Subsection 7.2.1 Find a solution to \((x^2-1)y'' + y' + y = 0\) around \(x = 1\)
First things first, note that the point \(x=1\) is a regular singular point.
Question 7.2.1. Why is \(x=1\) a regular singular point?
In this case we have \(P(x) = x^2-1\text{,}\) \(Q(x) = 1\text{,}\) and \(R(x) = 1\text{.}\) So, \(q(x) = Q(x)/P(x) = 1/(x^2-1)\) and \(r(x) = R(x)/P(x) = 1/(x^2-1)\text{.}\)
Note that \(P(1) = 0\text{.}\) The two limits we need to calculate unfold as follows:
\begin{align*}
\lim_{x \to 1} (x-1) q(x) = \frac12 \amp \quad \text{and} \quad \lim_{x \to 1} (x-1)^2 r(x) = 0.
\end{align*}
Since these two limits are finite, \(x=1\) is a regular singular point.
Next, we observe the indicial equation: the root we want to use is \(\lambda_1 = 1/2\text{.}\)
Question 7.2.2. Why do we use \(\lambda_1 = 1/2\text{?}\)
From the previous question, we have that \(q_0 = 1/2\) and \(r_0 = 0\text{.}\) So, the indicial equation is
\begin{gather*}
0 = \lambda(\lambda - 1) + \frac12\lambda = \lambda^2 - \frac12\lambda = \lambda(\lambda - 1/2).
\end{gather*}
The bigger solution to the indicial equation is \(\lambda_1 = 1/2\text{.}\)
Note that we only need to find one solution to the ODE \((x^2 - 1)y'' + y' + y = 0\text{.}\) The solution we look for is of the form
\begin{align*}
y_1(x) \amp = (x-1)^{1/2}\sum_{n=0}^\infty a_n(x-1)^n = \sum_{n=0}^\infty a_n (x-1)^{n+1/2}.
\end{align*}
Before we continue, we should make things easier by using a substitution. Set \(t = x-1\) and \(Y_1(t) = y_1(x)\) and \(Y(t) = y(x)\text{.}\) This way, the ODE we need to solve is
\begin{align*}
t(t+2) Y''(t) + Y'(t) + Y(t) = 0 \amp
\end{align*}
and the solution we expect is
\begin{align*}
Y_1(t) \amp = \sum_{n=0}^\infty a_n t^{n+1/2}.
\end{align*}
Substitute this into our new ODE \(t(t+2) Y''(t) + Y'(t) + Y(t) = 0\) to find the recurrence relation
\begin{gather*}
\left(n^2 + \frac34\right)a_n + (n+1)(2n+3)a_{n+1} = 0 \qquad \text{for all $n \geq 0$}.
\end{gather*}
Question 7.2.3. How did you get that?
Here's the algebra that gets that recurrence relation:
\begin{align*}
0 \amp = (t^2 + 2t) \sum_{n=0}^\infty \left(n - \frac12\right)\left(n + \frac12\right) a_n t^{n-3/2} + \sum_{n=0}^\infty \left(n + \frac12\right)a_n t^{n-1/2} + \sum_{n=0}^\infty a_nt^{n+1/2},\\
\amp = \sum_{n=0}^\infty \left(n^2 - \frac14\right) a_n t^{n+1/2} + \sum_{n=0}^\infty 2\left(n^2-\frac14\right) a_n t^{n-1/2} + \sum_{n=0}^\infty \left(n + \frac12\right)a_n t^{n-1/2} + \sum_{n=0}^\infty a_nt^{n+1/2} \\
\amp = \sum_{n=0}^\infty \left[ n^2 + \frac34 \right] a_nt^{n+1/2} + \sum_{n=0}^\infty \big[2n^2+n\big] a_nt^{n-1/2}, \\
\amp = \sum_{n=0}^\infty \left[ n^2 + \frac34 \right] a_nt^{n+1/2} + \sum_{n=0}^\infty \big[(n+1)(2n+3)\big]a_{n+1}t^{n+1/2}\\
\amp = \sum_{n=0}^\infty \left[\left(n^2 + \frac34\right)a_n + (n+1)(2n+3)a_{n+1} \right]t^{n+1/2}.
\end{align*}
From this, we get the recurrence relation
\begin{gather*}
\left(n^2 + \frac34\right)a_n + (n+1)(2n+3)a_{n+1} = 0.
\end{gather*}
To finalize, one solution of \((x^2-1)y'' + y' +y = 0\) is
\begin{align*}
y_1(x) \amp = \sum_{n=0}^\infty a_n(x-1)^{n+1/2},
\end{align*}
where \(a_n\) satisfies
\begin{gather*}
\left(n^2 + \frac34\right)a_n + (n+1)(2n+3)a_{n+1} = 0.
\end{gather*}