seascorpion1
  • seascorpion1
Reduction of order to find a second Frobenius series solution. Question to follow as soon as I can type it in.....
Differential Equations
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
seascorpion1
  • seascorpion1
\[xy''-y=0\]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:\[n(n+1)b_{n+1}-b_n=-C(\frac{2n+1}{n!(n+1)!})\]
seascorpion1
  • seascorpion1
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?
seascorpion1
  • seascorpion1
@phi and @Directrix You've both helped me in the past, can you help me with this at all?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

seascorpion1
  • seascorpion1
@lochana is there any chance you could point me in the right direction with this problem? Thanks.
lochana
  • lochana
I am afraid. I haven't solved these kind of problems before. I need some time.:(
lochana
  • lochana
where does y1 come from? is it given?
lochana
  • lochana
https://en.wikipedia.org/wiki/Frobenius_method this might be useful
seascorpion1
  • seascorpion1
y1 is not given, I calculated it. The Wikipedia link didn't help as id didn't cover reduction of order.
seascorpion1
  • seascorpion1
@lochana
lochana
  • lochana
yes. hey, I need the original question. I am lost here:)
seascorpion1
  • seascorpion1
Thanks, I'll try to send it as an attachment later.
lochana
  • lochana
okay. @ganeshie8 may help you with this.
seascorpion1
  • seascorpion1
Thanks.
seascorpion1
  • seascorpion1
Original question is attached (finally). If anyone can help with the problems I listed in my earlier posts then that would be great, thanks.
seascorpion1
  • seascorpion1
@ganeshie8
anonymous
  • anonymous
Is \(H_n\) the \(n\)th harmonic number? As in \[H_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\cdots+\frac{1}{n}\]
seascorpion1
  • seascorpion1
Yes
anonymous
  • anonymous
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 \(b_0=1\).
seascorpion1
  • seascorpion1
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.
anonymous
  • anonymous
That was an unfortunate typo... But if you've figured it out, great!

Looking for something else?

Not the answer you are looking for? Search for more explanations.