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Math Help - Frobenius, reduction of order

  1. #1
    MHF Contributor alexmahone's Avatar
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    Frobenius, reduction of order

    Find the first 4 nonzero terms in a Frobenius series solution of the given differential equation. Then use the reduction of order technique to find the logarithmic term and the first 3 nonzero terms in a second linearly independent solution.

    x^2y''+x(1+x)y'-4y=0

    My solution:

    y''+\frac{1+x}{x}y'-\frac{4}{x^2}y=0

    p_0=1 and q_0=-4

    r(r-1)+r-4=0

    r^2-4=0

    r_1=2 and r_2=-2

    y=x^{-2}\sum_{n=0}^\infty c_nx^n=\sum_{n=0}^\infty c_nx^{n-2}

    \sum_{n=0}^\infty(n-2)(n-3)c_nx^{n-2}+\sum_{n=0}^\infty (n-2)c_nx^{n-2} +\sum_{n=0}^\infty (n-2)c_nx^{n-1}-4\sum_{n=0}^\infty c_nx^{n-2}=0

    \sum_{n=0}^\infty(n-2)(n-3)c_nx^{n-2}+\sum_{n=0}^\infty (n-2)c_nx^{n-2} +\sum_{n=1}^\infty (n-3)c_{n-1}x^{n-2}-4\sum_{n=0}^\infty c_nx^{n-2}=0

    For n=0,

    (6-2-4)c_0=0

    0\cdot c_0=0

    For n\ge 1,

    (n-2)(n-3)c_n+(n-2)c_n+(n-3)c_{n-1}-4c_n=0

    (n-2)^2c_n-4c_n+(n-3)c_{n-1}=0

    (n^2-4n)c_n+(n-3)c_{n-1}=0

    n(n-4)c_n=(3-n)c_{n-1}

    c_1=-\frac{2}{3}c_0

    c_2=-\frac{1}{4}c_1=\frac{1}{6}c_0

    c_3=0

    0\cdot c_4=0

    5c_5=-2c_4

    c_5=-\frac{2}{5}c_4

    c_6=-\frac{1}{4}c_5=\frac{1}{10}c_4

    c_7=-\frac{4}{21}c_6=-\frac{2}{105}

    y(x)=c_0x^{-2}\left(1-\frac{2x}{3}+\frac{x^2}{6}\right)+c_4x^{-2}\left(x^4-\frac{2x^5}{5}+\frac{x^6}{10}-\frac{2x^7}{105}+...\right)

    =c_0x^{-2}\left(1-\frac{2x}{3}+\frac{x^2}{6}\right)+c_4x^2\left(1-\frac{2x}{5}+\frac{x^2}{10}-\frac{2x^3}{105}+...\right)

    =\frac{c_0}{6}x^{-2}(6-4x+x^2)+c_4x^2\left(1-\frac{2x}{5}+\frac{x^2}{10}-\frac{2x^3}{105}+...\right)

    y_1(x)=x^{-2}(6-4x+x^2)

    y_2(x)=x^2\left(1-\frac{2x}{5}+\frac{x^2}{10}-\frac{2x^3}{105}+...\right)

    -------------------------------------------------------------------------------------------------

    The answer given in my textbook is:

    y_1(x)=x^2\left(1-\frac{2x}{5}+\frac{x^2}{10}-\frac{2x^3}{105}+...\right)

    y_2(x)=y_1(x)\cdot\left(-\frac{1}{4x^4}+\frac{1}{15x^3}+\frac{1}{100x^2}-\frac{13}{1750x}+...\right)

    Has the book got y_2(x) wrong?
    Last edited by alexmahone; November 10th 2011 at 11:25 PM.
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  2. #2
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    Re: Frobenius, reduction of order

    No. Since the characteristic values, 2 and -2, differ by an integer, you cannot get two series. That was why the problem said "use the reduction of order technique to find the logarithmic term and the first 3 nonzero terms in a second linearly independent solution."
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Frobenius, reduction of order

    Quote Originally Posted by HallsofIvy View Post
    No. Since the characteristic values, 2 and -2, differ by an integer, you cannot get two series. That was why the problem said "use the reduction of order technique to find the logarithmic term and the first 3 nonzero terms in a second linearly independent solution."
    Even when the exponents of the DE differ by an integer, there can be 2 linearly independent series solutions; it's just that only the series solution for the larger exponent is guaranteed.

    BTW, are you implying that my answer is wrong? I've verified that my y_1(x) satisfies the given DE.
    Last edited by alexmahone; November 13th 2011 at 05:44 AM.
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