Question

Find the first three nonzero terms in each of TWO solutions (not multiples of each other) about x = 0

Answer 1

\[xy''+2xy'+6e^{x}y=0\]
That silly e^x in there makes me unsure of how to proceed. I know:
\[e^{x} = \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\], but how can you get a proper recurrence relation with that getting in the way?

Answer 2

These always take a lot of time. You just need help with finding the recurrence relation?

Answer 3

e^x will really not cause problems fir us here since we will only need terms from it up to the power of x.

Answer 4

For the most part, yeah. I think if I get the recurrence relation I can manage the rest of the problem.

Answer 5

Right so, x = 0 is definitely a regular singular point. Thus, we have \[x^2y'' + 2x^2y' + 6xe^xy = 0\] and \[y = \sum_{n = 0}^{}a_nx^{n+r}\]

Answer 6

\[y = \sum_{n = 0}^{}a_nx^{n+r}\]
\[y' = \sum_{n = 0}^{}(n+r)a_nx^{n+r-1}\]
\[y'' = \sum_{n = 0}^{}(n+r)(n+r-1)a_nx^{n+r-2}\]

Answer 7

Plug into the DE, and we have \[x^2\sum_{n = 0}^{}(n+r)(n+r-1)a_nx^{n+r-2} + 2x^2\sum_{n = 0}^{}(n+r)a_nx^{n+r-1} + 6xe^x\sum_{n = 0}^{}a_nx^{n+r} = 0\]

Answer 8

Reducing down, we have \[\sum_{n = 0}^{}(n+r)(n+r-1)a_nx^{n+r} + \sum_{n = 0}^{}2(n+r)a_nx^{n+r+1} + \sum_{n = 0}^{}\sum_{k=0}^{n}\frac{6a_{n-k}}{k!}x^{n+r+1} = 0\]

Answer 9

What are you doing to get the double summation? I haven't seen that sort of a transformation before when doing these.

Answer 10

Well, I'm not entirely certain, but I believe that by having a_{n - k} , we maintain the proper coefficients for each power of x. This is the way my professor did it, just coincidence that we had the same problem. You can swap summations, that's a theorem somewhere.

Answer 11

I see. Yeah, this wasn't in the section introducing these kind of problems. I'm able to do another technique the book mentions to get one solution, but a second solution always comes out wrong or Im unable to do the recurrence relation or any of the simplification for it properly.

So how would you finish reducing that recurrence relation down to where we could set it to zero and start solving for the an's?

Answer 12

\[\sum_{n = 0}^{}(n+r)(n+r-1)a_nx^{n+r} + \sum_{n = 1}^{}2(n+r-1)a_{n-1}x^{n+r} + \sum_{n = 1}^{}\sum_{k=0}^{n-1}\frac{6a_{n-k-1}}{k!}x^{n+r} = 0\]

Answer 13

= 0

Answer 14

\[r(r-1)a_0 +\sum_{n = 1}^{}[(n+r)(n+r-1)a_n + 2(n+r-1)a_{n-1}+ \sum_{k=0}^{n-1}\frac{6a_{n-k-1}}{k!}]x^{n + r} = 0\]

Answer 15

Clearly, r = 0 or r = 1

Answer 16

Let's chose a_0 to be 1, then we can write out our terms.

Answer 17

\[a_0 = 1\]\[a_1 = \frac{-2(1)a_{1-1} - \sum_{k = 0}^{1-1}\frac{6a_{1-1-k}}{k!}}{(1+1)(1)} = -4\]

Answer 18

Jeeze, not to handy with my latex yet. I have to get off now, but I think you should be able to derive the relation from here. Just find a_2 and you'll see what a_n should be. I'll get back on in a bit to see if you still need help.

Answer 19

No, that's awesome, thank you ^_^ I should be fine with that first solution. A second one I may or may not be able to derive, we'll see. Thanks :)

Answer 20

Oh, and do you happen to know the name of the theorem or formula that allowed you to make that double summation?

Answer 21

Its something fundamental and doesn't really have a name, as far as I know. You can prove it though. As an example, consider\[\sum_{i = 0}^{n}\sum_{k = 0}^{n}x_{ik} = \sum_{i = 0}^{n}x_{i0}+x_{i1}+x_{i2}+ ... +x_{i n} =\] \[ (x_{00}+x_{01}+x_{02}+ ... +x_{0 n}) + (x_{10}+x_{11}+x_{12}+ ... +x_{1 n}) + ... + (x_{n0}+x_{n1}+x_{n2}+ ... +x_{n n})= \]
\[ (x_{00}+x_{10}+x_{20}+ ... +x_{n 0}) + (x_{01}+x_{11}+x_{21}+ ... +x_{n 1}) +...+(x_{0n}+x_{1n}+x_{2n}+ ... +x_{n n}) =\]
\[ \sum_{k = 0}^nx_{0k}+x_{1k}+x_{2k}+ ... +x_{n k} =\sum_{k = 0}^{n}\sum_{i = 0}^{n}x_{ik} \]

Answer 22

I suppose in this problem, we don't really apply the idea, just carry over e^x into the sum prior to writing it as a sum.

Answer 23

I'll have to look into that then. From what I saw, it looked like they were calling it a convolution for power series. Would you happen to know how to come up with a second solution then @RBauer4?
And sorry for the delay in response >.<

Answer 24

So, given the info above, Im still unsure how to come up with a 2nd solution. The supposed second solution is of the form:
\[y_{2}(x) = \alpha y_{1}(x)\ln|x| + \sum_{n=0}^{\infty}c_{n}x^{r_{2}+n}\]. In this case r_2 = 0, so I just need
\[y_{2}(x) = \alpha y_{1}(x)\ln|x| + \sum_{n=0}^{\infty}c_{n}x^{n}\]

The only time Ive been successful in the past is when I was able to get a general power series solution for y_1. In this case, I only can seemingly get a term-by-term answer, so I don't know how to get the 2nd solution, especially with the e^x in there again. Anyone have any ideas?