Series solution to DE about ordinary point

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SUMMARY

The discussion focuses on finding two power series solutions for the differential equation (DE) (x+2)y'' + xy' - y = 0 around the ordinary point x = 0. The user substituted y = Ʃ0inf cnxn into the DE, leading to two key equations: 2c2 - c0 = 0 and a recurrence relation involving three coefficients. The resolution of c0 allows for the determination of cn for n ≥ 2, while c1 remains arbitrary, representing the second solution. This method effectively illustrates the process of deriving power series solutions for DEs.

PREREQUISITES
  • Understanding of power series expansions
  • Familiarity with differential equations, specifically second-order linear DEs
  • Knowledge of recurrence relations in series solutions
  • Basic calculus, including differentiation and series manipulation
NEXT STEPS
  • Study the method of Frobenius for solving differential equations
  • Learn about the convergence of power series solutions
  • Explore examples of second-order linear differential equations with variable coefficients
  • Investigate the role of arbitrary constants in the general solution of DEs
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as researchers seeking to understand power series methods in solving DEs.

lordsurya08
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Homework Statement



Find two power series solutions of the DE

(x+2)y'' + xy' - y = 0

about the ordinary point x = 0 . Include at least first four nonzero terms for each of the solutions.

2. The attempt at a solution

I distributed the y'' term and substituted
y = Ʃ0inf cnxn
and its derivatives into the DE. I equated it to 0 and got two equations:

2c2 - c0 = 0

xn(cn+1*n(n+1) + cn+2*(n+1)(n+2)+ cn*(n-1)) = 0

The weird thing is that the second equation (the recurrence relationship) has three c terms in it, although the examples shown have two. How do I get c2 and c0? After that happens should I simply solve for c1 using the recurrence relationship with n = 0?
 
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You seem to have dropped the factor of 2 from the 2y'' term, so your two equations are slightly wrong.

Once ##c_0## is set, you can determine ##c_n## for ##n \ge 2## through the recurrence relation. That's your first solution.

The coefficient ##c_1## is still arbitrary. That corresponds to your second solution.
 

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