[NOTES/ODE-02004] Series Solution Case-IV

In this lecture we shall take up solution of an ordinary differential equation by the method of series solution. The example to be discussed is such that the difference of the roots of the indicial equation is an integer and some coefficient becomes indeterminate.\\ \linebreak Consider the differential equation \begin{equation} \label{E1} \frac{{\mathrm d^2}y}{\mathrm{d}x^2} + x^2 y = 0 \end{equation} Substituting \begin{equation} \label{E2} y = \sum_{n=0}^{\infty}a_nx^{n+c} \end{equation} in \eqref{E1} we get \begin{equation} \label{E3} \sum_{n=0}^{\infty}a_n(n+c)(n+c-1)x^{n+c-2}+ x^2 \sum_{n=0}^{\infty}a_nx^{n+c} = 0 \end{equation} or, \begin{equation} \label{E4} \sum_{n=0}^{\infty}a_n(n+c)(n+c-1)x^{n+c-2}+ \sum_{n=0}^{\infty}a_nx^{n+c+2} = 0 \end{equation} The lowest power of $ x $ in the right hand side of \eqref{E4} is $ x^{c-2} $. This gives \begin{equation} \label{E5} a_0 c (c-1) =0 \end{equation} Therefore the two values of c are $ c=0$ and $c=1 $. Equating the coefficients of $ x^{c-1},x^c,x^{c+1},x^{c+2}, \ldots $ to zero successively gives \begin{equation} \label{E6} a_1 c(c+1)=0 \end{equation} \begin{equation} a_2(c+1)(c+2)=0 \label{E7} \end{equation} \begin{equation} a_3(c+2)(c+3)=0 \label{E8} \end{equation} \begin{equation} a_4(c+4)(c+3)+ a_0=0\label{E9} \end{equation} The recurrence relation obtained by considering the coefficient of $ x^{m+c+2} $ is \begin{equation} \label{E10} a_{m+4}(c+m+4)(c+m+3)+ a_m = 0 \end{equation} The solution for $ c= 1 $ can be constructed as before. For the case $ c=0 $, however, we get from\eqref{E6} \begin{equation} a_1 . 0 = 0 \label{E11} \end{equation} Thus $ a_1 $ cannot be fixed and is indeterminate. In this case we proceed as before except that we retain both $ a_0 $ and $ a_1 $ as unknown parameters.\\ \subsubsection*{Case $c=0:$} Substituting $ c=0 $ from \eqref{E6} to \eqref{E10} we get \begin{eqnarray} \label{E12}a_2 = a_3=0;\quad a_4 = -\frac{a_0}{4.3}\\ a_{m+4}= -\frac{a_m}{(m+4)(m+3)}\label{E13} \end{eqnarray} Combining \eqref{E12} and \eqref{E13} we see that \begin{equation} \label{E14} a_2 = a_6 = a_10 \cdots 0 \end{equation} and \begin{equation} \label{E15} a_3 = a_7 = a_11 \cdots 0 \end{equation} Also \begin{equation} \label{E16} a_4 = - \frac{1}{4.3}a_0 ;\quad a_8 = - \frac{1}{8 . 7}a_4; \quad a_{12} = - \frac{1}{12. 11}a_8 \end{equation} \begin{equation} \label{E17} a_5 = - \frac{1}{5.4}a_1 ;\quad a_9 = - \frac{1}{9 . 8}a_5; \quad a_{13} = - \frac{1}{13. 12}a_9 \end{equation} Solving \eqref{E16} and \eqref{E17} we get \begin{equation} \label{E18} a_4 = - \frac{1}{4.3}a_0 ;\quad a_8 = - \frac{1}{8 . 7.4.3}a_0; \quad a_{12} = - \frac{1}{12. 11.8.7.4.3}a_0 \end{equation} \begin{equation} \label{E19} a_5 = - \frac{1}{5.4}a_1 ;\quad a_9 = - \frac{1}{9 . 8.5.4}a_1; \quad a_{13} = - \frac{1}{13. 12.9.8.5.4}a_1 \end{equation} The series solution in this case contains two parameters, which are not determined by the recurrence relations, and is given by \begin{equation} \label{E20} y(x)= a_0 y_1(x) +a_1 y_2(x) \end{equation} \begin{equation} \label{E21} y_1(x)= 1- \frac{x^4}{3.4}+\frac{x^8}{3.4.7.8}- \frac{x^{12}}{3.4.7.8.11.12}+\cdots \end{equation} \begin{equation} \label{E22} y_2(x)= x\left\{ 1- \frac{x^4}{4.5}+\frac{x^8}{4.5.8.9}- \frac{x^{12}}{4.5.8.9.12.13}+\cdots \right\} \end{equation} These two functions $ y_1(x) $ and $ y_2(x) $ represent two linearly independent solutions. What happens when one tries to construct the solution for the second value of $ c $ ? In this case we recover one of the above two solutions already obtained. This will now be demonstrated explicitly. Case $c=1:$ In this case we get \begin{equation} \label{E23} a_1=a_2 = a_3 =0 \end{equation} \begin{equation} \label{E24} a_{m+4}= - \frac{a_m}{(m+5)(m+4)} \end{equation} We therefore get \begin{equation} \label{E25} a_4= - \frac{1}{5.4}a_0; \quad a_8= - \frac{1}{9.8}a_4; \quad a_{12}= - \frac{1}{13.12}a_8 \end{equation} Compare the equations \eqref{E25} with \eqref{E17} . We now construct the series \begin{equation} \label{E26} y= x^c \sum a_n x^n \end{equation} and get \begin{equation} \label{E27} y_2(x)= a_0 x\left\{ 1- \frac{x^4}{4.5}+\frac{x^8}{4.5.8.9}- \frac{x^{12}}{4.5.8.9.12.13}+\cdots \right\} \end{equation} This solution coincides with $ y_2(x) $ of \eqref{E22} except for an overall constant. Hence the most general solution of the differential equation \eqref{E1} is given by \eqref{E20}