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 $
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 $
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}
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 $\begin{align} \frac{u''}{u'}&=\frac{-\frac1{(1-x)^2}-\frac1{1-x}}{\frac{x}{1-x}}=-\frac{2-x}{x(1-x)}=-\frac{2}x+\frac1{1-x} \\ \implies u'&=\frac1{x^2(1-x)}=\frac{1+x}{x^2}-\frac1{1-x} \\ \implies u&=-\frac1x+\ln|x(1-x)| \end{align}$ so that the second basis solution is $ y_2=\frac{x\ln|x(1-x)|-1}{(1-x)^2} $