Subsection 4.3.1 When \(x=0\) is a Regular Singular Point of the Equation
In this case, the solution is
\begin{equation*}
y = \sum\limits_{\lambda =0}^{\infty}a_{\lambda}x^{\lambda+k}
\end{equation*}
and on substituting of \(y{,}y'\) and \(y"\) in the given differential equation we get the identity. The index \(k\) will be determined by the quadratic equation obtained by equating the coefficient of lowest power of \(x\) to zero. The quadratic equation in \(k\) is called the Indicial equation. Thus, we will get two values of \(k\text{.}\) The series solution of equation (4.3.1) will depend upon the nature of the roots of the indicial equation.
Case I. When roots (\(k_{1}{,}k_{2}\)) are distinct and not differing by an integer, then the complete solution is
\begin{equation*}
y=C_{1}(y)_{k_{1}}+C_{2}(y)_{k_{2}}
\end{equation*}
Case II. When roots are equal (\(k_{1}=k_{2}\)), then
\begin{equation*}
y=C_{1}(y)_{k_{1}}+C_{2}(\frac{dy}{dk})_{k_{1}}.
\end{equation*}
Case III. When roots are distinct and differ by an integer, then there arise two possibilities
- If some of the coefficients of y series become infinite at \(k=k_{2} \text{,}\) then replace \(a_{o}\) by \(b_{o}(k-k_{2})\text{.}\) The complete solution is then\begin{equation*} y=C_{1}(y)_{k_{1}}+C_{2}(\frac{dy}{dk})_{k_{2}} \end{equation*}[on taking \(k=k_{1}\text{,}\) we obtain a solution which is a constant multiple of that obtained for \(k_{1}\text{.}\)]
- When the coefficient of y series does not become infinite for any \(k\text{,}\) then the complete solution is\begin{equation*} y=C_{1}(y)_{k_{1}}+C_{2}(y)_{k_{2}} \end{equation*}
