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Thread: Solution of y''-(1+x)y=0 using power series ..

  1. #1
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    Solution of y''-(1+x)y=0 using power series ..

    Hello

    Obtain the solution of the following differential equation as a power series about the origin :

    $\displaystyle y''-(1+x)y=0$

    Solution:


    Let $\displaystyle \displaystyle y=\sum_{n=0}^{\infty} a_n x^n$
    then :

    $\displaystyle \displaystyle y'=\sum_{n=1}^{\infty} n a_n x^{n-1}$

    $\displaystyle \displaystyle y''=\sum_{n=2}^{\infty} n (n-1) a_n x^{n-2}$

    Rewrite the equation as : $\displaystyle y''-y -xy=0$

    and substitute the values of y & y'' gives:

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

    or:

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

    My problem is in the last summation
    I want to make the summations start with the same n and the powers of the x's are n in order to get the recursive relation, but I can not do that .

    the last summation makes my life hard

    any help?
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  2. #2
    A Plied Mathematician
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    Try breaking off the first term of the last summand thus:

    $\displaystyle \displaystyle \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} a_n x^n - a_{0}x-\sum_{n=1}^{\infty} a_n x^{n+1}=0,$ then re-indexing gives

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

    You might need to break off other terms as well, in order to get all the powers of x to be the same. Those are your two tricks: re-indexing and breaking off terms. You should be able to get all the starting indices and powers of x to be the same in all three summations, which would enable you to combine them into one summation.
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  3. #3
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    I'm not sure why Ackbeet put the last sum to power $\displaystyle x^{n+2}$. I would think you would want them all as $\displaystyle x^n$ so you could combine the sums. One way of doing that is, recognizing that you want to make "$\displaystyle x^{n+1}$" look like "$\displaystyle x^j$", set j= n+1. Then n= j- 1 so that $\displaystyle a_n= a_{j-1}$. Also when n= 0, j= 1 so the sum $\displaystyle \sum_{n=0}^\infty a_nx^{n+1}$ becomes $\displaystyle \sum_{j= 1}^\infty a_{j-1}x^j$. Relabel "j" to be "n" and you have
    $\displaystyle \displaystyle \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} a_n x^n - a_{0}x-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0$
    Since the last sum does not start until n= 1, as Ackbeet suggests, do n= 0 separately:
    With n= 0,
    $\displaystyle 2a_2- a_0= 0$
    and then
    $\displaystyle \displaystyle \sum_{n=1}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=1}^{\infty} a_n x^n - a_{0}x-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0$
    so that for each n, greater than or equal to 1,
    $\displaystyle (n+2)(n+1)a_{n+2}- a_n- a_{n-1}= 0$
    or
    $\displaystyle a_{n+2}= \frac{a_n+ a_{n-1}}{(n+2)(n+1)}$.

    Presumably, you have some "initial conditions" that will give you $\displaystyle a_0$ ($\displaystyle y(0)= a_0$) and $\displaystyle a_1$ ($\displaystyle y'(0)= a_1$). Then the first equation tells you that $\displaystyle a_2= a_0/2$ so that you have $\displaystyle a_0$, $\displaystyle a_1$, and $\displaystyle a_2$. Then you can find $\displaystyle a_3$ by taking n= 1 in the last equation, find $\displaystyle a_4$ by taking n= 2, etc. With luck, you might be able to find a general formula but that is rare.
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  4. #4
    Senior Member
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    I may be wrong, but for n=1 (as there is term $\displaystyle -a_0x $) it must be
    written separately

    $\displaystyle
    n=0: \; \; 2a_2-a_0=0
    $

    $\displaystyle
    n=1: \; \; 6a_3-a_1-2a_0=0
    $

    $\displaystyle
    n>=2: \; \; (n+2)(n+1)a_{n+2}-a_n-a_{n-1}=0
    $
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  5. #5
    MHF Contributor

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    The "$\displaystyle a_0x$" term is my mistake. Being basically lazy, I copied the LaTex version of the equation from Ackbeet's post but, being basically careless, I copied the wrong equation! I intended to copy the first equation before he had separated that part out. What I should have had was:
    $\displaystyle \displaystyle \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} a_n x^n - \sum_{n=1}^{\infty} a_{n-1} x^{n}=0$

    I just copied it again, but this time I took out the "$\displaystyle -a_0x$" part!
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  6. #6
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    Thanks all.
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