So I have this problem and it's making me crazy, I'm pretty sure it's simple but I'm getting stuck at a certain part.
$$x^2y''+xy'+2xy = 0$$ at x = 0.
So, x = 0 is a singular point (also regular, already did the math), so we need to use the Frobenius method to solve this.
I got:
$$ y = \sum_{n=0}^{\infty } A_{n} x^{r + n}$$
$$y' = \sum_{n=0}^{\infty } {(r+n)}A_{n} x^{r + n-1}$$
$$y'' = \sum_{n=0}^{\infty } {(r+n-1)(r+n)}A_{n} x^{r + n-2}$$
Putting the terms on the equation:
$$\sum_{n=0}^{\infty } {(r+n-1)(r+n)}A_{n} x^{r + n} + \sum_{n=0}^{\infty } {(r+n)}A_{n} x^{r + n} + \sum_{n=0}^{\infty } 2A_{n} x^{r + n + 1}$$
$$\sum_{n=0}^{\infty } {(r+n-1)(r+n)}A_{n} x^{r + n} + \sum_{n=0}^{\infty } {(r+n)}A_{n} x^{r + n} + \sum_{n=1}^{\infty } 2A_{n-1} x^{r + n}$$
And finally:
$$\sum_{n=0}^{\infty } {(r+n-1)(r+n)}A_{n} x^{r + n} + \sum_{n=0}^{\infty } {(r+n)}A_{n} x^{r + n} + \sum_{n=1}^{\infty } 2A_{n-1} x^{r + n}$$
From this we get the indicial equation:
$$((r-1)(r)+r) = 0$$
Giving me $r=0$. Don't really think this is possible for singular points.
Thank you very much!
Why do you a-priori reject the case $r=0$ ?
Case $r=0$ : $$\sum_{n=1}^{\infty } n^2 A_{n} x^{n} + \sum_{n=1}^{\infty } 2A_{n-1} x^{n}=0$$ $$A_n=-\frac{2}{n^2}A_{n-1} \quad\to\quad A_n=\frac{(-1)^n 2^n}{(n!)^2}A_0 \qquad \qquad n\geq 1$$ $$y(x)=A_0\sum_{n=0}^{\infty}\frac{(-1)^n 2^n}{(n!)^2}x^n =c_1 J_0(2\sqrt{2x}) \qquad\qquad \text{Bessel function of first kind}.$$ where $c_1=A_0=$ any constant.
Of course, this method makes to loose the solutions which are not expandable in series of the form assumed $y = \sum_{n=0}^{\infty } A_{n} x^{r + n}$ , especially those which are singular at $x=0$ . This is the case of the solutions of the form : $$y=c_2Y_0(2\sqrt{2x})\qquad\qquad \text{Bessel function of second kind}.$$