Question

Alright, dealing with Power Series Solutions of ODE's, first problem on the subject. Posted below in a moment.

Answer 1

\[\text{Find a power series solution of} \ y''+x^2y=0 \ \text{expanded about} \ x_{0}=0.\]

Answer 2

The first step, if I remember correctly, is to transform every term above into summations/Epsilon terms.

Answer 3

"Epsilon terms"?

Answer 4

(Sorry, I meant sigma; The capitol sigma term things that are summation signs. Either way, help would be appreciated.)

Answer 5

\[\begin{align*}y&=\sum_{k=0}^\infty a_kx^k\\\\
y'&=\sum_{k=0}^\infty ka_kx^{k-1}=\sum_{k=1}^\infty ka_kx^{k-1}\\\\
y''&=\sum_{k=1}^\infty k(k-1)a_kx^{k-2}=\sum_{k=2}^\infty k(k-1)a_kx^{k-2}
\end{align*}\]
Substitute these into your equation:
\[\begin{align*}0&=\sum_{k=2}^\infty k(k-1)a_kx^{k-2}+x^2\sum_{k=0}^\infty a_kx^k\\
&\text{simply subbing}\\\\

&=\sum_{k=2}^\infty k(k-1)a_kx^{k-2}+\sum_{k=0}^\infty a_kx^{k+2}\\
&\text{distributing the }x^2\\\\

&=\sum_{k+4=2}^\infty (k+4)(k+4-1)a_{k+4}x^{k+4-2}+\sum_{k=0}^\infty a_kx^{k+2}\\
&\text{shifting the index so that powers match up}\\\\

&=\sum_{k=-2}^\infty (k+4)(k+3)a_{k+4}x^{k+2}+\sum_{k=0}^\infty a_kx^{k+2}\\\\

&=2a_2+6a_3x+\sum_{k=0}^\infty \big[(k+4)(k+3)a_{k+4}+a_k\big]x^{k+2}\\
&\text{pulling out the first two terms from the first series}
\end{align*}\]
From here you have that \(a_2=a_3=0\).

Answer 6

Also, the equation is in standard form at the beginning and doesn't appear to have any singular points and thus the only points we're dealing with are ordinary, is this correct?

(I'm trying to rummage through the theory to make sure I get it)

Answer 7

Okay, so if I understand right, I can equate every coefficient of x to zero, so then I'll get what resembled maybe a quadratic or something set to zero expressed in k that I have to solve for.

Answer 8

If I understand correctly, the recurrence relation is:



Is this right? @siths\[a_{k+4}=-\frac{a_k}{(k+3)(k+4)}\]

Answer 9

^I would've stopped right there and left it as is, but the answer for the recurrence relation is expressed as the above term with the indices (and all terms with k) minus an additional two, so a_k+2 on the LHS, etc. Why, or how, should I have known, to shift like that?

Answer 10

This is really cool stuff. I've actually never had much exposure to solving ODEs with the power series method, but now that I see this I'm probably going to read a bit about it because it seems interesting.

Answer 11

Yeah, I'm not too worried about figuring this out because at most one problem like this will be on the final and I'll be able to get partial credit up until this point, but I'd like to find out nonetheless. Gonna tag some other folks.

Answer 12

@ganeshie8 , could you help me out on this?

Answer 13

I'm thinking the shifting near the end maybe has to do with the fact that, in the final series term, you have x^(k+2)?

Answer 14

@Zarkon , are you good with Power Series Solutions to ODE's?

Answer 15

@sourwing , are you good with solving ODE's with Power Series Solutions?

Answer 16

@phi

Answer 17

I just solved this using some instructions I found and I got what you were saying about the indices being less.

Answer 18

So, in order for us to "solve" it, our x has to be shifted to solely the kth power (and all other terms with k as appropriate), in general?

Answer 19

I'm unsure is Sith's is equivalent or not

Answer 20

Yeah I assumed a solution of the form:\[y(x)=\sum_{n=0}^{\infty} a _{0}x^{n}\]

Answer 21

I then differentiated twice and plugged into your equation y'' - x^2y = 0
\[\sum_{n=2}^{\infty}a_{n}n(n-1)x^{n-2}+ x^{2} \sum_{n=0}^{\infty}a_{n}x^{n}=0\]

Answer 22

i meant y'' + x^2y = 0, sorry. I typed the series right

Answer 23

But yeah, then you carried through Sith's process of solving it and then shifted it at the end? (I'll wait and see, heh.)

Answer 24

Then I change indices to try and combine the sums. I also move the x^2 into the second sum. I then use the following to match up things: m = n -2\[\sum_{m=0}^{\infty}a_{m+2}(m+2)(m+1)x^{m}+ \sum_{n=0}^{\infty}a_{n}x^{n+2} =0\]\[\sum_{m=0}^{\infty}a_{m+2}(m+2)(m+1)x^{m}+ \sum_{n=2}^{\infty}a_{n}x^{n} =0\]

Answer 25

And we're allowed to do that because m is a dummy variable, right?

(I don't have a real, effective understanding of the idea of, "It's a dummy variable", but I mechanically think I get how to make it work in the context of Power Series Solutions.)

Answer 26

This gives me:\[\sum_{k=0}^{\infty}[a_{k+2}(k+2)(k+1)+a_{k}]x^{k}=0\]
Which gives the relation:\[a_{k+2}=-\frac{a_{k}}{(k+2)(k+1)}\]

Answer 27

Yeah, I'm just renaming the index as something equivalent, but something that will allow us to have x^k form. whether x^n or x^m it's still x to something.

Answer 28

Alright, cool. The best way right now I can analogize or think about dummy variables in this context is that it doesn't matter what their actual value is in and of itself, but how its value relates to other things, sort of like how Gravitational Potential Energy can arbitrarily selected to have some point have some convenient value, but what matters is how two separate points behave in relation to one another

Answer 29

Alright, cool. I'm going to lie down for 10 minutes, but in case anybody is interested for their own sake, here is the given answer: http://i.imgur.com/Hodw2Bt.png

Answer 30

That is what I got
If y(0)=1, y'(0)=0

\[ y(x)= 1-\frac{x^4}{12}+\frac{x^8}{672}-\frac{x^{12}}{88704}+\frac{x^{16}}{21288
960}-\frac{x^{20}}{8089804800}+\cdots\]

If y(0)=0, y'(0)=1
\[y(x)=x-\frac{x^5}{20}+\frac{x^9}{1440}-\frac{x^{13}}{224640}+
\cdots\]

Answer 31

The general solution will be a linear combination of both