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{\Large\bf Solution Procedure for Homogeneous Equations}

\vspace{.3in}

We are concerned with the second order linear homogeneous equation
\begin{equation}
   y''(x) + p(x)y'(x) + q(x)y(x) = 0   \label{eq:homo}
\end{equation}
on some interval \( [a,b] \).

\vspace{.2in}

If 
$y_{1}(x)$ and $y_{2}(x)$
 both solve \ref{eq:homo} and a \& b are
real numbers, then let 
$ v(x) = ay_{1}(x) + by_{2}(x) $.  
Substituting in \ref{eq:homo} gives
\[ v''(x) + p(x)v'(x) + p(x)q(x) = \]
\[ [ay''_{1}(x) +by''_{2}(x)] + p(x)[ay'_{1}(x) + by'_{2}(x)]
   + q(x)[ay_{1}(x) + by_{2}(x)] = \]
\[ a[y''_{1}(x) + p(x)y'_{1}(x) + q(x)y_{1}(x)]
   + b[y''_{2}(x) + p(x)y'_{2}(x) + q(x)y_{2}(x)] = 0.\]

This shows that any linear combination of solutions to the homogeneous
equation is again a solution.  In fact, if 
$y_{1}(x)$ and $y_{2}(x)$
 are two {\em linearly independent} solutions of the homogeneous equation,
then \underline{any} 
solution, 
$v(x)$
, of the homogeneous equation \ref{eq:homo} can be expressed as 
\[ v(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) \]
 where 
$c_{1}$ and $c_{2}$
 are constants.  Thus to find the general solution of a second order
linear homogeneous ODE, one needs only to find two linearly linearly 
independent solutions.

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