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goalie2012 Group Title

use the frobenius method to solve xy''-y'+2y=0. find index "r" and recurrence relation. compute the first 5 terms(a0-a4) using the recurrence relation for each solution and index r.

  • one year ago
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  1. goalie2012 Group Title
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    @oldrin.bataku do you know this method?

    • one year ago
  2. oldrin.bataku Group Title
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    Let me see...

    • one year ago
  3. primeralph Group Title
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    @goalie2012 Are you very familiar with summations?

    • one year ago
  4. goalie2012 Group Title
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    I know that's how this problem should be solved, and I more or less know them, but I still get messed up on some of them. Like this one...

    • one year ago
  5. primeralph Group Title
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    Cool, @oldrin.bataku should be able to explain it to you. I skip a lot of steps when explaining stuff like that.

    • one year ago
  6. oldrin.bataku Group Title
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    $$xy''-y'+2y=0$$Observe we may rewrite this in standard form:$$y''-\frac1xy'+\frac2xy=0$$which has a very apparent singular point at \(x=0\). With a little further attention we can see it's a regular singular point.

    • one year ago
  7. oldrin.bataku Group Title
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    The method of Frobenius assumes, then, a solution of the form \(y=\sum\limits_{n=0}^\infty a_nx^{r+n}\) (note our expansion about the singular point).

    • one year ago
  8. goalie2012 Group Title
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    @oldrin.bataku I've tried what I can think of, but I'm not really sure where to go from here. I'm still getting used to and figuring these things out.

    • one year ago
  9. goalie2012 Group Title
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    @oldrin.bataku @primeralph how do I find/get rid of the r value?

    • one year ago
  10. primeralph Group Title
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    Find the constant number that shifts the power from n to n+r.

    • one year ago
  11. goalie2012 Group Title
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    I'm trying to work it through with my book, but it's not helping. It says to find an indicial equation. still not really sure what to do...

    • one year ago
  12. goalie2012 Group Title
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    @oldrin.bataku I'm so lost. Tried to so something and got a mess. you explained the others very well, could you explain this one to if you're still here somewhere?

    • one year ago
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