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{\Large\bf Second-Order Linear Ordinary Differential Equations}

\vspace{.3in}

A linear second-order ODE has the form:
$$
S(t)y''(t) + P(t)y'(t) + Q(t)y(t) = F(t).
$$
On any interval where S(t) is not equal to 0, the above equation
can be divided by S(t) to yield
$$
y''(t) + p(t)y'(t) + q(t)y(t) = f(t)
$$
The equation is called homogeneous if f(t)=0. Otherwise, it is
called nonhomogeneous.

\vspace{.3in}

{\bf Existence and Uniqueness}

\vspace{.2in}

A second-order differential equation is accompanied by initial
conditions or boundary conditions. Initial conditions are in
the form y(t0)=y0 and y'(t0)=y'0. Boundary conditions
might be of the form: y(t0)=a and y(t1)=b.

For the initial value problem, the existence and uniqueness
theorem states that if p(t), q(t) and f(t) are continuous on
some interval (a,b) containing t0, then there exists a unique
solution y(t) to the ode in the whole interval (a,b).

\vspace{.3in}

{\bf Procedure for Solving Linear Second-Order ODE}

\vspace{.2in}

The procedure for solving linear second-order ode has two steps
\begin{itemize}
\item[(1)] Find the general solution of the homogeneous problem:
$$
y''+p(t)y'+q(t)y=0
$$
According to the theory for linear differential equations, the
general solution of the homogeneous problem is
$$
y_H(t)=C_1y_1(t)+C_2y_2(t)
$$
where C1 and C2 are constants and y1 and y2 are any two
linearly independent solutions to the homogeneous equation.

\item[(2)] Find a particular solution of the nonhomogeneous
problem:
$$
y''+p(t)y'+q(t)y=f(t)
$$
The particular solution is any solution of the nonhomogeneous
problem and is denoted yp(t).
\end{itemize}

The general solution of the full nonhomogeneous problem is
$$
y(t)=C_1y_1(t)+C_2y_2(t)+y_p(t)
$$
The key point to note is that {\it all possible} solutions 
to a linear second-order ode can be obtained from two
linearly independent solutions to the homogeneous problem
and any particular solution.

Here is an example. Consider the ode

$$
y''+3y'+2y=6e^t
$$

The homogeneous equation is 

$$
y''+3y'+2y=0
$$

It can be shown that y_1=exp(-t) and y_2=exp(-2t) are solutions
to the homogeneous equation. Plug these expressions into the
ode and verify! 

A particular solution of the nonhomogeneous equation is exp(t).
Hence, the general solution of the ode is

$$
y(t)=C_1e^{-t}+C_2e^{-2t}+e^t
$$

where C1 and C2 are constants.

\vspace{.3in}

{\bf Linear Dependence}

\vspace{.2in}

Two functions linearly independent if they are not multiples
of each other. For example, exp(-t) and exp(-2t) are linear
independent. On the other hand, t+3 and 7t+21 are linearly
dependent.

\vspace{.3in}

{\bf Techniques for Solving Homogeneous Linear Second-Order ODE}

\vspace{.2in}

Certain classes of homogeneous linear second-order ode can
be solved analytically. We will consider two classes:

\begin{itemize}

\item[(1)] Constant-coefficient ode: 
$$
y''+py'+qy = 0
$$
p(t) and q(t) are constants.

\item[(2)] Euler-Cauchy ode:
$$
y''+\frac{p}{t}y'+\frac{q}{t^2}y=0
$$
p and q are constants.
\end{itemize}

\vspace{.3in}

{\bf Techniques for Determining a Particular Solution}

\vspace{.2in}

There are two principal techniques for determining
a particular solution:
\begin{itemize}
\item[(1)] The Method of Undetermined Coefficients
\item[(2)] Variation of Parameters
\end{itemize}

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