Section 7.2 An Involved Example
Subsection 7.2.1 Find a solution to
First things first, note that the point
Question 7.2.1. Why is
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.
Question 7.2.2. Why do we use
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 \(\lambda_1 = 1/2\text{.}\)
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*}