I am referring to Carl Bender's Advanced mathematics methods for scientists and Engineers.
Well, actually I know how to solve it....However, if I choose to do a so called "powerful" method,which is Frobenius, I would encounter some trouble....
Qns [3.26] Perform a local analysis of solutions to $(x-1)y''-xy'+y=0$ at $x=1$. Use the results of this analysis to prove that a Taylor series expansion of any solution about $x=0$ has an infinite radius of convergence.
I tried different methods.
Methods I: Using standard Frobenius, Trial $y=\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ x }^{ n+\alpha } } $ and we could identify that ${p_0}=0$,${q_0}=-1$ . if we choose to divide the $(x-1)$ through out and make it standard form like $y''+ \frac {p(x) }{(x-1) }+ \frac{q(x)}{{(x-1)}^{2}}$.
Hence I would get some recurrence relations between ${a}_{n}$s.This could be done.
After getting recurrence relation, by putting $n=0$, I get the equation about $\alpha$ is ${\alpha}^{2}-\alpha-1=0$. Solving I get $\alpha = \frac{1\pm \sqrt{5}}{2}$
Method II: If I try to do like this, by adding a $y'$ to both side, that is, converting the differential equation like this.$(x-1)y''-(x-1)y'+y=y'$.Using the Antazs above.
Solving for $\alpha$, I get $\alpha = 0 $ or $2$. It seems well,quite interesting . as this is not expected. It seems I arrive at the different solutions.
I checked on mathematica, it said two solutions are $y=x$ and $y=e^x$, both up to a multiplication constant.
It seems that my first method may never arrive at that answer.....
Secondly, another problem is that this qns asked me to show that the radius of convergence is $\infty $. I suspect if my solutions are those given by mathematica, this seems to be applied immediately.... I don't even need to answer that.....I don't know if I understand correctly...
Before considering the methods involving infinite series, let try something simpler : $$(x-1)y''-xy'+y=0$$ An obvious particular solution is visible at first sight : $y=x$ So, it is expected to obtain an ODE of separable kind with the change of function $y(x)=xf(x)$ , then $y'=f+xf'$ and $y''=2f'+f''$ $$(x-1)(2f'+f'')-x(f+xf')+xf=0$$ After simplification : $$\frac{f''}{f'}=\frac{x^2-2x+2}{x(x-1)}$$ The integration leads to : $$\ln(f')=x+\ln|x-1|-2\ln|x|+c$$ $$f'=C\frac{x-1}{x^2}e^x$$ The antiderivative is obvious : $$f=c_1\frac{e^x}{x}+c_2$$ Finally : $$y=xf=c_1e^x+c_2x$$ .
. Method with infinite series $y=\sum_{n=0}^\infty a_n x^n$
$y'=\sum_{n=1}^\infty na_n x^{n-1}$ and $y''=\sum_{n=2}^\infty n(n-1)a_n x^{n-2}$
$$(x-1)\sum_{n=2}^\infty n(n-1)a_n x^{n-2}-x\sum_{n=1}^\infty na_n x^{n-1}+\sum_{n=0}^\infty a_n x^n=0$$
$$\sum_{n=2}^\infty n(n-1)a_n x^{n-1}-\sum_{n=2}^\infty n(n-1)a_n x^{n-2}- \sum_{n=1}^\infty na_n x^{n}+\sum_{n=0}^\infty a_n x^n=0$$ Here is an important point : All the powers of $x$ must be the same i.e.: $x^n$ :
$$\sum_{n=1}^\infty (n+1)(n)a_{n+1} x^{n}-\sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^{n}-\sum_{n=1}^\infty na_n x^{n}+\sum_{n=0}^\infty a_n x^n=0$$
The reccurence Relationship is : $$(n+1)(n)a_{n+1}-(n+2)(n+1)a_{n+2}-na_n+a_n=0$$ $$a_{n+2}=\frac{n}{n+2}a_{n+1}-\frac{n-1}{(n+1)(n+2)}a_n$$ which can be reduced to : $$a_{n+1}=\frac{a_n}{n+1}$$ You see how to continue.
Note that the particular solution $e^x$ will finally appear on the form of infinite series.