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Thread: find the good series solution to the ODE

  1. #1
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    find the good series solution to the ODE

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

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

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

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

    If we substitute in (1), we have

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

    Let $\displaystyle a_2 = 0$ then the relation is given by :

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

    or

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

    Let's find the first terms:

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

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

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

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

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

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

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

    etc.

    But I can't find a better solution than

    $\displaystyle 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; Jul 19th 2011 at 03:07 AM. Reason: removing the \br
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  2. #2
    MHF Contributor chisigma's Avatar
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    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 ...

    $\displaystyle 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...

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

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

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

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

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

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

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

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

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

    Of course You can proceed to higher orders...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    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 ...

    $\displaystyle 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...

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

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

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

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

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

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

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

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

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

    Of course You can proceed to higher orders...

    Kind regards

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

    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 $\displaystyle c_2 e^{-t^3/3})$ but I can't figure out how with what I have. The Taylor Series Expansion of that is probably something like $\displaystyle c_2\sum_{n=0}^\infty \frac{-1}{3}\frac{t^{3n}}{(3n)!}$
    Last edited by Fractalus; Jul 19th 2011 at 03:16 AM.
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