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Math Help - Power series - general solution

  1. #1
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    Power series - general solution

    Assuming a power series solution of the form






    Show that


    2(dy/dx) + xy = x

    has general solution y(x) = 1 + Ae^[(x^2)/4] where A is a constant.


    I know I have to substitute y = a(0) + a(1) x + a(2) x^2 + ....
    into the equation and then equate the powers of x on both sides. After that solve for the sequence a(n). but i don't seem to be getting what I should? If anyone could help me that would be brilliant! xxx
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  2. #2
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    Hello,

    y=\sum_{n=0}^\infty a_n x^n=a_0+ \sum_{n=1}^\infty a_n x^n

    hence \frac{dy}{dx}=\sum_{n=1}^\infty na_n x^{n-1}

    Write your differential equation :
    2 \frac{dy}{dx}+xy=x


    \begin{aligned}<br />
2 \frac{dy}{dx}+xy &=2 \sum_{n=1}^\infty na_nx^{n-1}+x \sum_{n=0}^\infty a_nx^n \\<br />
&=\sum_{n=1}^\infty 2na_nx^{n-1}+\sum_{n=0}^\infty a_nx^{n+1} \end{aligned}

    Now change the indices so that you get x^n in the series :

    =\sum_{n=0}^\infty 2(n+1)a_{n+1} x^n+\sum_{n=1}^\infty a_{n-1}x^n

    Now make the starting indice be the same for the two series :

    =2(0+1)a_{0+1}+\sum_{n=1}^\infty 2(n+1)a_{n+1} x^n+\sum_{n=1}^\infty a_{n-1}x^n

    =2a_1+\sum_{n=1}^\infty \left(2(n+1)a_{n+1}+a_{n-1}\right) x^n

    You want this to be equal to x.
    Hence :

    x=2a_1+\sum_{n=1}^\infty \left(2(n+1)a_{n+1}+a_{n-1}\right) x^n

    Make appear the factors of x.

    x=2a_1+(4a_2+a_0)x+\sum_{n=2}^\infty \left(2(n+1)a_{n+1}+a_{n-1}\right) x^n


    By identification, we get :

    \left\{\begin{array}{lll} a_1=0 \\ 4a_2+a_0=1 \\ 2(n+1)a_{n+1}+a_{n-1}=0 \end{array} \right.


    Solve for the recursive relation :P
    Note that if n is odd, then the coefficient of the power series is 0, since a1=0.

    I've already made a big part of the working, so now it's you to go ! Good luck
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