DE: Lower Bound for radius of convergence

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SUMMARY

The discussion centers on determining the radius of convergence for the differential equation Prb: (x^4 + 4*x^2 + 16)y" + 4(x - 1)y' + 6xy = 0. The polynomial P(x) = x^4 + 4*x^2 + 16 has roots at -1 ± 3^(1/2)*i and 1 ± 3^(1/2)*i, while Q(x) and R(x) have roots at 1 and 0, respectively. The radius of convergence for the power series solution centered at 0 is conclusively determined to be 2, as it is the shortest distance from 0 to the complex roots of P.

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BobMarly
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Prb:(x^4+4*x^2+16)y"+4(x-1)y'+6xy=0
P=(x^4+4*x^2+16) Q=4(x-1) R=6x
P=0 for - 1 - 3^(1/2)*i
1 - 3^(1/2)*i
- 1 + 3^(1/2)*i
1 + 3^(1/2)*i
Q=0 for 1
R=0 for 0
Do we ignore Q & R, plotting P, then find shortest distance which would equal 2?
 
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Basically, yes. If P were never 0, all solutions to the differential equation would be analytic. Because P is 0 at those four complex numbers, we can take the solution to be 0 away from them. The distance from 0 to those four points is 2 so the radius of convergence of a power series for the solution, centered at 0 wil have radius of convergence 2.
 

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