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Math Help - 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 :

    y''-(1+x)y=0

    Solution:


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

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

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

    Rewrite the equation as : y''-y -xy=0

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

    \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 \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 \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 \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 x^{n+2}. I would think you would want them all as x^n so you could combine the sums. One way of doing that is, recognizing that you want to make " x^{n+1}" look like " x^j", set j= n+1. Then n= j- 1 so that a_n= a_{j-1}. Also when n= 0, j= 1 so the sum \sum_{n=0}^\infty a_nx^{n+1} becomes \sum_{j= 1}^\infty a_{j-1}x^j. Relabel "j" to be "n" and you have
    \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,
    2a_2- a_0= 0
    and then
    \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,
    (n+2)(n+1)a_{n+2}- a_n- a_{n-1}= 0
    or
    a_{n+2}= \frac{a_n+ a_{n-1}}{(n+2)(n+1)}.

    Presumably, you have some "initial conditions" that will give you a_0 ( y(0)= a_0) and a_1 ( y'(0)= a_1). Then the first equation tells you that a_2= a_0/2 so that you have a_0, a_1, and a_2. Then you can find a_3 by taking n= 1 in the last equation, find 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 -a_0x ) it must be
    written separately

    <br />
n=0: \; \; 2a_2-a_0=0<br />

    <br />
n=1: \; \; 6a_3-a_1-2a_0=0<br />

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

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    The " 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 \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 " -a_0x" part!
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  6. #6
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    Thanks all.
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