The given equation is $$xy''-(2x-1)y'+(x-1)y=0$$ Now, I haven't learned anything about the Frobenius method beyond what I've read from the Wikipedia article on it, so I expect I might have done something incorrectly.

I substituted $y=\displaystyle\sum_{k=0}^\infty a_kx^{k+r}$ and I end up with the following: $$r(r-1)a_0x^{r-2}+r(r+1)a_1x^{r-1}-2ra_0x^{r-1}+ra_0x^{r-2}+(r+1)a_1x^{r-1}-a_0x^{r-1}+\sum_{k=2}^\infty\bigg[(k+r)(k+r-1)a_k-2(k+r-1)a_{k-1}+(k+r)a_k+a_{k-2}-a_{k-1}\bigg]x^{k+r-2}$$ What's bothering me is the indicial equation... My attempt at solving gives me $r=0$ for the $x^{r-2}$ term, but I am unsure about the second term. I was under the impression the roots for it would be $$r=\frac{(a_0-a_1)\pm\sqrt{a_0^2-3a_1a_0}}{a_1}$$ but I haven't come across any other examples that have roots written in terms of $a_i$.

  • $\begingroup$ Did you try to do this with : r=0 from the start? Because that's a two dimensional space of solutions so you won't get anything about $a_0$ or $a_1$, but it looks like you're gonna get an interesting relation between three consecutive terms of the expansion $\endgroup$ – mvggz Nov 5 '14 at 15:59
  • $\begingroup$ I get: $ (k+2)^2*a_{k+2}-(2k+3)a_{k+1} +a_k =0 $ for $k \geq 0$ $\endgroup$ – mvggz Nov 5 '14 at 16:05
  • $\begingroup$ I don't know much about the method, but if you're interested in the solution, you may want to note since all 3 coefficients sum to $0$, it should be trivial to show that $e^x$ is a solution. $\endgroup$ – Mike Nov 5 '14 at 16:21
  • $\begingroup$ Duplicate of math.stackexchange.com/questions/1006832/…. Is this equality true keynote=user170231 $\endgroup$ – dustin Nov 6 '14 at 14:26

$\textbf{If you are open to other methods}$

due to your following statement $\textit{"I haven't learned anything about the Frobenius method"}$

so if you need to just solve it you could use this trick(of sorts)

$$ xy'' - (2x-1)y' + (x-1)y = xy'' - xy' - (x-1)y' +(x-1)y = 0 $$


$$ x\left(y''-y'\right) - (x-1)\left(y'-y\right) = 0 $$ then let $v = y' - y$


$$ x\dfrac{dv}{dx} - (x-1)v = 0 $$

solve for v then you have

$$ y' - y = v(x) $$


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