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    Second Order Differential Equation - Power Series Solution

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    Hi there, I have a question regarding Differential Equations which can be located here http://none-questions.blogspot.com/2011/09/d-q7.html. can someone please take a look? Thanks.

    © BrainMass Inc. brainmass.com September 29, 2022, 2:27 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/second-order-differential-equation-power-series-solution-426637

    SOLUTION This solution is FREE courtesy of BrainMass!

    We wish to solve the following differential equation using power series:

    xy'' - 2y + 9x^5 y = 0.

    First write out y(x) as a power series, i.e.

    y(x) = sum_{k=0}^infty(c_k x^k).

    Then we have

    y'(x) = sum_{k=1}^infty[k c_k x^(k-1)]
    and

    y''(x) = sum_{k=2}^infty[k(k-1) c_k x^(k-2)]

    whence

    0 = xy'' - 2y + 9x^5 y
    = sum_{k=2}^infty[k(k-1) c_k x^(k-1)] - 2 sum_{k=1}^infty[k c_k x^(k-1)]
    + 9 sum_{k'=0}^infty[c_k' x^(k'+5)].

    Now let k' = k-6. Then we have

    0 = sum_{k=1}^infty{[k(k-3) c_k + 9 c_(k-6)] x^(k-1)}.

    Since the sum is zero, each coefficient multiplying x^(k-1) must be zero, i.e.

    k(k-3) c_k + 9 c_(k-6) = 0 for all k.

    Thus we have

    c_k = -9 c_(k-6) / [k(k-3)]

    for all k.

    Assume c_0 is nonzero. Then we have

    c_(6k) = -9 c_(6k-6) / [(6k)(6k-3)]
    = -c_(6k-6) / [(2k)(2k-1)]
    = +c_(6k-12) / [(2k)(2k-1)(2k-2)(2k-3)]
    ...
    = (-1)^k c_0/(2k)!

    Thus we have the particular solution

    y_1(x) = sum_{k=0}^infty(c_k x^k) = c_0 sum_{k=0}^infty[(-1)^k x^(6k)/(2k)!] = c_0 cos(x^3).

    To find a linearly independent solution, assume c_3 is nonzero as proceed as before. We have

    c_(6k+3) = -9 c_(6k-3) / [(6k+3)(6k)]
    = -c_(6k-3) / [(2k+1)(2k)]
    = +c_(6k-9) / [(2k+1)(2k)(2k-1)(2k-2)]
    ...
    = (-1)^k c_3/(2k+1)!

    Thus we have the linearly independent solution

    y_2(x) = sum_{k=0}^infty(c_k x^k) = c_3 sum_{k=0}^infty[(-1)^k x^(6k+3)/(2k+1)!] = c_3 sin(x^3).

    Thus the general solution of the differential equation is

    y(x) = A cos(x^3) + B sin(x^3)

    where A and B are arbitrary constants.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com September 29, 2022, 2:27 pm ad1c9bdddf>
    https://brainmass.com/math/calculus-and-analysis/second-order-differential-equation-power-series-solution-426637

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