ODE - Power Series Convergence

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SUMMARY

The discussion centers on solving the differential equation (1-4x²)y'' + 34xy' - 70y = 0 using power series methods. The recurrence relation derived is a_{n+2} = (2(-7+n)(-5+2n)/((n+1)(n+2)))a_n. The first part of the problem was successfully solved for y1 with a_0 = 0 and a_1 = 1, resulting in a polynomial of degree 7. The second part involves finding the radius of convergence for y2, where a_0 = 1 and a_1 = 0, which presents challenges, particularly in applying the ratio test correctly.

PREREQUISITES
  • Understanding of power series and their convergence
  • Familiarity with differential equations, specifically second-order linear equations
  • Knowledge of recurrence relations and their applications
  • Proficiency in applying the ratio test for series convergence
NEXT STEPS
  • Study the application of the ratio test in detail, focusing on series derived from recurrence relations
  • Explore the concept of radius of convergence for power series
  • Investigate the methods for solving second-order linear differential equations using power series
  • Review examples of finding patterns in coefficients of power series solutions
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Students and educators in mathematics, particularly those focusing on differential equations and series solutions, as well as anyone seeking to deepen their understanding of power series convergence.

thejinx0r
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Homework Statement



Solve
[tex](1-4x^2)y''+34x\cdot y'-70y=0[/tex]

Homework Equations



Basically, I found the recurrence relationship to be:
[tex]a_{n+2}=\frac{2 (-7 + n) (-5 + 2 n)}{(n+1)(n+2)}a_n}[/tex]



Now, I solve for y1 where y1 had a_0=0 and a_1 = 1. It is a simple polynomial of degree 7.
That was the first part.


The second part wants me to find the radius of convergence of y2, where y2 has a_0=1 and a_1 = 0.
In this case, all the odd coefficients are equal to 0. But I am having a hard time trying to find the radius of convergence. I actually don't have a clue here.
I tried doing the ratio test directly on the recurrence term, but led me to 1/x^2, but that didn't work, and I think I did wrong to.

I tried finding the pattern and trying to find what the series actually was, but couldn't. :S
So, now I'm stuck.

(P.S. For the patternn, I got a_n = [tex]\frac{2*(-7)(-5)(..)(2n-9)(-5)(-1)(3)(7)(4n-9)}{(2n)!}[/tex]
But it appears to be invalid when I do the ratio test on it according to webworks.
 
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thejinx0r said:

Homework Statement



Solve
[tex](1-4x^2)y''+34x\cdot y'-70y=0[/tex]

Homework Equations



Basically, I found the recurrence relationship to be:
[tex]a_{n+2}=\frac{2 (-7 + n) (-5 + 2 n)}{(n+1)(n+2)}a_n}[/tex]



Now, I solve for y1 where y1 had a_0=0 and a_1 = 1. It is a simple polynomial of degree 7.
That was the first part.


The second part wants me to find the radius of convergence of y2, where y2 has a_0=1 and a_1 = 0.
In this case, all the odd coefficients are equal to 0. But I am having a hard time trying to find the radius of convergence. I actually don't have a clue here.
I tried doing the ratio test directly on the recurrence term, but led me to 1/x^2, but that didn't work, and I think I did wrong to.
No, the ratio test will work fine. If it led you to 1/x^2, then you have it upside down! Look at [tex]\frac{a_{2(n+1)}x^{2(n+1)}}{a_{2n}x^{2n}}= \frac{a_{2n+2}}{a_{2n}}x^2[/itex]. For what values of x is that less than 1?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I tried finding the pattern and trying to find what the series actually was, but couldn't. :S<br /> So, now I'm stuck.<br /> <br /> (P.S. For the patternn, I got a_n = [tex]\frac{2*(-7)(-5)(..)(2n-9)(-5)(-1)(3)(7)(4n-9)}{(2n)!}[/tex]<br /> But it appears to be invalid when I do the ratio test on it according to webworks. </div> </div> </blockquote>[/tex]
 

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