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Section 5.2 Examples of Regular Singular Points

Subsection 5.2.1 Find the regular singular points of the ODE \(x^2 y'' + x y' + x^2 y = 0\text{.}\)

Matching this up with \(P(x) y'' + Q(x) y' + R(x) y = 0\text{,}\) we see that \(P(x) = x^2\text{,}\) \(Q(x) = x\text{,}\) and \(R(x) = x^2\text{.}\) Note that all of these are (technically) power series in \(x\text{.}\)

Next, we find the singular points of the ODE, which starts by setting \(P(x) = x^2 = 0\text{.}\) Thus, the only possible singular point is \(x = 0\text{.}\)

Next, we check if \(x = 0\) is regular. As in the method, define \(q(x) = Q(x)/P(x) = x/x^2 = 1/x\) and \(r(x) = R(x)/P(x) = x^2/x^2 = 1\text{.}\) Then, we'll compute the following limits:

\begin{gather*} \lim_{x \to 0} x q(x) = \lim_{x \to 0} x \cdot \frac1x = 1, \end{gather*}

and

\begin{gather*} \lim_{x \to 0} x^2 r(x) = \lim_{x \to 0} x^2 \cdot 1 = 0. \end{gather*}

Since both of these limits are finite, \(x = 0\) is a regular singular point.

Subsection 5.2.2 Find the regular singular points of \((x-1)^2(x+1)y'' + xy' + (x-1)y = 0\)

Try this one yourself! We have \(P(x) = (x-1)^2 (x+1)\text{,}\) \(Q(x) = x\text{,}\) and \(R(x) = x-1\) in this case.

Start by setting \(P(x) = 0\) and solving for \(x\) to find the possible singular points of the ODE. You'll see that \(x = 1, -1\text{.}\)

Next, find \(q(x) = Q(x)/P(x)\) and \(r(x) = R(x)/P(x)\text{.}\) Calculate the two limits

\begin{gather*} \lim_{x \to 1} (x - 1)q(x) \qquad \text{and} \qquad \lim_{x \to 1} (x-1)^2 r(x), \end{gather*}

and the two limits

\begin{gather*} \lim_{x \to -1} (x + 1)q(x) \qquad \text{and} \qquad \lim_{x \to -1} (x+1)^2 r(x). \end{gather*}

Which of these limits are finite?

Answer

\(x = -1\) is a regular singular point and \(x = 1\) is not.