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Thread: 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.

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

    My solution:

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

    $\displaystyle p_0=1$ and $\displaystyle q_0=-4$

    $\displaystyle r(r-1)+r-4=0$

    $\displaystyle r^2-4=0$

    $\displaystyle r_1=2$ and $\displaystyle r_2=-2$

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

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

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

    For $\displaystyle n=0$,

    $\displaystyle (6-2-4)c_0=0$

    $\displaystyle 0\cdot c_0=0$

    For $\displaystyle n\ge 1$,

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

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

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

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

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

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

    $\displaystyle c_3=0$

    $\displaystyle 0\cdot c_4=0$

    $\displaystyle 5c_5=-2c_4$

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

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

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

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

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

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

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

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

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

    The answer given in my textbook is:

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

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

    Has the book got $\displaystyle y_2(x)$ wrong?
    Last edited by alexmahone; Nov 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 $\displaystyle y_1(x)$ satisfies the given DE.
    Last edited by alexmahone; Nov 13th 2011 at 05:44 AM.
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