Results 1 to 3 of 3

Math Help - find the good series solution to the ODE

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    13

    find the good series solution to the ODE

    y''+t^2y'+2ty = 0, y(0) = 1, y'(0) = 0

    This is what I did:
    P(t) = 1, Q(t) = t^2, R(t) = 2t

    So, t_0 is such that P(t_0) \neq 0 \forall t_0 \in \mathbb{R}. Therefore, t_0 is an ordinary point for all real value of t_0.

    We want a solution like y= \sum_{n=0}^\infty a_nt^n

    If we substitute in (1), we have

    \sum_{n=2}^\infty n(n-1)a_nt^{n-2} + t^2\sum_{n=1}^\infty na_nt^{n-1} + 2t\sum_{n=0}^\infty a_nt^n
    = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}t^n + \sum_{n=1}^\infty na_nt^{n+1} + \sum_{n=0}^\infty 2a_nt^{n+1}
    = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}t^n + \sum_{n=0}^\infty  na_nt^{n+1} + \sum_{n=0}^\infty 2a_nt^{n+1} because 0*a_0*t^0 = 0 doesn't change anything.
    = \sum_{n=1}^\infty (n+2)(n+1)a_{n+2}t^n + 2a_2 + \sum_{n=1}^\infty (n-1)a_{n-1}t^n + \sum_{n=1}^\infty 2a_{n-1}t^n
    = \sum_{n=1}^\infty \left( (n+2)(n+1)a_{n+2}  + (n-1)a_{n-1} + 2a_{n-1} \right) t^n + 2a_2
    = \sum_{n=1}^\infty \left( (n+2)(n+1)a_{n+2}  + (n+1)a_{n-1} \right) t^n + 2a_2 = 0

    Let a_2 = 0 then the relation is given by :

     a_{n+2} = -\frac{1}{n+2}a_{n-1}, n \geq 1

    or

     a_{n+3} = -\frac{1}{n+3}a_n, n\geq 0

    Let's find the first terms:

    n=0 : a_3 = -\frac{1}{3}a_0

    n=1 : a_4 = -\frac{1}{4}a_1

    n=2 : a_5 = -\frac{1}{5}a_2 = 0

    n=3 : a_6 = -\frac{1}{6}a_3 = -\frac{1}{6}\frac{-1}{3}a_0 = \frac{1}{3 \cdot 6} a_0

    n=4 : a_7 = -\frac{1}{8}a_4 = -\frac{1}{7}\frac{-1}{4}a_1 = \frac{1}{4 \cdot 7} a_1

    n=5 : a_8 = -\frac{1}{8}a_5 = -\frac{1}{8}\frac{-1}{5}a_2 = 0

    n=6 : a_9 = -\frac{1}{9}a_6 = -\frac{1}{9}\frac{-1}{6}a_0 = \frac{-1}{3 \cdot 6 \cdot 9} a_0

    etc.

    But I can't find a better solution than

    y = a_0\left( 1 - \frac{t^3}{3} + \frac{t^6}{3\cdot 6} - \frac{t^9}{3  \cdot 6 \cdot 9} + \cdots + \frac{(-1)^n}{3 \cdot 6 \cdot 9 \cdots  (3n)}t^{3n} + \cdots \right )  + a_1\left( t - \frac{t^4}{4} + \frac{t^7}{4\cdot 7}  -  \frac{t^{11}}{4 \cdot 7 \cdot 11} + \cdots + \frac{(-1)^n}{4 \cdot 7  \cdot 11 \cdots (3n+1)}t^{3n+1} + \cdots \right)

    Can I improve it?

    Best Regards,
    Fractalus
    Last edited by Fractalus; July 19th 2011 at 03:07 AM. Reason: removing the \br
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: find the good series solution to the ODE

    Your solution seems to be 'all right'!...A simple approach however for finding a solution of the ODE ...

     y^{''} + t^{2}\ y^{'} + 2\ t\ y = 0\ , \ y(0)= 1\ y^{'}(0)=0 (1)

    ... is to search for a search of the coefficients of the Taylor expansion...

    y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}\ t^{n} (2)

    The initial conditions give us y(0) and y^{'}(0). The other derivatives are derived from (1)...

    y^{(2)} = -t^{2}\ y^{'} -2\ t\ y \implies y^{(2)} (0)= 0 (3)

    y^{(3)} = -t^{2}\ y^{''} -4\ t\ y^{'}-2\ y \implies y^{(3)} (0)= -2 (4)

    y^{(4)} = -t^{2}\ y^{(3)} -6\ t\ y^{''}-6\ y^{'} \implies y^{(4)} (0)= 0 (5)

    y^{(5)} = -t^{2}\ y^{(4)} -8\ t\ y^{(3)}-12\ y^{''} \implies y^{(5)} (0)= 0 (6)

    y^{(6)} = -t^{2}\ y^{(5)} -10\ t\ y^{(4)}-20\ y^{(3)} \implies y^{(6)} (0)= 40 (7)

    Now You are able to write the Taylor expasion till to the order 6...

    y(x)= 1 - \frac{x^{3}}{3} + \frac{x^{6}}{18} + ... (8)

    Of course You can proceed to higher orders...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    13

    Re: find the good series solution to the ODE

    Quote Originally Posted by chisigma View Post
    Your solution seems to be 'all right'!...A simple approach however for finding a solution of the ODE ...

     y^{''} + t^{2}\ y^{'} + 2\ t\ y = 0\ , \ y(0)= 1\ y^{'}(0)=0 (1)

    ... is to search for a search of the coefficients of the Taylor expansion...

    y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}\ t^{n} (2)

    The initial conditions give us y(0) and y^{'}(0). The other derivatives are derived from (1)...

    y^{(2)} = -t^{2}\ y^{'} -2\ t\ y \implies y^{(2)} (0)= 0 (3)

    y^{(3)} = -t^{2}\ y^{''} -4\ t\ y^{'}-2\ y \implies y^{(3)} (0)= -2 (4)

    y^{(4)} = -t^{2}\ y^{(3)} -6\ t\ y^{''}-6\ y^{'} \implies y^{(4)} (0)= 0 (5)

    y^{(5)} = -t^{2}\ y^{(4)} -8\ t\ y^{(3)}-12\ y^{''} \implies y^{(5)} (0)= 0 (6)

    y^{(6)} = -t^{2}\ y^{(5)} -10\ t\ y^{(4)}-20\ y^{(3)} \implies y^{(6)} (0)= 40 (7)

    Now You are able to write the Taylor expasion till to the order 6...

    y(x)= 1 - \frac{x^{3}}{3} + \frac{x^{6}}{18} + ... (8)

    Of course You can proceed to higher orders...

    Kind regards

    \chi \sigma
    Thanks chisigma,
    I computed a solution with Mathematica and he gave me that:
    y(t) = c_2 e^{-t^3/3}-\frac{(c_1 e^{-t^3/3} t \Gamma(\frac{1}{3}, \frac{-t^{3}}{3})}{(3^{2/3} (-t^3)^{1/3})}

    Taken from:

    http://www.wolframalpha.com/input/?i=y%27%27+%2B+t^2y%27+%2B+2ty+%3D+0

    I think that I should have been able to find c_2 e^{-t^3/3}) but I can't figure out how with what I have. The Taylor Series Expansion of that is probably something like c_2\sum_{n=0}^\infty \frac{-1}{3}\frac{t^{3n}}{(3n)!}
    Last edited by Fractalus; July 19th 2011 at 03:16 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 30th 2011, 07:47 PM
  2. [SOLVED] help find critical points/good one
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 6th 2011, 04:54 PM
  3. Replies: 7
    Last Post: August 12th 2011, 07:05 AM
  4. Replies: 1
    Last Post: March 24th 2011, 08:19 AM
  5. Good Book Covering Series Of Functions, Advance Analysis
    Posted in the Advanced Math Topics Forum
    Replies: 8
    Last Post: July 18th 2007, 06:36 PM

Search Tags


/mathhelpforum @mathhelpforum