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{\Large\bf Constant-Coefficient Homogeneous ODE}

\vspace{.3in}

A constant-coefficient homogeneous second-order ode can be put
in the form
$$
y''(t) + py'(t) + qy(t) = 0 
$$
where p and q are constants. Recall that the general solution is 
$$
y(t)=C_1y_1(t)+C_2y_2(t)
$$
where C_1 and C_2 are constants and y_1(t) and y_2(t) are any
two linearly independent solutions of the ode. Our goal is
to find two linearly independent solutions of the ode. 

For reasons that become clear below, we try a solution of the
form y=exp(rt), where r is an unknown constant. We must find r.
If y=exp(rt), then y'=r(exp(rt)) and y''=r^2(exp(rt)).
Substituting these into the ode, we have 
$$
r^2e^{rt}+pre^{rt}+qe^{rt}=e^{rt}(r^2+pr+q)=0
$$
This equation is satisfied if exp(rt)=0 or r^2+pr+q=0.
The condition exp(rt)=0 is satisfied only if rt is negative
infinity. This is considered a degenerate case and is
neglected. Hence, r must satisfy the equation
$$
F(r)=r^2+pr+q=0
$$
F(r) is called the characteristic polynomial. Solving the
original ode is reduced to solving an algebraic equation. 

Assuming the coefficients p and q are real numbers, there
are three cases to consider: 

\begin{itemize}
\item Characteristic polynomial has distinct roots 
\item Characteristic polynomial has a double root
\item Characteristic polynomial has complex conjugate roots  
\end{itemize}

\vspace{.3in}

{\bf Characteristic Polynomial has Distinct Roots}

\vspace{.2in}

Suppose that the characteristic polynomial has two distinct,
real roots (call them a and b). Then exp(at) and exp(bt) are 
linearly independent solutions to the original ode and
the general solution to the ode is:
$$
y(t) = C_1e^{at}+C_2e^{bt}
$$

Consider the following example:
$$
y''-3y'-18y=0
$$
The characteristic polynomial is r^{2} - 3r -18 = (r - 6)(r +3),
which has roots 6 and -3. 
$$
y(t)=C_1e^{6t}+C_2e^{-3t}
$$

\vspace{.3in}

{\bf Characteristic Polynomial has a Double Root}

\vspace{.2in}

Suppose that the characteristic polynomial has a double root
(call it a). y_1=exp(at) is one solution to differential equation.
We need a second linearly independent solution to the ode to
get the general solution. Using the technique of reduction of
order, it can be shown that texp(at) is also a solution of the
ode. The general solution is 
$$
y(t)=C_1e^{at}+C_2te^{at}
$$

Consider the following example: 
$$
y'' + 6y' + 9y = 0.
$$
The characteristic polynomial is r^2 + 6r + 9 = (r + 3)^2,
which has a double root -3. The general solution is 
$$
y(t) = C_1e^{-3t} + C_2te^{-3t}
$$

\vspace{.3in}

{\bf Complex-Conjugate Roots}

\vspace{.2in}

Suppose that the characteristic polynomial has complex
roots a+ib and a-ib, where a and b are real. These are distinct
roots, so the the general solution can be written:
$$
y(t) = C_1 e^{(a+ib)t}+C_2e^{(a-ib)t} 
$$
The problem with writing the solution in this form is
that it involves complex-valued functions. It is possible
to re-express the general solution in terms of two linearly
independent real-valued functions. 

Recall, Euler's identity:
$$
e^{iy}=\cos(y)+i\sin(y)
$$
Using this identity, we have:
$$
e^{(a+ib)t}=e^{at}e^{ibt}=e^{at}(\cos(bt)+i\sin(bt))
$$
Similarly, we have
$$
e^{(a-ib)t}=e^{at}e^{-ibt}=e^{at}(\cos(bt)-i\sin(bt))
$$

Substituting, these two expressions into the general solution
we have
$$
y(t) = C_1e^{at}(\cos(bt)+i\sin(bt)) + C_2e^{at}(\cos(bt)-i\sin(bt))
$$

Choose C_1=0.5 and C_2=0.5. This yields the solution:
$$
y(t)=e^{at}\cos(bt)
$$
Now choose C_1=i0.5 and C_2=-i0.5. This yields the solution:
$$
y(t)=e^{at}\sin(bt)
$$
Both of these functions are solutions to the original ode. In
addition, they are linearly independent, since they are not
multiples of each. Hence, any solution to the ode can be
expressed in terms of these function. So we can write: 
$$
y(t) = D_1e^{at}\cos(bx) + D_2e^{at}\sin(bt)
$$
for real numbers D_1 and D_2 are constants.

Consider the example:

$$
y'' - 6y' + 13 = 0
$$

The characteristic polynomial is $r^{2} - 6r + 13$.  Using the
quadratic formula, we find that the roots are 3 + 2i and 3 - 2i.
Hence, the general solution can be written:
$$
y(t)=D_1e^{3t}\cos(2t)+D_2e^{3t}\sin(2t)
$$

\end{document}
