Simple Frobenius problem without recurrence relation? I am just learning frobenius method in my 'math methods in physics' class. The first problem i am trying to solve is $$ x^2y''-xy'+n^2y=0$$ (where n is a constant).
I know that i have to plug in the maclaurin expansion of y which results in me getting this summation
$$y(z) = \sum_{k = 0}^{ \infty} a_kx^{(k+s)}[n^2-(k+s)+(k+s)(k+s-1)] = 0$$
[
by asserting that $$ a_0 \neq 0 $$ is non-zero i solved for s and got $$ s=1 \pm \sqrt{1-n^2} $$
Now i am stuck on how to continue to get the coefficients a_1, a_2, and so on. I remember seeing an example that solved for a recurrence relation but since every term had the same power of x when we plugged it back in we cant do that method.
How do i continue?
 A: Given
$$\tag 1 x^2y''-xy'+n^2y=0$$ 
where n is a constant, solve the DEQ by the method of Frobenius.
Method I Euler-Cauchy
We see that this is a Euler-Cauchy equation, so lets choose:
$$y = x^r \rightarrow y' = r x^{r-1} \rightarrow y'' = r(r-1)x^{r-2}$$
Plugging this back into $(1)$, yields:
$$x^2(r)(r-1)x^{r-1} - x(r)x^{r-1} + n^2 x^r = (r(r-1) + r + n^2)x^r = 0 \rightarrow r_{1,2} = 1 \pm \sqrt{1-n^2}$$
This gives us the linear combination of a solution as:
$$y(x) = c_1 x^{1 + \sqrt{1-n^2}} + c_2x^{1 - \sqrt{1-n^2}}$$
Method II Frobenius Method
Using the Method of Frobenius, we have:
$$\tag 2 y = x^s \sum_{k=0}^\infty a_k~x_k$$
This gives us:
$$\tag 3 y' = \sum_{k=0}^\infty (k+s)~ a_k~x^{k+s-1}$$
$$\tag 4 y'' = \sum_{k=0}^\infty (k+s)(k+s-1)~a_k~x^{k+s-2}$$
Now, in order to calculate the coefficients, lets set up a table of $x$ powers for each term in the DEQ as:
$$\begin{array}{c|c|c|c|c} 
\text{} & x^s & x^{s+1} & x^{s+2} & \ldots & x^{k+s} \\ \hline
x^2 y'' & s(s-1)~a_0 & (s+1)s~a_1 & (s+2)(s+1)~a_2 & \ldots & (k+s)(k+s-1)~a_k \\ \hline
-x y' & -s~a_0 & -(s+1)~a_1 & -(s+2)~ a_2 & \ldots & -(k+s)~a_k\\ \hline
n^2 y & n^2~a_0 & n^2~a_1 & n^2~a_2 & \ldots & n^2~a_k \\ \hline
\end{array}$$
The total coefficient of each power of $x$ must be zero. If we sum the terms in column $x^s$ for $a_0$, we get:
$$(s^2 -2s+ n^2)a_0 = 0  \rightarrow s^2 -2s+ n^2 = 0 \rightarrow s_{1,2} = 1 \pm ~\sqrt{1-n^2}$$
This equation for $s$ is called the indicial equation and we solve it to find our values of $s$.
Next, if we sum the general column for $a_k$, we get (regardless of substituting $s_{1,2}$:
$$(k^2+2 k s-2 k+s^2-2 s + n^2)~a_k = 0 \rightarrow a_k = 0$$
Since $n$ is a constant.
So, from $(1)$, we have:
$$y(x) = x^s \sum_{n=0}^\infty a_n~x_n = a_0 x^s \rightarrow y = c_1 x^{1+\sqrt{1-n^2}} + c_2x^{1-\sqrt{1-n^2}}$$
