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$C = p^{2}$

When the constant is positive, the separable equations are
$\displaystyle X^{\prime\prime} - p^{2} X$ $\textstyle =$ $\displaystyle 0,$ (2.26)
$\displaystyle \ddot{T} - c^{2}p^{2} T$ $\textstyle =$ $\displaystyle 0.$ (2.27)

Equations (2.26) and (2.27) have constant coefficients and so can be solved in the usual way. For example, for (2.26), we set

\begin{displaymath}
X(x) = e^{\lambda x}, \qquad \Rightarrow \qquad
X^{\prime\prime}(x) = \lambda^{2}e^{\lambda x} = \lambda^{2}X(x).
\end{displaymath}

Substituting into (2.26) we get

\begin{displaymath}
\lambda^{2}X - p^{2}X = 0, \qquad \Rightarrow \qquad \lambda^{2} =
p^{2}.
\end{displaymath}

Thus,

\begin{displaymath}
\lambda = \pm p,
\end{displaymath}

and the solution to (2.26) is
\begin{displaymath}
X(x) = A e^{p x} + B e^{-p x}.
\end{displaymath} (2.28)

To satisfy the boundary conditions in $x$ we find that the only solution is the trivial solution with

\begin{displaymath}
A = 0 \qquad \hbox{and} \qquad B = 0.
\end{displaymath}



Prof. Alan Hood
2000-10-30