Find general solution from recurrence relation, done most of work

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Homework Help Overview

The problem involves solving a second-order linear homogeneous differential equation given by \(2xy'' + y' + xy = 0\) with an initial condition at \(x_0 = 0\). Participants are tasked with finding the indicial equation, recurrence relation, and series solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive the recurrence relation and series solution but expresses uncertainty about how to proceed from the recurrence relation to obtain the series solutions \(y_1\) and \(y_2\). They note a potential pattern in the coefficients and question how to formalize it.
  • Some participants suggest examining the coefficients more explicitly to identify patterns, particularly regarding odd and even terms in the series.
  • Others propose that the coefficients for odd terms may be zero, while even terms could relate to factorial expressions.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the recurrence relation and its implications for the series solutions. Some guidance has been provided regarding the explicit calculation of terms, but no consensus has been reached on the final forms of the solutions.

Contextual Notes

Participants are working under the constraints of deriving a series solution from a recurrence relation, with specific values for \(a_0\) and \(a_1\) being discussed. The initial conditions and the nature of the differential equation are central to the exploration.

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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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