Frobenius, reduction of order

• November 10th 2011, 08:01 PM
alexmahone
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 $n\ge 1$,

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

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

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

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

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

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

$c_3=0$

$0\cdot c_4=0$

$5c_5=-2c_4$

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

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

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

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

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

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

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

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

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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?
• November 13th 2011, 05:27 AM
HallsofIvy
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."
• November 13th 2011, 05:32 AM
alexmahone
Re: Frobenius, reduction of order
Quote:

Originally Posted by HallsofIvy
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.