Solutions of D.E - Bessel Function

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SUMMARY

The discussion focuses on verifying that the Bessel function of index 0 is a solution to the differential equation xy" + y' + xy = 0. The Bessel function is expressed as a series from n=0 to infinity, with specific derivatives calculated for y' and y". The user encounters difficulties in simplifying the resulting series to zero when substituting these derivatives back into the differential equation. The thread also references additional insights from a related discussion on Physics Forums.

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  • Understanding of Bessel functions, specifically Bessel function of index 0.
  • Familiarity with differential equations and their solutions.
  • Knowledge of series expansions and convergence.
  • Proficiency in calculus, particularly in computing derivatives of series.
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  • Study the properties and applications of Bessel functions in mathematical physics.
  • Learn about series solutions to differential equations, focusing on power series methods.
  • Explore the concept of convergence in infinite series and its implications for solutions.
  • Review advanced calculus techniques for simplifying complex series expressions.
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Mathematicians, physicists, and engineering students who are working with differential equations and Bessel functions, particularly those seeking to understand series solutions and their applications in physical problems.

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Hello, I hope someone can show me where I got stuck/wrong.

Verify that the Bessel function of index 0 is a solution to the differential equation xy" + y' + xy = 0.

Note that my "<= 1" DOES NOT mean less than or equal to 1 but an arrow pointing to the left... it is said to be "equation 1".

It is given that the bessel function of index 0 is:

y = (from n= 0 to infinity){(-1)^n x^(2n) / [(n!)^2 2^(2n)] } <=0

getting the 1st derivative (w/ respect to x) is

y' = (from n= 1 to infinity){(-1)^n x^(2n-1) / [(n!)(n-1)! 2^(2n - 1)] } <= 1

Can I say this?

y' = (from n= 0 to infinity){(-1)^(n+1) x^(2n+1) / [(n!)(n+1)! 2^(2n + 1)] } <= 2

going back to <= 1,

y" = (from n= 1 to infinity){(-1)^n (2n -1)x^(2n-2) / [(n!)(n-1)! 2^(2n - 1)] } <= 3 or

y" = (from n= 0 to infinity){(-1)^(n+1) (2n + 1)x^(2n) / [(n!)(n+1)! 2^(2n + 1)] } <= 4


Substituting y", y' and y with <=4, <=2 and <=0 respectively to the given differential equation... but my can't simplify it to 0. There are terms in my series that cause me not to cancel them out to become zero.. where are my mistakes?
 
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Thank you very much.. I understood about the solutions...
 

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