1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Series solutions to ODE

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve for
    [tex]xy'' + y' +αy + βxy = 0[/tex]

    α and β are constants

    3. The attempt at a solution
    What I initially had in mind was:
    [tex]xy'' + y' +αy + βxy = x²y'' + xy' +αxy + βx²y = 0[/tex]
    [tex]y = \sum_{n=0}^\infty a_n x^{n}[/tex]
    [tex]xy = \sum_{n=0}^\infty a_n x^{n+1} = \sum_{n=1}^\infty a_{n-1} x^{n} = a_0x + \sum_{n=2}^\infty a_{n-1} x^{n}[/tex]
    [tex]x²y = \sum_{n=0}^\infty a_n x^{n+2} = \sum_{n=2}^\infty a_{n-2} x^{n}[/tex]
    [tex]y' = \sum_{n=1}^\infty na_n x^{n-1}[/tex]
    [tex]xy' = \sum_{n=1}^\infty na_n x^{n} = a_1x + \sum_{n=2}^\infty na_n x^{n}[/tex]
    [tex]y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}[/tex]
    [tex]xy'' = \sum_{n=2}^\infty n(n-1)a_n x^{n}[/tex]
    [tex]a_0x + a_1x +\sum_{n=2}^\infty[n(n-1)a_n + na_n + αa_{n-1} + βa_{n-2}]x^{n} = 0 [/tex]
    [tex]a_1 = - a_0[/tex]
    Recurrence relation for n ≥ 2:
    [tex]a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n(n-1)+n}[/tex]
    [tex]a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n²}[/tex]
    [tex]a_2 = \frac {-(αa_{1}+βa_{0})}{4} = \frac {αa_{0}-βa_{0}}{4} = \frac {(α-β)a_{0}}{4}[/tex]
    [tex]a_3 = \frac {-(αa_{2}+βa_{1})}{9} = \frac {-α\frac {(α-β)a_{0}}{4}+βa_{0}}{9} = \frac {(4β + αβ - α^{2})a_0}{36}[/tex]
    [tex]a_4 = \frac {-(αa_{3}+βa_{2})}{16} =\frac {(9αβ - 9β^{2} + 4αβ - α^{3} + α^{2}β)a_0}{576} [/tex]
    Sadly I can't see how to proceed from here. Did I mess anything up?

    Apparently this has to be solved with Frobenius, I'll edit this thread later with another attempt at this.
     
    Last edited: Apr 10, 2016
  2. jcsd
  3. Apr 10, 2016 #2
    Frobenius attempt:
    [tex]y = \sum_{n=0}^\infty a_n x^{n+r}[/tex]
    [tex]y' = \sum_{n=0}^\infty a_n(n+r) x^{n+r-1}[/tex]
    [tex]y'' = \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r-2}[/tex]
    [tex]x²y'' = \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r} = a_0r(r-1)x^{r} + a_1r(r+1)x^{r+1} + \sum_{n=2}^\infty a_n(n+r)(n+r-1) x^{n+r}[/tex]
    [tex]xy' = \sum_{n=0}^\infty a_n(n+r) x^{n+r} = a_0rx^{r} + a_1(r+1)x^{r+1} + \sum_{n=2}^\infty a_n(n+r) x^{n+r}[/tex]
    [tex]xy = \sum_{n=0}^\infty a_n x^{n+r+1} = a_0x^{r+1} + \sum_{n=2}^\infty a_{n-1} x^{n+r}[/tex]
    [tex]x²y = \sum_{n=0}^\infty a_n x^{n+r+2} = \sum_{n=2}^\infty a_{n-2} x^{n+r}[/tex]
    [tex]a_0r(r-1)x^{r} + a_1r(r+1)x^{r+1} + a_0rx^{r} + a_1(r+1)x^{r+1} + αa_0x^{r+1} + \sum_{n=2}^\infty[a_n(n+r)(n+r-1) + a_n(n+r) + αa_{n-1} + βa_{n-2}]x^{n+r} [/tex]
    [tex]a_n = \frac {-(αa_{n-1}+βa_{n-2})}{(n+r)^{2}}[/tex]
    [tex]a_0r(r-1)x^{r} + a_1r(r+1)x^{r+1} + a_0rx^{r} + a_1(r+1)x^{r+1} + αa_0x^{r+1} = 0[/tex]
    [tex]a_0r(r-1) + a_1r(r+1)x + a_0r + a_1(r+1)x + αa_0x = 0[/tex]
    [tex]a_0[r(r-1) + r+αx ] + a_1[r(r+1)x +(r+1)x] = 0[/tex]
    [tex]a_0[r²+αx ] + a_1x[(r+1)^{2}] = 0[/tex]
    Then I get stuck. Is there any other method that can solve this kind of ODE?
     
  4. Apr 10, 2016 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You want to collect powers of ##x##:
    $$a_0 r^2 + [\alpha a_0 +(r+1)^2 a_1]x = 0.$$ For this to hold true for all values of ##x##, you require each term to vanish. That lets you solve for ##r## and ##a_1##. (I didn't check your algebra, but I didn't see any obvious mistakes.)
     
  5. Apr 10, 2016 #4
    $$a_0 r^2 + [\alpha a_0 +(r+1)^2 a_1]x = 0$$
    $$a_0r^2 = 0$$
    $$r = 0$$
    $$\alpha a_0 +(r+1)^2 a_1 = 0$$
    $$a_1 = \frac{-\alpha a_0}{(r+1)^2} = -\alpha a_0$$
    $$a_n = \frac {-(αa_{n-1}+βa_{n-2})}{(n+r)^{2}}$$
    $$a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n^{2}}$$
    $$a_2 = \frac {(α^{2}+β)a_0}{4}$$
    $$a_3 = \frac {(5αβ -α^{3})a_0}{9*4}$$
    $$a_4 = \frac {(α^{4} -6α^{2}β -β^{2})a_0}{16*9*4}$$
    The denominators seems to follow the pattern:
    $$n^{2}*(n-1)^{2}*(n-2)^{2}*...*9*4$$
    But I can't see any with the numerators.
    I'll double check everything.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Series solutions to ODE
  1. Series Solution to ODE (Replies: 10)

  2. Series Solution of ODE (Replies: 2)

Loading...