Finding a power series solution to a differential equation?

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SUMMARY

The discussion focuses on finding the power series solution to the differential equation \(y'' - xy' - y = 0\) centered at \(x_0 = 1\). The correct recurrence relation derived from the power series is \((n+2)a_{n+2} - a_{n+1} - a_n = 0\). The user initially struggled with the calculations but later identified an error in their manipulation of the series terms, particularly in the expression for \(xy'\). The final corrected form of the equation confirms the recurrence relation provided in the textbook.

PREREQUISITES
  • Understanding of power series representation, specifically \(y = \sum_{n=0}^\infty a_n(x-1)^n\)
  • Familiarity with differential equations, particularly second-order linear equations
  • Knowledge of recurrence relations and their application in series solutions
  • Proficiency in manipulating series and summation indices
NEXT STEPS
  • Study the derivation of power series solutions for different types of differential equations
  • Learn about the method of Frobenius for solving differential equations with regular singular points
  • Explore the implications of the recurrence relation in determining coefficients \(a_n\)
  • Investigate the convergence criteria for power series solutions in differential equations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations and series solutions, as well as anyone looking to deepen their understanding of recurrence relations in mathematical contexts.

Eclair_de_XII
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Homework Statement


"Find the recurrence relation in the power series solution for ##y''-xy'-y=0## centered about ##x_0=1##."

Homework Equations


##y=\sum_{n=0}^\infty a_nx^n##
Answer as given in book: ##(n+2)a_{n+2}-a_{n+1}-a_n=0##

The Attempt at a Solution


##y=\sum_{n=0}^\infty a_n(x-1)^n##
##xy'=x\sum_{n=0}^\infty na_n(x-1)^{n-1}=x\sum_{n=1}^\infty na_n(x-1)^{n-1}=x\sum_{n=0}^\infty (n+1)a_{n+1}n(x-1)^{n}##
##y''=\sum_{n=0}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=0}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=2}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=0}^\infty a_{n+2}(n+2)(n+1)(x-1)^{n}##

##y''-xy'-y=0=\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}-x(n+1)a_{n+1}-a_n](x-1)^n##

##(n+2)(n+1)a_{n+2}=x(n+1)a_{n+1}+a_n##

I don't know what I'm doing wrong. It's not matching the answer given in the back of the book.
 
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Never mind, I found what I did wrong.

##xy'=x\sum_{n=0}^\infty na_n(x-1)^{n-1}=[1+(x-1)]⋅\sum_{n=0}^\infty na_n(x-1)^{n-1}=\sum_{n=1}^\infty na_n(x-1)^{n-1}+\sum_{n=0}^\infty na_n(x-1)^{n}=\sum_{n=0}^\infty (n+1)a_{n+1}(x-1)^{n}+\sum_{n=1}^\infty na_n(x-1)^{n}##
 

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