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Solve the ODE y'' + (3x)/(1+x^2)y' + 1/(1+x^2)y = 0

  • #1
I need to determine the fundamental set of solutions of

[tex]y''+\frac{3x}{1+x^2}y'+\frac{1}{1+x^2}y=0[/tex]

in form of a power series centered around [tex]x_0=0[/tex].

When I expanded the y's in power series, I am unable to bring the 1/1+x^2 inside the infinite sum. Can anyone help?
 

Answers and Replies

  • #2
Physics Monkey
Science Advisor
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Try multiplying the whole equation by [tex] 1 + x^2 [/tex].
 
  • #3
saltydog
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Treadstone 71 said:
I need to determine the fundamental set of solutions of
[tex]y''+\frac{3x}{1+x^2}y'+\frac{1}{1+x^2}y=0[/tex]
in form of a power series centered around [tex]x_0=0[/tex].
When I expanded the y's in power series, I am unable to bring the 1/1+x^2 inside the infinite sum. Can anyone help?
Multiply throughout by [itex](1+x^2)[/itex] first to get:

[tex](1+x^2)y^{''}+3xy^{'}+y=0[/tex]

So wouldn't that first term just be:

[tex]\sum n(n-1)a_n x^{n-2}+\sum n(n-1)a_nx^n[/tex]

right?
 
  • #4
60
0
There is much to this problem after you get passed your issue, however; I will restrain my response to your question. First off, multiply everything by 1+x^2. Now after subsituting your power series expression for y, y', and y'' you will be left with only 2 powers of x: x^n-2 and x^n. I imagine your question begins here although now in different form. You must make a change of index on the summation containing x^n-2, try m=n-2. After this substitution you will have an expression with a sum with x^m and one with x^n. Since "m" is a dummy index let m=n. It seems circular but it is correct and required. Now you will have an expression in which the only powers of x that appear are x^n. Grouping the coefficients and setting them equal to zero is your next step which I will let you do on your own. It is interesting to note that in order to do this problem you are switching the "shift" of the powers of x to the coefficients an. Not only is this helpful, but it is mandatory when solving the original DE since you must have different an's from which to find your recurrence relations.
 
  • #5
If I make a substitution m=n-2, at some point I'll have a term in the form of:

[tex]\sum_{n=0}(n+2)(n+1)a_{n+2}x^{n+2}[/tex]

Are you saying that at this point I can reset the index of x to n and leave a_n the way it is?
 
  • #6
60
0
That is not what you will have. The coefficients look correct but its not x^(n+2) it is x^n. Recall: you had x^(n-2), you let m=n-2, therefore
x^((m+2)-2)=x^m with the coefficients you have. Then subsititute back and let m=n.
 
  • #7
Ok, got it, I think:

[tex]\sum_{n=0}^{\infty}x^n[a_{n+2}(n+2)(n+1)+a_n[n(n-1)+3n+1]][/tex]

This implies that

[tex]a_{n+2}(n+2)(n+1)+a_n(n+1)^2=0[/tex]
 
Last edited:
  • #8
60
0
yes, now find your recurence relation and your done.
 

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