Frobenius Method
This method is a series solution about a regular singular point.
Neat site: 7.3: Singular Points and the Method of Frobenius - Mathematics LibreTexts
Let x₀ be a regular singular point of $$a_0(x)y’’+a_1(x)y’++a_2(x)y=0$$
Solution is in the form $$y(x)=(x-x_{0})^{r}\sum\limits {n=0}^{\infty} a{n}(x-x_{0})^{n}$$
Process to find the y #
1. Substitute y,y’,y’’ in the differential equation #
$$ \begin{align*} y(x)=\sum\limits {n=0}^{\infty}a{n}(x-x_{0})^{n+r}\\ \\ y’(x)=\sum\limits {n=0}^{\infty}(n+r)a{n}(x-x_{0})^{n+r-1}\\ \\ y’’(x)=\sum\limits {n=0}^{\infty}(n+r)(n+r-1)a{n}(x-x_{0})^{n+r-2} \end{align*} $$
2. Equating the coefficient of the lowest power to 0 #
We obtain a quadratic equation in ‘r’ known as an indicial equation. This quadratic equation has two roots $r_{1},r_{2}$ known as indicial roots
This is to make $a_{0}\ne0$
3. Equating the coeffs of remaining powers of (x-x_0) to 0 #
We obtain a recurrence relation relating the coefficients $a_n$
Substitution of the indicial roots gives the values of $a_n$
4. General solution #
$$ y=A_1y_1(x)+A_2y_2(x) $$
$y_1,y_2$ are linearly independent
By the above three steps we obtain y₁, but to obtain y₂ we have cases based on indicial roots.
Cases based on indicial roots #
Generally $r_1>r_2$
If they are distinct and do not differ by an integer #
$$ \begin{align*} y_1=(x-x_{0})^{r_{1}}\sum\limits {n=0}^{\infty} a{n}(x-x_{0})^{n}\\ \\ y_2=(x-x_{0})^{r_{2}}\sum\limits {n=0}^{\infty} b{n}(x-x_{0})^{n} \end{align*} $$
If they have a double root #
$r_1=r_2=r$
$$ \begin{align*} y_1=(x-x_{0})^{r}\sum\limits {n=0}^{\infty} a{n}(x-x_{0})^{n}\\ \\ y_2=(x-x_{0})^{r+1}\sum\limits {n=0}^{\infty} b{n}(x-x_{0})^{n}+(\ln (x-x_0))y_{1} \end{align*} $$
If they are distinct and differ by an integer #
Take $r_1$ to be the smaller root because it is sometimes possible to obtain the general solution using the smaller root alone. $$ \begin{align*} y_1=(x-x_{0})^{r_1}\sum\limits {n=0}^{\infty} a{n}(x-x_{0})^{n}\\ \\ y_2=(x-x_{0})^{r_2}\sum\limits {n=0}^{\infty} b{n}(x-x_{0})^{n}+C(\ln (x-x_0))y_{1} \end{align*} $$
C may be 0 sometimes
Can also be found by #
$$ y_2(x)=\dd{y}{r} | _{r=r_1} $$
If they are complex #
$$ \begin{align*} y=(x-x_{0})^{r}\sum\limits {n=0}^{\infty} a{n}(x-x_{0})^{n}\\\ \end{align*} $$
$y_1,y_2$ can be found by taking the real and imaginary parts of y