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\title{Math 1b.  Lecture 34 \\
Second-Order Homogeneous Differential Equations with Constant Coefficients (II)}

\author{T. Judson}

\date{Spring 2006}


\begin{document}

\maketitle


\section{Goals}

\begin{itemize}


\item
To be able to solve equations of the form $ax'' + bx' + cx = 0$, when the roots of the characteristic equations and repeated and real or complex.



\end{itemize} 

\section{The Complex Case}

Now let us solve the initial value problem
\begin{eqnarray*}
x'' + 4 x' + 5 x & = & 0 \\
x(0) & = & 3 \\
x'(0) & = & -2.
\end{eqnarray*}
Again, we will assume that our solution has the form $x(t) = e^{rt}$.  Substituting this function into our differential equation, we find that
$$
0
=
x'' + 4 x' + 5 x
=
r^2 e^{rt} + 4r e^{rt} + 5 e^{rt}
=
(r^2 + 4r + 5) e^{rt}.
$$
As with the real case, $r^2 + 4r + 5 = 0$ or
$$
r = \frac{-4 \pm \sqrt{-4}}{2} = - 2 \pm i.
$$
Thus, we can find a solution
$$
x(t) 
=
e^{(-2 + i)t}
=
e^{-2t} e^{it}
=
e^{-2t} (\cos t + i \sin t).
$$
The real and imaginary parts of this solution are also solutions to our differential equation.  Therefore, the general solution to our equation is
$$
x(t) = c_1 e^{-2t} \cos t + c_2 e^{-2t} \sin t.
$$
Applying our initial conditions, we can quickly calculate $c_1 =3$ and $c_2= 4$.  Hence, the solution to our initial value problem is
$$
x(t) = 3 e^{-2t} \cos t + 4 e^{-2t} \sin t.
$$

If  $b^2 - 4ac < 0$, the differential equation 
$$
a x'' + b x' + c x = 0
$$
has a general solution 
$$
x(t) = c_1 e^{\alpha t} \cos \beta t + c_2 e^{\alpha t} \sin \beta t,
$$
where $\alpha \pm i \beta$ are the roots of $a r^2 + br + c = 0$.  


\section{Repeated Roots}


Finally, we will consider the equation
$$
x'' + 2x' + x = 0.
$$
If we choose $e^{rt}$ as our guess, we find
\begin{eqnarray*}
x'' + 2 x' + x
& = &
r^2 e^{rt} + 2r e^{rt} + e^{rt} \\
& = &
e^{rt} (r+ 1)^2 \\
& = &
0.
\end{eqnarray*}
Thus, $r = -1$ and we have a solution $x_1(t) = e^{-t}$.  

In order to find a general solution to $x'' + 2x' + x = 0$, we must find a second solution that is not a multiple of $x_1(t) = e^{-t}$.  Since we already know that $c x_1(t)$ is a solution to our differential equation, we will try to generalize this observation by replacing $c$ with a function $v(t)$ and then try to determine $v(t)$ so that $v(t) x_1(t)$ is a solution to $x'' + 2x' + x = 0$.
Indeed, if 
$$
x(t) = v(t) x_1(t) = v(t) e^{-t},
$$
then
$$
x'(t) 
=
v(t) x_1'(t) + v'(t) x_1(t)
=
-v(t) e^{-t} + v'(t) e^{-t}
$$
and
\begin{eqnarray*}
x''(t)
& = &
v''(t) x_1(t) + 2v'(t) x_1'(t) + v(t) x_1''(t) \\
& = &
v''(t) e^{-t} - 2 v'(t) e^{-t} + v(t) e^{-t}.
\end{eqnarray*}
Consequently,
\begin{eqnarray*}
x'' + 2 x' + x
& = &
[v'' e^{-t} - 2 v' e^{-t} + v e^{-t}]
+
2 [-v e^{-t} + v' e^{-t}] 
+
[v e^{-t}] \\
& = &
e^{-t} v'' \\
& = & 
0,
\end{eqnarray*}
and $v'' = 0$.  Therefore, $v = c_1 t + c_2$.  Letting $c_1 = 1$ and $c_2 = 0$, we can assume that $v(t) = t$, and the second solution to our equation is $x = t e^{-t}$.  Hence, the general solution to  $x'' + 2x' + x = 0$ is
$$
x(t) = c_1 e^{-t} + c_2 t e^{-t},
$$
which agrees with the solution that we would obtain from solving the system equivalent to this second-order differential equation.





\section{Worksheet Problems}

\begin{enumerate}

\item
Find the general solution of the equation $y'' + 4y = 0$

\item
Find the general solution of the equation $y'' + 2y' + 17y = 0$

\item
Find the general solution of the equation $y'' - 6y' + 9y = 0$


\item
Solve the initial value problem
\begin{eqnarray*}
y'' +2 y' + 3y & = & 0 \\
y(0) & = & 1 \\
y'(0) & = & 0
\end{eqnarray*}



\item
Solve the initial value problem
\begin{eqnarray*}
y'' - 4y' + 13y & = & 0 \\
y(0) & = & 4 \\
y'(0) & = & 0
\end{eqnarray*}


\item
Solve the initial value problem
\begin{eqnarray*}
y'' + 10y' + 25y & = & 0 \\
y(0) & = & 2 \\
y'(0) & = & -1
\end{eqnarray*}




\end{enumerate}





\section*{References}

\begin{itemize}



\item
\S 31.6 in
Robin J. Gottlieb. 
{\it Calculus: An Integrated Approach to Functions and Their Rates of Change}, preliminary edition. 
Addison Wesley, Boston, 2002.
ISBN 0--201--70929--5.


\end{itemize}



\section*{Notes}

\subsection*{May 3, 2006}



\end{document}

