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\noindent
Math 325

\noindent
Spring, 2001

\bigskip

\begin{center}
SOLVING FIRST ORDER LINEAR CONSTANT COEFFICIENT EQUATIONS
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\bigskip

In section 2.1 of Boyce and DiPrima, you learned how to solve a 
first order linear ordinary differential equation using an integrating factor
(typically called 
\begin{math}
  \mu
\end{math}).  
Such an equation has the form 
\begin{math}
  y'+p(t)y=g(t).
\end{math}
This method works for any first order linear ODE.

However, if the equation happens to be constant coefficient and the
function 
\begin{math}
  g
\end{math}
is of a particularly simple form, there is another way to think about
the problem.  The equation has the form 
\begin{equation}
\label{eqn}
  y'+ay=g(t)
\end{equation}
where 
\begin{math}
  a
\end{math}
is a constant.  You can think of this as a special case of an 
\begin{math}
  n
\end{math}th
order linear inhomogeneous ODE (with 
\begin{math}
  n=1
\end{math}).
If you think of it that way, you can solve it the same way you solve higher
order constant coefficient linear ODEs.  Here's a sketch.

\emph{Step 1}
Solve the corresponding homogeneous equation
\begin{equation}
\label{homogeneous}
  y'+ay=0
\end{equation}
by looking for a solution of the form
\begin{math}
  y=Ce^{rt}.
\end{math}
You find that 
\begin{math}
  r=-a.
\end{math}
So the general solution to (\ref{homogeneous}) is
\begin{displaymath}
  y_c= Ce^{-at}.
\end{displaymath}


Now, back to the original equation, (\ref{eqn}).  The general
solution will be of the form 
\begin{displaymath}
  y=y_c+y_p
\end{displaymath}
where 
\begin{math}
  y_p
\end{math}
is a particular solution, that is, one solution you will find somehow.
Step 2 will apply if 
\begin{math}
  g(t)
\end{math}
is of a particularly nice form.
Suppose 
\begin{displaymath}
  g(t)=p(t)e^{-at}
\end{displaymath}
where 
\begin{math}
  p(t)
\end{math}
is a polynomial of degree 
\begin{math}
  k.
\end{math} 

\emph{Step 2} Use the 
\textbf{method of undetermined coefficients}.
Look for a particular solution of the form 
\begin{displaymath}
  y_p=t(A_0t^k+A_1t^{k-1}+\dots A_{k-1}t + A_k)e^{-at},
\end{displaymath}
that is,  
\begin{math}
  t
\end{math}
times a general polynomial of degree 
\begin{math}
  k,
\end{math}
with undetermined coefficients which you need to determine, times an
exponential.  (You need the factor of 
\begin{math}
  t
\end{math}
in front because the exponential term solves the homogeneous equation 
(\ref{homogeneous}).)  Plug 
\begin{math}
  y_p
\end{math}
into the original equation (\ref{eqn}).  Then equate corresponding
terms.
This will give you 
\begin{math}
  k+1
\end{math}
equations for the 
\begin{math}
  k+1
\end{math}
undetermined coefficients 
\begin{math}
  A_0,\dots,A_k.
\end{math}
Solve these equations to determine the coefficients.  Now you have
found 
\begin{math}
  y_p.
\end{math}

(You can actually handle somewhat more general forms of 
\begin{math}
  g(t),
\end{math}
any form that can be handled for 
\begin{math}
  n
\end{math}th
order equations by the method of undetermined coefficients, but this
is the form of 
\begin{math}
  g(t)
\end{math}
which comes up when you are solving a system of the form 
\begin{math}
  \mathbf{x}'=J\mathbf{x}
\end{math}
where 
\begin{math}
  J
\end{math}
is a matrix in Jordan canonical form.)

\emph{Step 3}
The general solution to (\ref{eqn}) is
\begin{displaymath}
  y=Ce^{-at}+t(A_0t^k+A_1t^{k-1}+\dots A_{k-1}t + A_k)e^{-at}
\end{displaymath}
where 
\begin{math}
  A_0,\dots,A_k
\end{math}
are the coefficients you found in Step 2.

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