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Math Help - power series solutions

  1. #1
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    power series solutions

    xy'=y
    I have to find a power series solution to this equation of the form \sum a_{n}x^{n}
    Then solve the equation directly.

    I have worked out by studying the equation that the solution is nx but i can't work out how to obtain that answer.

    My working so far is:
    xy'-y=0
    y'-\frac{y}{x}=0
    \sum na_{n}x^{n-1} - \frac{1}{x}\sum a_{n}x^{n}=0
    (both of sums go from n=0 to \infty I wasn't sure how to write those in LaTex.)

    Thanks in advance
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  2. #2
    Behold, the power of SARDINES!
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    I solve the wrong equation
    Last edited by TheEmptySet; June 12th 2009 at 08:20 AM. Reason: I solved the wrong equation. Thanks Moo
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  3. #3
    Moo
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    Hello,

    For this kind of problems, you should write the starting indice
    It's important.

    Let y=\sum_{n\geq 0} a_nx^n be the solution of the equation.

    y=a_0+\sum_{n\geq 1}a_nx^n. Then y'=\sum_{n\geq 1}na_nx^{n-1}

    \begin{aligned}xy'=y &\Longrightarrow x\sum_{n\geq 1} na_nx^{n-1}=\sum_{n\geq 0} a_nx^n \\<br />
&\Longrightarrow \sum_{n\geq 1} na_nx^n=a_0+\sum_{n\geq 1} a_nx^n \\<br />
&\Longrightarrow a_0+\sum_{n\geq 1} \{a_nx^n-na_nx^n\}=0 \end{aligned}

    Let x=0 to get a_0=0

    Differentiate successively, and let x=0 to get a_n-na_n=0 \quad \quad \forall n\geq 1

    Thus a_n(n-1)=0 \quad \forall n\geq 1

    So either n=1, either a_n=0

    So all the possible solutions are :
    - a_n=0 \quad \forall n\geq 0 ---> y=0 is a solution

    - a_n=0 \quad \forall n\geq 2 and a_0=0. And a_1 can be whatever you want. So y=kx is a solution.


    @ Tessy : you got the wrong equation
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  4. #4
    Behold, the power of SARDINES!
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    Thanks Moo I solve the wrong equation
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