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Math Help - solve differential equation by series

  1. #1
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    solve differential equation by series

    4zy''+2y'+y=0

    find series solution about z=0 and identify these solution in closed form
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    Quote Originally Posted by szpengchao View Post
    4xy''+2y'+y=0

    find series solution about x=0 and identify these solution in closed form
    If you divide through by x you get 4y'' + \frac{2}{x}y'+\frac{1}{x}y = 0. The coefficient function for y' is Laurent series of order 1 and same for y. This means the equation is solvable through Frobenius method. The Frobenius equation would be 4k(k-1)+2k = 0 so k=0,2/3. These do not differ by integers so you can solve this equation on (0,\infty) by letting y = x^{k}\sum_{n=0}^{\infty} a_n x^n where k=0,2/3. For each value you will get a new linearly independent solution.
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    Hello,

    Quote Originally Posted by szpengchao View Post
    4zy''+2y'+y=0

    find series solution about z=0 and identify these solution in closed form
    Let y(z)=\sum_{n \ge 0} a_n z^n

    You will have to determine a_n ~

    y'(z)=\sum_{n \ge 1} n a_n z^{n-1}

    y''(z)=\sum_{n \ge 2} n(n-1) a_n z^{n-2}


    \implies 4zy''(z)+2y'(z)+y(z)=0

    4 \sum_{n \ge 2} n(n-1) a_n z^{n-1}+2 \sum_{n \ge 1} n a_n z^{n-1}+\sum_{n \ge 0} a_n z^n=0

    Putting all the powers to z^n, by changing the indices ~

    In the first and second ones, n-1 becomes n. Then the series will start at 1 and 0..

    4 \sum_{n \ge 1} n(n+1) a_{n+1} z^n+2 \sum_{n \ge 0} (n+1) a_{n +1} z^n+\sum_{n \ge 0} a_n z^n=0

    Now, starting the series at the same point :

    4 \sum_{n \ge 1} n(n+1) a_{n+1} z^n+2 \sum_{n \ge {\color{red}1}} (n+1) a_{n+1} z^n+{\color{red}a_1}+\sum_{n \ge {\color{red}1}} a_n z^n+{\color{red}a_0}=0

    Therefore :

    a_0+a_1+\sum_{n \ge 1} \left[4n(n+1)a_{n+1}+2(n+1)a_{n+1}+a_n\right] z^n=0

    --> a_0+a_1=0

    And :

    4n(n+1)a_{n+1}+2(n+1)a_{n+1}+a_n=0

    Find the induction formula
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