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Differential equation with power series help

  1. Aug 13, 2011 #1
    1. The problem statement, all variables and given/known data
    (1+2x^2)y''+6xy'+2y=0
    1. find the power series solutions of the equation near x0=0...show the recurrence relation for an, derive a formula for an in terms of a0 and a1, and show the solution in the form y=a0y1(x)+a1y2(x)
    2.what is the lower bound for the radii convergence of the power series solutions in part 1
    3.what is the lower bound for the power series solutions of the same equation near x0=-1


    2. Relevant equations



    3. The attempt at a solution
    I attached the word doc I typed it up in because that will probably be easier to read.
    (1+〖2x〗^2 ) y^''+6xy^'+2y=0
    Let…
    y=∑_(n=0)^∞▒〖a_n x^n 〗
    y^'=∑_(n=1)^∞▒〖〖na〗_n x^(n-1) 〗
    y^''=∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗
    (1+〖2x〗^2 ) ∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗+6x∑_(n=1)^∞▒〖〖na〗_n x^(n-1) 〗+2∑_(n=0)^∞▒〖a_n x^n 〗=0
    ∑_(n=2)^∞▒〖2n(n-1) a_n x^n+∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗〗+∑_(n=1)^∞▒〖〖6na〗_n x^n+∑_(n=0)^∞▒〖〖2a〗_n x^n 〗〗=0
    ∑_(n=2)^∞▒〖2n(n-1) a_n x^n+∑_(n=0)^∞▒〖(n+2)(n+1)a_(n+2) x^n 〗〗+∑_(n=1)^∞▒〖〖6na〗_n x^n+∑_(n=0)^∞▒〖〖2a〗_n x^n 〗〗=0
    Since the first term is 0 for both n=0, 1, we can start the summation at n=0.
    Since the third term is 0 for n=0, we can start the summation at n=0, also.
    ∑_(n=0)^∞▒〖2n(n-1) a_n x^n+∑_(n=0)^∞▒〖(n+2)(n+1)a_(n+2) x^n 〗〗+∑_(n=0)^∞▒〖〖6na〗_n x^n+∑_(n=0)^∞▒〖〖2a〗_n x^n 〗〗=0
    ∑_(n=0)^∞▒〖[2n(n+1) a_n+(n+2)(n+1) a_(n+2)+6na_n+2a_n ] x^n 〗=0
    ∑_(n=0)^∞▒〖[2(n^2+4n+1) a_n+(n+2)(n+1) a_(n+2) ] x^n 〗=0
    2(n^2+4n+1) a_n+(n+2)(n+1) a_(n+2)=0 n=0,1,2,…
    〖(a)recurrence relation→a〗_(n+2)=(-2(n^2+4n+1))/((n+2)(n+1)) a_n n=0,1,2,…
    n=0 a_2=(-2)/2 a_0 a_2=-a_0
    n=1 a_3=(-12)/6 a_1 a_3=-〖2a〗_1
    n=2 a_4=(-13)/6 a_2 a_4=13/6 a_0
    n=3 a_5=(-11)/5 a_3 a_5=22/5 a_1
    I am having trouble creating a formula in terms of a_0 and a_1. More specifically, I don’t know how to create a general summation for this case because of the weird numbers in the numerator. I am pretty sure that the denominator for the odd case is (2k+1)! and the even case would be (2k)!
    For part(c), I have y(x)=a_0 (-1+13/6 x^2+⋯)+a_1 (-2x+22/5 x^3+⋯)
    As for the lower bound for radii convergence of the power series solutions from the last part, I tried to solve this but I`m not sure if I am approaching it correctly since I didn’t know how to find the power series from the last part. I found p(x) and q(x)…
    p(x)=6x/(1+〖2x〗^2 ) and q(x)=2/(1+〖2x〗^2 )
    My problem with this is that 1+〖2x〗^2 does not have any zeroes. Does that mean that the radius of convergence is infinity?
    For the final part which asks for the lower bound for the power series solutions of the same equation near x_0=-1, I used the same p(x) and q(x) as I did in the last part. But since the denominator has no zeroes I feel like the lower bound is infinity also. The questions about radii of convergence really confuse me so any help would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Aug 13, 2011 #2

    vela

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    I didn't see an obvious mistake in your algebra, but I think the recurrence relation you found isn't correct. I used Mathematica to find the solution to the differential equation, and it found the following two solutions:
    \begin{align*}
    y_1(x) &= \frac{1}{\sqrt{1+2x^2}} \\
    y_2(x) &= \frac{1}{\sqrt{1+2x^2}}\frac{\sinh^{-1}(\sqrt{2}x)}{\sqrt{2}}
    \end{align*}
    The first solution has the series expansion[tex]y_1(x) = 1-x^2+\frac{3x^4}{2}-\frac{5x^6}{2}+\frac{35x^8}{8} - \frac{63x^{10}}{8} + \cdots[/tex]The coefficients you found don't match up with those.
     
  4. Aug 13, 2011 #3
    Thanks for your help! Do you know where I went wrong though? I did the problem multiple times and I get the same recurrence relation everytime
     
  5. Aug 13, 2011 #4

    vela

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    I just spotted the error. When you combined the individual sums and factored xn out, you changed the 2n(n-1) in the first term to 2n(n+1).
     
  6. Aug 13, 2011 #5
    I got what I believe to be the correct recurrence relation...
    an+2=[itex]\frac{-2(n+1)}{(n+2)}[/itex]an
    could you help get me started in how to find the radii of convergence?
     
  7. Aug 13, 2011 #6

    vela

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    Did you find an expression for an yet?
     
  8. Aug 13, 2011 #7
    I found
    y1(x)=a0(1-x2+(3/2)x4-(5/2)x6...)+a1(x-(4/3)x3+(32/15)x5-(128/35)x7...)
     
  9. Aug 13, 2011 #8

    vela

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    How about applying the ratio test to the two series? I was thinking to do this, you'd need a closed form expression for an, which is why I asked the earlier question, but I don't think you actually do.

    Do you have some theorems about a lower bound on the radius of convergence? I ask because typically you're asked what the radius of convergence actually is rather than to find some bound.
     
  10. Aug 13, 2011 #9
    But don`t I need the solution in the form of a summation? this is where i am really struggling. I don`t understand how to do that
     
  11. Aug 13, 2011 #10

    vela

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    The problem did ask you to find a formula for an. I take it that's where you're stuck right now and you haven't gotten to trying to find the radius of convergence yet.

    I usually write out the first few terms without simplifying much and look for a pattern. For example, you have
    \begin{align*}
    a_0 &= a_0 \\
    a_2 &= -2\frac{1}{2}a_0 = (-2)^1\frac{1}{2} a_0 \\
    a_4 &= -2\frac{3}{4}a_2 = (-2)^2\frac{1\cdot 3}{2 \cdot 4} a_0 \\
    a_6 &= -2\frac{5}{6}a_4 = (-2)^3\frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} a_0
    \end{align*}
    How would you express a2k in a general form?
     
  12. Aug 13, 2011 #11
    This helps a lot. I think I am trying to combine the numbers and simplify too quickly instead of letting it go so I can find a pattern.
    a2k=(-2)k[itex]\frac{(2k+1)!}{(2k)!}[/itex]
     
  13. Aug 13, 2011 #12

    vela

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    That's not quite right. Neither the top nor bottom is a factorial since the top is missing the even factors and the bottom is missing the odd factors. Also, the last factor in the numerator is (2k-1), not (2k+1). Look at a6 for example. There you have 2k=6 but the last factor in the numerator is 5 = (2k-1).

    Are you familiar with double-factorial notation? Or you could also use some tricks like
    [tex]2 \cdot 4 \cdot 6 \cdots (2k) = (2 \times 1) \cdot (2 \times 2) \cdot (2 \times 3) \cdots (2 \times k) = 2^k k![/tex]
    and
    [tex]1 \cdot 3 \cdot 5 \cdots (2k-1) = \frac{1\cdot 2\cdot 3 \cdot 4 \cdots (2k-1)\cdot(2k)}{2\cdot 4\cdots (2k)} = \frac{(2k)!}{2^k k!}[/tex]
     
  14. Aug 13, 2011 #13
    I didn`t even think about double factorials but yes my professor just went over them in class the other day. Using them I have...
    a2k=((-2)k[itex]\frac{(2k-1)!!}{(2k)!!}[/itex])a0
    and likewise...
    a2k+1=((-2)k[itex]\frac{(2k)!!}{(2k-1)!!}[/itex])a1
    so in power series form, I have...
    (i attached the file with the form. i was struggling to type it into the forum with the latex formatting. I didn`t know how to do the summation symbol properly)
     

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    Last edited: Aug 13, 2011
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