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Math Help - Closed-form of a power series solution to an ODE

  1. #1
    Senior Member chella182's Avatar
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    Closed-form of a power series solution to an ODE

    Seek a power series solution about x=0 of the equation

    (1+x^2)y''+2xy'-2y=0

    as far as the term in x^6. Then find a closed-form of the solution (i.e. write the solution using simple functions, in this case arctan, rather than infinite series) by using

    \arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}+...+(-1)^n\frac{x^{2n+1}}{2n+1}+....

    It's the closed-form bit that's confusing me, 'cause my answer looks nothing like the arctan thing. The answer I got was

    y=A(1+x^2-\frac{1}{3}x^4+\frac{2}{5}x^6+...) where A is just a constant.



    Can anyone help?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chella182 View Post
    Seek a power series solution about x=0 of the equation

    (1+x^2)y''+2xy'-2y=0

    as far as the term in x^6. Then find a closed-form of the solution (i.e. write the solution using simple functions, in this case arctan, rather than infinite series) by using

    \arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}+...+(-1)^n\frac{x^{2n+1}}{2n+1}+....

    It's the closed-form bit that's confusing me, 'cause my answer looks nothing like the arctan thing. The answer I got was

    y=A(1+x^2-\frac{1}{3}x^4+\frac{2}{5}x^6+...) where A is just a constant.



    Can anyone help?
    Rewrite your series as:

    y=A+Ax(x-\frac{1}{3}x^3+\frac{2}{5}x^5+...)

    CB
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  3. #3
    Senior Member chella182's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Rewrite your series as:

    y=A+Ax(x-\frac{1}{3}x^3+\frac{2}{5}x^5+...)

    CB
    But that still doesn't look like arctan with the \frac{2}{5}... unless I'm missing something really obvious here...
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  4. #4
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    Quote Originally Posted by chella182 View Post
    But that still doesn't look like arctan with the \frac{2}{5}... unless I'm missing something really obvious here...
    Re-check your answer for the coefficient of x^5.
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  5. #5
    Senior Member chella182's Avatar
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    Quote Originally Posted by Opalg View Post
    Re-check your answer for the coefficient of x^5.
    I did do, the \frac{2}{5} fits with everything I've done. Obviously wrong then, I'll just leave this question 'cause I'm clearly not gonna get it right now haha.
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  6. #6
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    Quote Originally Posted by chella182 View Post
    I did do, the \frac{2}{5} fits with everything I've done. Obviously wrong then, I'll just leave this question 'cause I'm clearly not gonna get it right now haha.
    I can't comment on where you went wrong because you haven't shown the working. But when I looked for a power series solution y = \textstyle\sum a_nx^n I found that a_0 = a and a_1=b can be arbitrary. Then a_n=0 for all odd n\geqslant3, a_2=a, a_4 = -\tfrac13a, and a_6 = -\tfrac35a_4 = \tfrac15a.
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  7. #7
    Senior Member chella182's Avatar
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    Yeah well I got wrong last time I showed my working if I remember rightly 'cause this site's not meant for checking work or something, + it would take an age to type it out in latex with all of the sum signs etc.

    I got a_n=0 for all odd n, but then I got a_0=a_2 and then 12a_4+4a_2=0, so 12a_4+4a_0=0 so rearranging that I got that a_4=\frac{-a_0}{3}.

    I then got 30a_6+36a_4=0, and subbing in a_4=\frac{-a_0}{3} I got 30a_6-12a_0=0, getting that a_6=\frac{2}{5}a_0.



    Obviously it's just some little mistake somewhere, cheers though. I actually think I can see my mistake... according to me 4\times3=30 haha! Oh well, sorted now. Thanks again.
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