How can I find the indicial equation of $x(x-1)y''+3y'-2y=0$? I tried the method of Frobenius but I keep getting lost in the algebra. Is there any other way to get the indicial equation?
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$\begingroup$ $y''+\dfrac{3}{x(x-1)}y'-\dfrac{2}{x(x-1)}y=0$ then $p(x)=\dfrac{3}{x(x-1)}$ and $q(x)=-\dfrac{2}{x(x-1)}$. $\endgroup$– NosratiOct 29, 2017 at 4:05
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$\begingroup$ Could you give me a little more? I got that far but I'm doing the algebra for the Frobenius method and I simply can't get it to work. $\endgroup$– John SmithOct 29, 2017 at 4:07
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$y''+\dfrac{3}{x(x-1)}y'-\dfrac{2}{x(x-1)}y=0$ then $p(x)=\dfrac{3}{x(x-1)}$ and $q(x)=-\dfrac{2}{x(x-1)}$. The equation has two regular singular points $x=0$ and $x=1$. For $x=0$ we see $$p_0=\lim_{x\to0}xp(x)=\lim_{x\to0}\dfrac{3}{x-1}=-3$$ and $$q_0=\lim_{x\to0}x^2q(x)=\lim_{x\to0}\dfrac{-2x}{x-1}=0$$ then the indicial equation is $r(r-1)+p_0r+q_0=0$ or $r^2-4r=0$ shows $r=0$ and $r=4$. In this case $4-0\in\mathbb{Z}$ so $r=4$ gives a solution and let $y=x^4(a_0+a_1x+a_2x^2+\cdots)$ to find the answer. Do like this with point $x=1$.
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$\begingroup$ Hey man thanks so much. Your wonderful explanation really saved me!!!! $\endgroup$ Oct 29, 2017 at 4:31
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2$\begingroup$ Why do we use x and x^2? What's the motivation for that? Why not some other expression? $\endgroup$– T A OMar 8, 2022 at 16:54
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$\begingroup$ @TAO from the top of en.wikipedia.org/wiki/Frobenius_method: "...[the ODE] will not be solvable with regular power series methods if either $p(z)/z$ or $q(z)/z^2$ are not analytic at $z = 0$. The Frobenius method enables one to create a power series solution to such a differential equation, provided that $p(z)$ and $q(z)$ are themselves analytic at 0 or, being analytic elsewhere, both their limits at 0 exist (and are finite)." (assuming a regular singular point at $z=0$). Check the article and links within for details on the reasoning why, but this is the logic behind the limit check. $\endgroup$– ZxvAug 11 at 3:42