Frobenius Method to solve $x(1 - x)y'' - 3xy' - y = 0$ So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.
$$ x(1 - x)y'' - 3xy' - y = 0 $$
 A: The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.
Let $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$. Then we get that $$y'(x) = \sum_{n=0}^{\infty} na_n x^{n-1}$$
$$3xy'(x) = \sum_{n=0}^{\infty} 3na_n x^{n}$$
$$y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}$$
$$xy''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-1} = \sum_{n=0}^{\infty} n(n+1)a_{n+1} x^{n}$$
$$x^2y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n}$$
The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$
Plugging in the appropriate series expansions, we get that
$$\sum_{n=0}^{\infty} \left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_n\right)x^n = 0$$
Hence, we get that
$$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n \implies a_{n+1} = \dfrac{n+1}{n}a_n$$
First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.
Hence, the solution is given by
$$y(x) = a_1 \left(x+2x^2 + 3x^3 + \cdots\right)$$
This power series is valid only within $\vert x \vert <1$. In this region, we can simplify the power series to get
\begin{align}
y(x) & = a_1 x \left(1 + 2x + 3x^2 + \cdots \right)\\
& = a_1 x \dfrac{d}{dx} \left(x + x^2 + x^3 + \cdots \right)\\
& = a_1 x \dfrac{d}{dx} \left(\dfrac{x}{1-x}\right)\\
& = a_1 \dfrac{x}{(1-x)^2}
\end{align}
A: The order reduction method seeks a second basis solution in the form $y=y_1u$, where $y_1(x)=\frac{x}{(1-x)^2}$ is the already found basis solution.
$$
x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0  
\implies \frac{u''}{u'}=\frac{3y_1-2(1-x)y_1'}{(1-x)y_1}
$$
Insert $y_1(x)=\dfrac1{(1-x)^2}-\dfrac1{1-x}$, $y_1'=\dfrac2{(1-x)^3}-\dfrac1{(1-x)^2}$, $y_1''=\dfrac{6}{(1-x)^4}-\dfrac{2}{(1-x)^3}$
into that formula to find
\begin{align}
\frac{u''}{u'}&=\frac{-\frac1{(1-x)^2}-\frac1{1-x}}{\frac{x}{1-x}}=-\frac{2-x}{x(1-x)}=-\frac{2}x+\frac1{1-x}
\\
\implies u'&=\frac1{x^2(1-x)}=\frac{1+x}{x^2}+\frac1{1-x}
\\
\implies  u&=-\frac1x+\ln|x|-\ln|1-x|
\end{align}
so that the second basis solution is
$$
y_2=\frac{x\ln|x|-x\ln|1-x|-1}{(1-x)^2}
$$
