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Thread: Solve ODE by power series

  1. #1
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    Solve ODE by power series

    Hi everyone,
    I faced a problem in solve a second order ODE by power series. Hope someone help me out there:
    Here is the question:
    y'' - 2y' + y = 0

    This is my solution (I have checked many times, probably nothing wrong):
    Let
    y = \sum\limits_{n = 0}^\infty {{c_n}{x^n}}
    y' = \sum\limits_{n = 1}^\infty {n{c_n}{x^{n - 1}}}
    y'' = \sum\limits_{n = 2}^\infty {n(n - 1){c_n}{x^{n - 2}}}

    Then,
    \sum\limits_{n = 2}^\infty {n(n - 1){c_n}{x^{n - 2}}} - 2\sum\limits_{n = 1}^\infty {n{c_n}{x^{n - 1}}} + \sum\limits_{n = 0}^\infty {{c_n}{x^n}} = 0

    After shift the index,
    \sum\limits_{n = 0}^\infty {(n + 2)(n + 1){c_{n + 2}}{x^n}} + \sum\limits_{n = 0}^\infty { - 2(n + 1){c_{n + 1}}{x^n}} + \sum\limits_{n = 0}^\infty {{c_n}{x^n}} = 0
    \sum\limits_{n = 0}^\infty {(n + 2)(n + 1){c_{n + 2}}{x^n} - 2(n + 1){c_{n + 1}}{x^n} + {c_n}{x^n}} = 0
    \sum\limits_{n = 0}^\infty {[(n + 2)(n + 1){c_{n + 2}} - 2(n + 1){c_{n + 1}} + {c_n}]{x^n}} = 0

    Therefore a recurrence relation, for all n > 0:
    (n + 2)(n + 1){c_{n + 2}} - 2(n + 1){c_{n + 1}} + {c_n} = 0
    {c_{n + 2}} = \frac{2}{{(n + 2)}}{c_{n + 1}} - \frac{1}{{(n + 2)(n + 1)}}{c_n}

    To find {c_n} terms in only {c_0} & {c_1},
    For n = 0,
    {c_2} = \frac{2}{2}{c_1} - \frac{1}{{2 \cdot 1}}{c_0}
    {c_2} = \frac{2}{{2!}}{c_1} - \frac{1}{{2!}}{c_0}
    {c_2} = \frac{1}{{1!}}{c_1} - \frac{1}{{2!}}{c_0}

    For n = 1,
    {c_3} = \frac{2}{3}{c_2} - \frac{1}{{3 \cdot 2}}{c_1}
    {c_3} = \frac{2}{3}\left[ {\frac{2}{{2!}}{c_1} - \frac{1}{{2!}}{c_0}} \right] - \frac{1}{{3!}}{c_1}
    {c_3} = \left[ {\frac{{2 \cdot 2}}{{3!}} - \frac{1}{{3!}}} \right]{c_1} - \frac{2}{{3!}}{c_0}
    {c_3} = \frac{3}{{3!}}{c_1} - \frac{2}{{3!}}{c_0}
    {c_3} = \frac{1}{{2!}}{c_1} - \frac{2}{{3!}}{c_0}

    For n = 2,
    {c_4} = \frac{2}{4}{c_3} - \frac{1}{{4 \cdot 3}}{c_2}
    {c_4} = \frac{2}{4}\left[ {\frac{3}{{3!}}{c_1} - \frac{2}{{3!}}{c_0}} \right] - \frac{1}{{4 \cdot 3}}\left[ {\frac{2}{{2!}}{c_1} - \frac{1}{{2!}}{c_0}} \right]
    {c_4} = \left[ {\frac{{3 \cdot 2}}{{4!}}{c_1} - \frac{{2 \cdot 2}}{{4!}}{c_0}} \right] - \left[ {\frac{2}{{4!}}{c_1} - \frac{1}{{4!}}{c_0}} \right]
    {c_4} = \left[ {\frac{{3 \cdot 2}}{{4!}} - \frac{2}{{4!}}} \right]{c_1} - \left[ {\frac{{2 \cdot 2}}{{4!}} - \frac{1}{{4!}}} \right]{c_0}
    {c_4} = \frac{4}{{4!}}{c_1} - \frac{3}{{4!}}{c_0}
    {c_4} = \frac{1}{{3!}}{c_1} - \frac{3}{{4!}}{c_0}

    For n = 3,
    {c_5} = \frac{2}{5}{c_4} - \frac{1}{{5 \cdot 4}}{c_3}
    {c_5} = \frac{2}{5}\left[ {\frac{4}{{4!}}{c_1} - \frac{3}{{4!}}{c_0}} \right] - \frac{1}{{5 \cdot 4}}\left[ {\frac{3}{{3!}}{c_1} - \frac{2}{{3!}}{c_0}} \right]
    {c_5} = \left[ {\frac{{2 \cdot 4}}{{5!}}{c_1} - \frac{{2 \cdot 3}}{{5!}}{c_0}} \right] - \left[ {\frac{3}{{5!}}{c_1} - \frac{2}{{5!}}{c_0}} \right]
    {c_5} = \left[ {\frac{{2 \cdot 4}}{{5!}} - \frac{3}{{5!}}} \right]{c_1} - \left[ {\frac{{2 \cdot 3}}{{5!}} - \frac{2}{{5!}}} \right]{c_0}
    {c_5} = \frac{5}{{5!}}{c_1} - \frac{4}{{5!}}{c_0}
    {c_5} = \frac{1}{{4!}}{c_1} - \frac{4}{{5!}}{c_0}

    .....

    Since
    y = \sum\limits_{n = 0}^\infty {{c_n}{x^n}}
    y = {c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + {c_4}{x^4} + {c_5}{x^5} + ...
    y = {c_0} + {c_1}x + \left[ {\frac{1}{{1!}}{c_1} - \frac{1}{{2!}}{c_0}} \right]{x^2} + \left[ {\frac{1}{{2!}}{c_1} - \frac{2}{{3!}}{c_0}} \right]{x^3} + \left[ {\frac{1}{{3!}}{c_1} - \frac{3}{{4!}}{c_0}} \right]{x^4} + \left[ {\frac{1}{{4!}}{c_1} - \frac{4}{{5!}}{c_0}} \right]{x^5} + ...
    y = {c_0}\left[ {1 - \frac{1}{{2!}}{x^2} - \frac{2}{{3!}}{x^3} - \frac{3}{{4!}}{x^4} - \frac{4}{{5!}}{x^5} - ...} \right] + {c_1}x\left[ {1 + \frac{1}{{1!}}x + \frac{1}{{2!}}{x^2} + \frac{1}{{3!}}{x^3} + \frac{1}{{4!}}{x^4} + ...} \right]
    y = {c_0}\left[ {1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} } \right] + {c_1}x\left[ {\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}{x^n}} } \right]

    As far as I know,
    \sum\limits_{n = 0}^\infty {\frac{1}{{n!}}{x^n}} = {e^x}

    In fact, when I solve y'' - 2y' + y = 0 using charactheristic equation,
    {r^2} - 2r + 1 = 0
    {(r - 1)^2} = 0
    r = 1

    And the solution is
    y = ({c_0} + {c_1}x){e^x}

    Now, here my question:
    Is 1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = {e^x}?
    If yes, please can you help me proof it?
    If not, could you point out where my mistake?
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  2. #2
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    Re: Solve ODE by power series

    Why do you want a power series solution when it is incredibly easy to solve exactly?
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  3. #3
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    Solve ODE by power series

    Hi Prove It,
    This is the exercise from my lecturer told us to do

    The proving part is actually that I'm exploring my own
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  4. #4
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    Re: Solve ODE by power series

    The solution to $y^{\prime\prime}-2y^\prime + y=0$ is not just $y=c e^t$

    The characteristic polynomial has a repeated root at $s=1$ and thus the full solution is

    $y(t) = c_1 e^t + c_2 t e^t$

    see if your series solution, replicates this.
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  5. #5
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    Re: Solve ODE by power series

    Hi Romsek, thanks for reminding me,
    However, after cooperating with a Dr., we had figured out the possible way to explain my question:
    At first, I will say
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} \ne {e^x}
    And there is no fault in my solution above,

    The continuation is like this,
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = 1 - \sum\limits_{n = 2}^\infty {\left( {\frac{n}{{n!}} - \frac{1}{{n!}}} \right)} {x^n}
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = 1 - \left( {\sum\limits_{n = 2}^\infty {\left( {\frac{n}{{n!}}} \right)} {x^n} - \sum\limits_{n = 2}^\infty {\left( {\frac{1}{{n!}}} \right)} {x^n}} \right)
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = 1 - \sum\limits_{n = 2}^\infty {\left( {\frac{n}{{n!}}} \right)} {x^n} + \sum\limits_{n = 2}^\infty {\left( {\frac{1}{{n!}}} \right)} {x^n} + x - x
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = \left( {1 + x + \sum\limits_{n = 2}^\infty {\left( {\frac{1}{{n!}}} \right)} {x^n}} \right) - \left( {x + \sum\limits_{n = 2}^\infty {\left( {\frac{1}{{(n - 1)!}}} \right)} {x^n}} \right)
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = {e^x} - \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{(n - 1)!}}} \right)} {x^n}
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = {e^x} - \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{(n + 1 - 1)!}}} \right)} {x^{n + 1}}
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = {e^x} - x\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{n!}}} \right)} {x^n}
    1 - \sum\limits_{n = 2}^\infty {\frac{{n - 1}}{{n!}}{x^n}} = {e^x} - x{e^x}

    Therefore for this case,
    y = {c_0}({e^x} - x{e^x}) + {c_1}x{e^x}
    y = {c_0}{e^x} - {c_0}x{e^x} + {c_1}x{e^x}
    y = {c_0}{e^x} + ({c_1} - {c_0})x{e^x}

    Now, the solutions from 2 methods (Charactheristics Equation & Power Series) has finally equated:
    Charactheristics Equation Method:
    (Let me use {a_0} and {a_1} for the 2 constant for Charactheristics Equation Method, so that itís donít get confusing)
    y = {a_0}{e^x} + {a_1}x{e^x}

    And Power Series Method:
    y = {c_0}{e^x} + ({c_1} - {c_0})x{e^x}

    Where the second constant from power series is dependent from the first constant.
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