Find general solution from recurrence relation, done most of work

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SUMMARY

The discussion focuses on solving the differential equation \(2xy'' + y' + xy = 0\) using recurrence relations to find series solutions \(y_1\) and \(y_2\). The indicial equation derived is \(r + 2r^2 - 2 = 0\), yielding roots \(r = 0\) and \(r = \frac{1}{2}\). The recurrence relation is established as \(a_n = -\frac{a_{n-2}}{(r+n)(r+n-1)}\), leading to a pattern where \(a_{odd} = 0\) and \(a_{even} = \frac{1}{n!}\). The user successfully identifies the series coefficients and seeks clarification on transitioning from \(r = 0\) to \(r = \frac{1}{2}\) for the second solution.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with series solutions and recurrence relations.
  • Knowledge of indicial equations and their significance in solving differential equations.
  • Basic proficiency in manipulating factorials and series expansions.
NEXT STEPS
  • Study the derivation of series solutions for different types of differential equations.
  • Learn about the Frobenius method for solving linear differential equations around regular singular points.
  • Explore the implications of changing roots in indicial equations on series solutions.
  • Investigate the convergence of series solutions and their behavior near singular points.
USEFUL FOR

Students and researchers in mathematics or engineering, particularly those focusing on differential equations and series solutions. This discussion is beneficial for anyone looking to deepen their understanding of recurrence relations in the context of differential equations.

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Homework Statement


2xy" + y' + xy =0 xo=0
find the indicial equation, recurrence relation and series solution.


Homework Equations





The Attempt at a Solution


I've done most of the work but i don't know how to use a recurrence relation to obtain a y1 and y2 series solution.
this is what I've done:

xy= sum (n=2 to infinity) an-2 x^(r+n-1)
y'= sum(n=1 to infinity) (r+n) an x^(r+n-1)
2xy''= 2*sum (n=2 to infinity) (r+n) (r+n-1) an x^(r+n-1)

indicial equation:
rao + 2r(r-1) ao=0
r+ 2r^2-2 = 0
r= 0 , r=1/2

{(r+n) (r+n-1) an + (r+n) an + an-2}=0
an= - an-2/ (r+n)(r+n-1)

r=0:
ao=1 a1=0
a2= a0/2*1
a3= a1/ 3*2
a4= a2/4*3
a5= a3/5*4

what kind of pattern could i get out of this? the denominator looks like an n! but not exactly. Where can i go from here to get a y1 ?
 
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It seems to be that it should be easy to show, unless I'm mistaken, that
a_{odd} = 0
a_{even} = 1/n!
 
itunescape said:

Homework Statement


2xy" + y' + xy =0 xo=0
find the indicial equation, recurrence relation and series solution.


Homework Equations





The Attempt at a Solution


I've done most of the work but i don't know how to use a recurrence relation to obtain a y1 and y2 series solution.
this is what I've done:

xy= sum (n=2 to infinity) an-2 x^(r+n-1)
y'= sum(n=1 to infinity) (r+n) an x^(r+n-1)
2xy''= 2*sum (n=2 to infinity) (r+n) (r+n-1) an x^(r+n-1)

indicial equation:
rao + 2r(r-1) ao=0
r+ 2r^2-2 = 0
r= 0 , r=1/2

{(r+n) (r+n-1) an + (r+n) an + an-2}=0
an= - an-2/ (r+n)(r+n-1)

r=0:
ao=1 a1=0
a2= a0/2*1
a3= a1/ 3*2
a4= a2/4*3
a5= a3/5*4

what kind of pattern could i get out of this? the denominator looks like an n! but not exactly. Where can i go from here to get a y1 ?
The first thing you should have done is write out a few terms explicitely.

If a0= 1 then a2= a0/2*1= 1/2. If a1= 0, a3= a1/3*2= 0. a4= a2/(4*3)= (1/2)1/(3*4))= 1/(2*3*4), a5= a3/(5*4)= 0/(5*4)= 0, a6= a4/(6*5)= [1/(2*3*4)][1/(6*5)]= 1/(2*3*4*5*6).

Do you see now?
 
yes I do, thank you very much. So to get y2 all i have to do is change from r=0 to r=1/2 and keep ao=1 and a1=0 right?
 
Yes.
 

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