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Power Series Solution to Linear ODE

  1. Nov 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Let y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞) be a power series solution of

    (x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0

    Note that x=0 is an ordinary point.

    2. Relevant equations
    y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞)
    y'(x)=[itex]\sum[/itex](kckxk-1) (k=1 to ∞)
    y''(x)=[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞)

    3. The attempt at a solution

    (x2-1)[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞) +[itex]\sum[/itex](kckxk+2) (k=1 to ∞)+[itex]\sum[/itex]ckxk (k=0 to ∞) -2x=0

    Making all the series start with the x3 term
    2C2x0+[itex]\sum[/itex](k(k-1))ckxk (k=3 to ∞)- 2C2x0-6C3x1-12C4x2-[itex]\sum[/itex](k(k-1))ckxk-2 (k=5 to ∞) + [itex]\sum[/itex](kckxk+2) (k=1 to ∞)+C0x0+C1x1+C2x2+[itex]\sum[/itex]ckxk (k=3 to ∞)


    2C2x0+[itex]\sum[/itex](n(n-1))cnxn (n=3 to ∞)- 2C2x0-6C3x1-12C4x2-[itex]\sum[/itex](n+2)(n+1))cn+2xn (n=3 to ∞) + [itex]\sum[/itex]((n-2)cn-2xn) (n=3 to ∞)+C0x0+C1x1+C2x2+[itex]\sum[/itex]cnxn (n=3 to ∞)



    2C2x0-2C2x0-6C3x1-12C4x2+C0x0+C1x1+C2x2=0
    c0=0
    c3={c1-2}/{6}
    c4={c2}/{12}

    cn+2={n(n-1)cn+(n-2)cn-2+cn}/{(n+2)(n-1)} n=3,4,5...


    I started with c5={(7c3+c1)}/{10}
    but both of my c1 and c3 terms are tied up in the same equation, not really sure were to go from here
     
    Last edited: Nov 29, 2013
  2. jcsd
  3. Nov 29, 2013 #2

    vela

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    The y' term should be ##x^3 \sum_{k=1}^\infty kc_k x^{k-1} = \sum_{k=1}^\infty kc_k x^{k+2}##.

     
  4. Nov 29, 2013 #3
    Sorry, it would seem I made an error on the first line, but after that I think this term has been corrected
     
  5. Nov 29, 2013 #4

    vela

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    It seems you made some more algebra mistakes. After I fixed all of my mistakes, I had
    \begin{align*}
    -2c_2 + c_0 &= 0 \\
    -6c_3 + c_1 &= 2 \\
    2c_2 - 12 c_4 + c_2 &= 0
    \end{align*} and
    $$[n(n-1)+1]c_n - (n+2)(n+1)c_{n+2} + (n-2)c_{n-2} = 0$$ for the general recurrence relation.
     
  6. Nov 29, 2013 #5
    oh wow! so many mistakes! Thank you

    So for the recurrence relation, I want it in terms of cn+2

    $$ c_{n+2} = \frac{[n(n-1)+1]c_n + (n-2)c_{n-2}}{(n+2)(n+1)} = 0 $$

    and then I solve for n=3 giving me

    c5= $$ \frac{7c_3 c_1}{20} $$ ?
     
  7. Nov 29, 2013 #6

    vela

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    The =0 shouldn't be there on the end.

    Almost — perhaps just a typo. It should be
    $$c_5 = \frac{7c_3 + c_1}{20}.$$ Note that you know what ##c_3## is in terms of ##c_1##, so you can express ##c_5## in terms of ##c_1##.
     
  8. Nov 29, 2013 #7

    $$ c_5=\frac{8c_1 - 14}{20} $$
    which I think I'm supposed to see some sort of condensed version involving n! of some sort?

    and
    $$ c_6=\frac{13c_4 + 2c_2}{30} $$
    but I have two equations with C2 and one with both c2 and c4?
     
  9. Nov 29, 2013 #8

    vela

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    Good luck with that. :wink: I don't see anything straightforward, but I didn't look very hard.

    You can express ##c_2## in terms of ##c_0##. Use that to express ##c_4## in terms of ##c_0##. Then you can express ##c_6## in terms of ##c_0##. See how this is going?
     
  10. Nov 29, 2013 #9
    So all of my odd terms will be expressed in terms of c1 and all the even terms c0

    which means I need two solutions? I'm not very familiar with power series in general, I struggled with them in first year calc too
     
  11. Nov 29, 2013 #10

    vela

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    You need two independent solutions because you have a second-order differential equation, and you have two, one proportional to ##c_0## and one proportional to ##c_1##. There's also a solution that arises from the 2x term. That's the particular solution. The constants ##c_0## and ##c_1## are the arbitrary constants you generally have when you solve a differential equation with no boundary conditions.
     
  12. Nov 29, 2013 #11
    oh wow, the question actually asks for c1 to c4 and the radius of converge. which I assume doesn't mean I need to put the solutions into series form.. my apologies.


    The radius of convergence is |x-x0|<R but how do you find R?
     
  13. Nov 29, 2013 #12

    vela

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    You apply the ratio test to each series.
     
  14. Nov 29, 2013 #13
    oh. so I need to build two series equations, how can I do this? especially since the odd terms contain a weird fraction part?
     
  15. Nov 29, 2013 #14

    vela

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    Oh, I just saw that you were given initial conditions. You should apply those to get specific values for ##c_0## and ##c_1##.
     
  16. Nov 29, 2013 #15
    c0=1 and c1= 0? I just took the first term of the power series and it's derivative, unless I actually plug in 0 for x?
     
    Last edited: Nov 29, 2013
  17. Nov 29, 2013 #16

    vela

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    You have to plug in x=0 into the series for y and y' and solve for the constants.

    I'm not getting this to work out. I think we may have made a mistake somewhere. Let me look at this again for a bit.
     
  18. Nov 29, 2013 #17

    vela

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    OK, I was just confusing myself. Your work so far is fine.

    I have an idea that might help you find the radius of convergence. I'm not completely sure this method is valid, so you might want to run it by your professor. For the ratio test, you want to calculate the limit
    $$\lim_{n \to \infty} \left|\frac{c_{n+2}x^{n+2}}{c_n x^n}\right|.$$ The recurrence relation you have is
    $$c_{n+2} = \frac{n(n-1)+1}{(n+1)(n+2)} c_n + \frac{n-2}{(n+1)(n+2)} c_{n-2}.$$ If that second term on the righthand side weren't there, it would be straightforward to apply the ratio test. You could try arguing that in the limit ##n \to \infty##, the second term doesn't matter. Do you see why?
     
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