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Power Series

  1. Jul 27, 2007 #1
    y''+(x^2)y = 0
    I tried to solve this problem using Power Series.But i can't make the solution in the form of series that have only two constants(a0,a1)that is, there are a0,a1, a2, a3. So i just wonder how can i make it has two constants.
  2. jcsd
  3. Jul 27, 2007 #2
    You would have to show us how you got 4 different constants in order to show where the error occurred.
  4. Jul 27, 2007 #3
    Do you need to get a series solution? If not, if you substitue [tex]y'' = r'[/tex], it seems it becomes a simple seperation of variables problem.
  5. Jul 27, 2007 #4
    how can i post some method or my idea solving the problem.what programm needed to do that

    PS to..EugP , i want the series solution because my teacher want :smile:
  6. Jul 27, 2007 #5
    I was hoping you wouldn't, because I haven't learned series solutions yet :frown:
    So, sorry but I can't help.

    You don't need any program to show us your work, just use LaTeX:


    Good luck!
  7. Jul 27, 2007 #6
    thank you very much
  8. Jul 27, 2007 #7
    Reduction of order won't work, as you have the second derivative and the function.

    To solve it with power series, you must take [itex]y(x)=\sum_0^\infty a_n x^n[/itex], so

    [tex]y'(x)=\sum_{n=1}^\infty n a_n x^{n-1},[/tex]

    [tex]y''(x)=\sum_{n=2}^\infty n (n-1) a_n x^{n-2}.[/tex]

    If you plug it in the ode, you'll obtain the following equation:

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

    therefore, you'll have to change the index in the sums in order to have the same powers of [itex]x[/itex] in both sums. To do that, take [itex]n=m+4[/itex] in the first sum. The equation becomes

    [tex]\sum_{m=-2}^\infty (m+4) (m+3) a_{m+4} x^{m+2} + \sum_{n=0}^\infty a_n x^{n+2}=0.[/tex]

    Writing down the first two terms of the left sum and factorizing, we have

    [tex]2 a_2+6 a_3 x + \sum_{n=0}^\infty [(n+4) (n+3) a_{n+4}+a_n]x^{n+2} =0.[/tex]

    The only way the equation is going to be fullfilled is that [itex]a_2=a_3=0[/itex], since there is no constants or first order powers under the sum sign. Hence,

    [tex]\sum_{n=0}^\infty [(n+4) (n+3) a_{n+4}+a_n]x^{n+2} =0,[/tex]

    wich gives the recurrence relation


    Now, I've already done the hard part. You have to take it from here. To construct the first solution take [itex]a_0=1,a_1=0[/itex]. To construct the second solution take [itex]a_0=0,a_1=1[/itex] (why?). From the relationship above, it's easy to see that the condition [itex]a_2=a_3=0[/itex], makes a hole bunch of terms to be zero, so you only have to calculate two general terms.

    Good luck.
    Last edited: Jul 27, 2007
  9. Jul 27, 2007 #8
    oh i see,thank you.but u make little mistake after u take n=m+4 ,right? please check it.now i try to think why a0=0,a1=1 and vice versa
  10. Jul 27, 2007 #9
    What mistake? Could you point it out please?
  11. Jul 27, 2007 #10
    hmm,now the mistake disappear!!.Last time i saw that after u take n=m+4 the first sum u wrote the power of x =x^m which was wrong! but now it is x^m+2 ,that's right.i guess u may edit it already or my sight was wrong at that time.Anyway,thank you for your work.
    Last edited: Jul 27, 2007
  12. Jul 27, 2007 #11
    No problem :)
  13. Jul 28, 2007 #12
    i don't understand why you can take a0=0,a1=1 and vice versa
  14. Jul 28, 2007 #13


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    You can take them to be any thing you want- they will be the constants you get integrating. Taking them to be 1 and 0 (and then 0 and 1) means you can just multiply the solutions you get by C1 and C2 to get the general solution.

    It would help a lot if we could see what you know (or think you know) about this kind of problem. Is there any reason why you won't post your original solution to the problem as the first response asked?
  15. Jul 30, 2007 #14


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    just look at it. it says y'' is x^2 times y. how often does the degree of a derivative go up instead of down?

    well at least that says it is not a polynomial. but doesn't it also say it isnt a powers eries? i mean look at the lowest degree term. multiplying it by x^2 is quite different from differentiating it twice.

    try a laurent series. does anything help?

    just set y = a0 + a1X + a2X^2 +....., multiply by -x^2
    and set equal to y''.

    i guess thats what you did. but what happened? what puzzling thing occurred?
  16. Aug 2, 2007 #15
    I normally refer to Michael D. Greenberg, Advanced Engineering Mathematics for solving mathematical problem. It is suggested that we use the following change of variables
    t=x^2/2 , u =x^(-1/2)y
    to convert y''+(x^2)y = 0 into a Bessel equation.

    The solution will be
    y = x^(1/2) ( A J_1/4(x^2/2) + B J_-1/4(x^2/2) )
    where A and B are constants and
    J_1/4 (t) is the Bessel function of the first kind, of order 1/4.
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