Question

Reduction of order to find a second Frobenius series solution. Question to follow as soon as I can type it in.....

Answer 1

\[xy''-y=0\]and \[y_1=\sum_{n=1}^{\infty}\frac{x^n}{n!(n-1)!}\]so via reduction of order:\[y_2=Cy_1\ln(x)+\sum_{n=0}^{\infty}b_nx^n\]I've substituted y2 into the ODE and got the reccurence relation:\[n(n+1)b_{n+1}-b_n=-C(\frac{2n+1}{n!(n+1)!})\]

Answer 2

however I can't get the right substitution in the form bn=f(n)*cn (where f(n) is a term involving n and possibly C) to simplify the recurrence relation and then solve for y2. I've tried multiple substitutions and they all end up dividing by 0 meaning that my recurrence relation can not be in terms of b0 or c0. Can someone please explain the algorithm for identifying the substitution that simplifies the recurrence relation?

Answer 3

@phi and @Directrix You've both helped me in the past, can you help me with this at all?

Answer 4

@lochana is there any chance you could point me in the right direction with this problem? Thanks.

Answer 5

I am afraid. I haven't solved these kind of problems before. I need some time.:(

Answer 6

where does y1 come from? is it given?

Answer 7

https://en.wikipedia.org/wiki/Frobenius_method
this might be useful

Answer 8

y1 is not given, I calculated it. The Wikipedia link didn't help as id didn't cover reduction of order.

Answer 9

@lochana

Answer 10

yes. hey, I need the original question. I am lost here:)

Answer 11

Thanks, I'll try to send it as an attachment later.

Answer 12

okay. @ganeshie8 may help you with this.

Answer 13

Thanks.

Answer 14

Original question is attached (finally). If anyone can help with the problems I listed in my earlier posts then that would be great, thanks.

Answer 15

@ganeshie8

Answer 16

Is \(H_n\) the \(n\)th harmonic number? As in
\[H_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\cdots+\frac{1}{n}\]

Answer 17

Yes

Answer 18

Just a quick comment for starters (I don't have enough time at the moment to look into this in much detail, but I'll come back to this question later)...\[\begin{align*}H_n-H_{n-1}&=\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}+\frac{1}{n}\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}\right)\\[1ex]
&=\frac{1}{n}\end{align*}\]so you should expect the solution to the second recurrence to be given by
\[b_n=\frac{1}{n(n!)(n-1)!}\]with \(b_0=1\).

Answer 19

Thanks for getting back to me. Did you mean Hn+Hn-1 in your post above? I think I have now managed to work my way through it to the correct solution. Thanks again for your time on this. However, I do now have a new differential equation question which I'm stuck with. I've posted it in general maths (called 'ODE question to follow') and it relates to simplifying a power series substitution into an ODE. Any help would be brilliant, thanks.

Answer 20

That was an unfortunate typo... But if you've figured it out, great!