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

  1. Nov 17, 2005 #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?
     
  2. jcsd
  3. Nov 17, 2005 #2

    Physics Monkey

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    Try multiplying the whole equation by [tex] 1 + x^2 [/tex].
     
  4. Nov 17, 2005 #3

    saltydog

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    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?
     
  5. Nov 17, 2005 #4
    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.
     
  6. Nov 17, 2005 #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?
     
  7. Nov 17, 2005 #6
    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.
     
  8. Nov 17, 2005 #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: Nov 17, 2005
  9. Nov 17, 2005 #8
    yes, now find your recurence relation and your done.
     
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