c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} + c_{n-1} ) x^{n} = 0 \]
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.
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.
I think you made a slight mistake @experimentX though I'm no fan of power series myself
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)
oh ... i see that. i put up + instead of -
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
EDIT:: \[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - \sum_{n=0}^\infty c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - c_{n-1} x^{n} = 0 \\ \huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} - c_{n-1} ) x^{n} = 0 \]
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 \[ \huge y = c_0 e^{x^2 \over 2}\]
2 am here. gotta sleep ... i'll be watching this thread tomorrow morning :) gotta be careful with power series. this is quite annoying.
yes your solution seems right @experimentX sleep well, I still am gonna try to brush up myself :)
@MathSofiya I have confirmed @experimentX 's answer in a slightly different way if you would like to go over it, let me know
\[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](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.
at a very opportune time too :)
Sorry, women and hairstylist... I'm reading everything now
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 :)
sounds good, It's gonna take me a while to understand it. I'm gonna type questions as I come upon them.
\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\] because \[\sum_{n=0}^\infty a_nx^n x \] since the exponents are added that's why it's \[x^{n+1}\] correct?
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
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
for\[\sum_{n=1}^\infty na_nx^{n-1}\]what is the first term?
not quite, it's \(a_n\)
what about the first term of\[\sum_{n=0}^\infty(n+1)a_nx^n\]?
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
does this mean that we're trying to manipulate the way we write the sum without changing the value of it?
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
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?
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....
these are not short problems :P
{n=..... } are those the indices? I'm kinda rusty on the terminology
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.
that should say "we explicitly plugged in x=0..." *
by taking an \[a_1\] out of the summation how did that change the indices? nothing else seems to have been changed. \[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]
because we wrote out the first term explitily
ohh...so either way the first answer would be a_1 is the point again I guess
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
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
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.
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
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
well no, we need to differentiate between the two cases for n n only equals what you wrote under what condition for n ?
a_n only equals what you wrote for which n ?
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?
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
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
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=
yes, how did you know?
that does not illustrate the proper reasoning, and I think you know that ;)
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\[a_1=0\]
hence the problem we are now dealing with is really just\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]
so where were we....?
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
oh ok the series is not defined at n=0 and this is left \[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]
my goodness this is exhausting!
\[a_1+\cancel{\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n}^{\text{not defined}}=0\]\[a_1=0\]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
and yes, it is extremely exhausting I think most people who do DE's whould agree, but they handle non-constnat coefficient problems well
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
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.
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
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.....
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
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
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
what I screwed up on earlier by calling it \[\frac{a_0}{2^n(n!)}\]
ok I'll fix it...let's see here...
hint: We are only talking about even numbers, the odd subscripts are all zero. All even numbers can be written as...?
\[a_n= \frac{a_0}{2^{2k}(2nk!)}\] If this is right.....
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
? k=2 n=2k=4 check a_4
oh I see what you mean
\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\] uhm? what do you want me to check? the fact that it is not zero?
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 :)
divide it by half because it seems like double the answer so \[a_{2k}=\frac{a_0}{2^{k}(k)!}\]
now is that my y_p or something?
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...
it's my "y" final answer
basically, yes it is the key to the final answer
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
\[y=\sum_{k=0}^\infty \frac{a_0}{2^kk!}x^{2k}\]
and just to make things crystal clear that we are in fact right...
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
with \(C=a_0\) of course
experimentX was right too, but I thought I would try to practice through explanation myself :)
Amazing! I owe you half of my paychecks when I begin working as civil engineer. Dude, you are awesome. Thanks for helping me.
patience is a virtue...you have an abundance of that...oh yeah, sorry I forgot you don't like compliments :P
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...
Thank you. I'm off to watch an episode of the Big Bang Theory. Is that silly? LOL
Makes more sense than doing more series solution DE's to me! I'm off to do something unproductive too, see ya!
gotta take it away from experimentX... which I'll do, he has enough sorry @experimentX ;)
\[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](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\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{2^kk!}\]
you guys have patience of a saint ... usually i don't write more than lines.