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Thread: Series solution to ODE

  1. #1
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    Series solution to ODE

    I have an equation

    $\displaystyle y''-8y'+16y=0$

    And I need to find a power series solution for it. I can write this as

    $\displaystyle \sum{a_{n+2}(n+2)(n+1)x^n}-8\sum{a_{n+1}(n+1)x^n}+16\sum{a_{n}x^n}=0$

    Now, as I understand it, in order to solve this, I need to get rid of one of the a terms (either $\displaystyle a_{n}$ , $\displaystyle a_{n+1}$ or $\displaystyle a_{n+2}$ ), but I don't know how to do this...

    Any guidance is greatly appreciated
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  2. #2
    MHF Contributor chisigma's Avatar
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    The DE is...

    $\displaystyle y^{''} - 8 y^{'} + 16 y= 0$ (1)

    ... and we suppose its solutions are expandable is McLaurin series as...

    $\displaystyle y(x)= a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + ... $ (2)

    From (2) we derive first...

    $\displaystyle y^{'} (x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + ... $

    $\displaystyle y^{''} (x) = 2 a_{2} + 6 a_{3} x + ...$ (3)

    The (1) can be solved only if we know the 'initial conditions' that usuallly are expressed as $\displaystyle y(0)= y_{0}$ , $\displaystyle y^{'} (0) = y_{1} $. If now we impose these 'initial conditions' we obtain...

    $\displaystyle a_{0} = y_{0}$

    $\displaystyle a_{1} = y_{1}$

    Now we can valuate the succesive terms $\displaystyle a_{n}$ inserting (2) and (3) in (1)...

    $\displaystyle 2 a_{2} - 8 a_{1} + 16 a_{0}= 0 \rightarrow a_{2} = 4 a_{1} - 8 a_{0} $

    $\displaystyle 6 a_{3} - 16 a_{2} + 16 a_{1} = 0 \rightarrow a_{3} = 8 a_{1} - \frac{64}{3} a_{0}$

    ... and so on...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    The initial conditions for this equation are y(0)=6 and y''(0)=0.

    I'm a bit more confused now though. Where am I supposed to plug those in?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by aspotonthewall View Post
    The initial conditions for this equation are y(0)=6 and y''(0)=0.

    I'm a bit more confused now though. Where am I supposed to plug those in?
    No problems!... in that case the 'initial conditions' permit us to write...

    $\displaystyle a_{0}=6$

    $\displaystyle a_{2}=0$

    ... and the other $\displaystyle a_{n}$ can be found as follows...

    $\displaystyle 2 a_{2} - 8 a_{1} + 16 a_{0} = 0 \rightarrow a_{1} = 2 a_{0} = 12$

    $\displaystyle 6 a_{3} - 16 a_{2} + 16 a_{1} = 0 \rightarrow a_{3} = -\frac{8}{3} a_{1} = -32 $

    ... and so one...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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