\magnification=\magstep1
\centerline{Mathematics 304, section 1 -- Ordinary Differential Equations}
\centerline{Exam 2 Solutions}
\centerline{November 22, 2004}
\bigskip
\noindent
I.  A)  With $a = 10$, we have $A = \pmatrix{-1 & 10 \cr -1 & 0\cr}$
which has trace $T = -1$ and determinant $D = 10$.  Since
$D > T^2/4$, we are in the region of the trace-determinant
plane corresponding to systems with {\it spiral sinks} at the origin.

B)  With $a = -2$, we have $T = -1$, $D = -2$.  The characteristic
polynomial is $\lambda^2 + \lambda - 2 = (\lambda + 2)(\lambda - 1)$
so the eigenvalues are $\lambda = -2, 1$.  For $\lambda = -2$,
$$\ker(A + 2I) = \ker \pmatrix{1 & -2 \cr -1 & 2\cr} = 
{\rm span}\pmatrix{2\cr 1\cr}.$$  
 For $\lambda = 1$,
$$\ker(A - I) = \ker \pmatrix{-2 & -2 \cr -1 & -1\cr} = 
{\rm span}\pmatrix{-1\cr 1\cr}.$$
So the general solution looks like 
$$X = c_1 e^{-2t}\pmatrix{2\cr 1\cr} + c_2 e^t \pmatrix{-1\cr 1\cr},$$
where $c_1$ and $c_2$ are arbitrary constants.  Since the signs of 
the eigenvalues are opposite, the origin is a {\it saddle point}
for this system.

C)  As $a$ varies over the real numbers, the motion in 
the trace-determinant plane takes us ``up'' the vertical
line with $T = -1$.  The systems in the family have
saddle points if $a < 0$ (like $a = -2$ in part B), 
a line of equilibria if $a = 0$,
an ordinary sink if $0 < a < 1/4$, a sink from a repeated
real root if $a = 1/4$, and spiral sinks for all $a > 1/4$
(like $a = 10$ in part A).
\bigskip
\noindent
II.  A) By the usual linearity of differentiation, We have 
$$\eqalign{
&m(x_{p,1}+x_{p,2})'' + b(x_{p,1}+x_{p,2})' + k(x_{p,1}+x_{p,2})\cr
&=m x_{p,1}''+mx_{p,2}'' + bx_{p,1}'+bx_{p,2}' + kx_{p,1}+kx_{p,2}\cr
&=(m x_{p,1}''+ bx_{p,1}'+kx_{p,1})+(mx_{p,2}'' +bx_{p,2}' + kx_{p,2})\cr
&=g_1(t)+g_2(t)\cr}$$
Hence $x = x_{p,1}(t) + x_{p,2}(t)$ is a particular
solution of $m x'' + b x' + k x = g_1(t) + g_2(t)$.

B) The characteristic polynomial of $x'' + 4x' + 3x = 0$
is $\lambda^2 + 4\lambda + 3 = (\lambda + 1)(\lambda + 3)$
so $\lambda = -1, -3$.  The general solution of the 
homogeneous equation is 
$$x_h = c_1 e^{-t} + c_2 e^{-3t}$$
where $c_i$ are arbitrary constants.  Next we find
particular solutions for $g_1(t) = e^{7t}$ and 
$g_2(t) = 4e^{-t}$.  For $g_1(t)$ (not a solution of 
the homogeneous equation), we take $x_{p,1}(t) = Ae^{7t}$.
Then
$$x_{p,1}''+ 4x_{p,1}'+3x_{p,1} = (49+28+3)A e^{7t} = 80Ae^{7t} = e^{7t}$$
implies $A = 1/80$.  For $g_2(t)$ (which {\it is} a solution of 
the homogeneous equation), we take $x_{p,2}(t) = Bte^{-t}$.
Then
$$x_{p,2}''+ 4x_{p,2}'+3x_{p,2} = 0\cdot te^{-t} + 2Be^{-t} = 4e^{-t}$$
implies $B = 2$.  By part A, our general solution looks
like
$$x = x_h + x_{p,1} + x_{p,2} = 
c_1 e^{-t} + c_2 e^{-3t} + {1\over 80} e^{7t} + 2 t e^{-t}$$  
The final step is to solve for $c_1,c_2$ from the initial 
conditions.  We get
$$\eqalign{x(0) = 0 &= c_1 + c_2 + {1\over 80}\cr
          x'(0) = 1 &= -c_1 - 3c_2 + {7\over 80} + 2\cr}$$
Solving simultaneously, 
$$c_1 = {-45\over 80} = {-9\over 16}, \quad c_2 = {11\over 20}.$$
\bigskip
\noindent
III.  Resonance occurs in solutions of an undamped, 
sinusoidally forced harmonic 
oscillator equation when the frequency of the forcing 
term is the same as the natural frequency of the solutions
of the homogeneous equation.  For $4 x'' + 400 x = 0$,
we have $4 \lambda^2 + 400 = 0$, which implies $\lambda = \pm 10 i$.
The natural frequency is $10$, so resonance occurs for 
$\omega = 10$.  The {\it reason} this happens is that
the particular solution of the equation with the 
forcing term is
$$x_p = A t \cos(\omega t)$$
for some $A$.
The extra factor of $t$ means that the solutions are not
bounded as $t \to +\infty$.

\bigskip
\noindent
IV.  For matrices as described in the problem, 
there are one block for $\lambda = 1$, and 2
blocks for $\lambda = 2$.  So there
are exactly {\it two} possible canonical forms:
$$\pmatrix{1& 0& 0& 0\cr
           0& 2& 1& 0\cr
           0& 0& 2& 0\cr
           0& 0& 0& 2\cr}, \quad
\pmatrix{1& 1& 0& 0\cr
           0& 1& 0& 0\cr
           0& 0& 2& 0\cr
           0& 0& 0& 2\cr}$$
\bigskip
\noindent
V.  The equilibrium points of the system are found by
setting the right sides equal to zero and solving
for $x,y$.  From $-y(1-y) = 0$, we get $y = 0$ or $y = 1$.
Similarly, $x = 0$ or $x = -1$.  All combinations
of an $x$ and $y$ value are possible, so there are
four equilibria in all:
$$(0,0), (0,1), (-1,0), (-1,1).$$
\bigskip
\noindent
{\it Extra Credit}
\bigskip
The given graph shows a system with {\it centers} at $(0,0),(-1,1)$
and {\it saddle points} at $(0,1),(-1,0)$.  This {\it 
cannot} be the system (2) because the linearization of 
(2) at the origin is
$$X' = \pmatrix {0 & -1\cr -2 & 0\cr}X.$$
This linear system has a {\it saddle point}  at the 
origin (since the determinant is 
negative).  Moreover, our general theorem says this is
one case where the nonlinear system has the same type
of equilibrium as the linearization.  Hence (2) also
has a saddle point at $(0,0)$.  
\bye 
