\magnification=\magstep1
\overfullrule=0pt
\leftline{MATH 304 -- Ordinary Differential Equations}
\leftline{Final Exam Solutions}
\bigskip
\noindent
I. 
\item{A)}  By the Chain Rule, and since $y = \phi(x)$ solves 
${dy\over dx} = f(y)$,
$$\psi'(x) = \phi'(x - x_0)\cdot 1 = f(\phi(x-x_0)) = f(\psi(x))$$
In addition, $\psi(x_0) = \phi(x-x_0) = \phi(0) = \alpha$.
\item{B)}  The equilibria are the roots of $f(x) = 0$, at 
$x = -2, 3/2$ (approximately), and $4$.
\item{C)}  The solution of the IVP with $\alpha = 1$ is always
decreasing since $f(y) < 0$ for all $y$ with $-2 < y < 3/2$.
It satisfies $\lim_{x\to\infty} \phi(x) = -2$.
\item{D)}  The solution of the IVP with $\alpha = 2$ is always
increasing since $f(y) > 0$ for all $y$ with $3/2 < y < 4$.
It satisfies $\lim_{x\to\infty} \phi(x) = 4$.
\bigskip
\noindent
II. \item{A)}  After dividing by $t^3$, the equation has the form
$${dz \over dt} + {-4\over t} z = t^5\sin(t)$$
which is {\it first order linear}.  The integrating factor
is $e^{\int -4/t\ dt} = t^{-4}$, so 
$$z = ct^4 + t^4\int{1\over t^4} t^5 \sin(t)\ dt = ct^4 + t^4\int t \sin(t)\ dt$$
Integrating by parts, we have
$$z = ct^4 + t^4(\sin(t) - t\cos(t))$$
\item{B)}  This is a {\it homogeneous} equation.  Letting $v = \theta/s$, 
we get ${d\theta\over ds} = s{dv\over ds} + v$, so
$$ s{dv\over ds} + v = {4 - 3v\over 2 - v}$$
After some algebra:
$${dv\over ds} = {4-5v+v^2\over 2-v} = {(v-1)(v-4)\over 2-v}$$
which is separable:
$$\int {2-v\over (v-1)(v-4)}\ dv = \int {1\over s} ds$$
To integrate the left side, use the entries from the table of integrals
in the textbook for rational functions with a factorable quadratic
downstairs ($b = \beta = 1, a = -4, \alpha = -1$):
$${-2\over 3}\ln\left|{v-1\over v-4}\right| + {1\over 3}(-4\ln|v-4| + \ln|v-1|)
=\ln|s| + c$$
This can be simplified to yield
$$-2\ln|v-4| - \ln|v-1| = 3\ln|s| + c$$
so after exponentiating 
$${(v-4)^2 (v-1)} = c/s^3$$
Substituting $v = \theta/s$ yields the final answer.
\item{C)}  Rearranging to 
$$(\cos(y) - e^x)dx + (-y^2 - x\sin(y))dy = 0$$
the equation is seen to be {\it exact}.  Solutions
$$x\cos(y) - e^x - y^3/3 = c$$
\bigskip
\noindent
III.  
\item{A)}  The time constant is $1/k = 3/2$, hence $k = 2/3$.
Putting $t = 0$ at 1 am, the outside temperature
can be written as $M(t) = 55 - 15\cos(\pi t/12)$.  Hence by Newton's
Law of Heating and Cooling, the differential equation is:
$${dT\over dt} = {2\over 3}(55 - 15\cos(\pi t/12) - T)$$
Rearranging, we have
$${dT\over dt} + {2\over 3}T = {2\over 3}(55 - 15\cos(\pi t/12)),$$
which is {\it first order linear}.  Using the entry from the table
of integrals for $\int e^{au}\cos(nu)\ du$, the solution is:
$$\eqalign{
T &= ce^{-2t/3} + e^{-2t/3}\int e^{2t/3}{2\over 3}(55 - 15\cos(\pi t/12))\ dt\cr
&=ce^{-2t/3} + 55 - 10 e^{-2t/3} \int e^{2t/3} \cos(\pi t/12))\ dt\cr
&=ce^{-2t/3} + 55 - {10\over {4\over 9} + {\pi^2\over 144}}
({2\over 3}\cos(\pi t/12) + {\pi\over 12}\sin(\pi t/12))\cr
&\doteq ce^{-2t/3} + 55 - 12.99\cos(\pi t/12) - 5.10\sin(\pi t/12)\cr}$$
\item{B)}  After the exponential term dies off, the solution is approximately
$$T = 55 - 12.99\cos(\pi t/12) - 5.10\sin(\pi t/12)$$
Using the identity $A\cos(\omega t) + B\sin(\omega t) =
C\cos(\omega t - \phi)$ where $C = \sqrt{A^2 + B^2}$ and 
$\phi = \arctan(B/A)$, we
see that the maximum temperature is attained 
when $\pi t/12 - \arctan(5.10/12.99) = \pi$, or 
$$t \doteq {12\over\pi}(\pi + .37) \doteq 13.41$$
This means approximately 13.41 hours after 1am, or 2:24 pm (and every
24 hours thereafter).
\bigskip
\noindent
IV.  We have $y_2 - y_1 = 3x$ and $y_1 - y_3 = 3x^2 - x$ are
solutions of the homogeneous equation $y'' + py' + q = 0$.
They are linearly independent because they are polynomials of
two different degrees.  Hence by Abel's Identity,
$$Ce^{-\int p(x)\ dx} = W[3x,3x^2-x] = 9x^2$$
for some constant $C$.
Taking logarithms and differentiating, we find
$$p(x) = {-2\over x}$$
\bigskip
\noindent
V.  
\item{A)}  $y = c_1 e^{-4x} + c_2 x e^{-4x}$
\item{B)}  The particular solution should have the form
$A x^2 e^{-4x} + Bx^2 + Cx + D$.  Solving for $A,B,C,D$ yields
$$y_p = {3\over 2} x^2 e^{-4x} + {1\over 16} x^2 + {-5\over 16}x + {35\over 128}$$
\item{C)}  If the 16 is replaced by $16 + \epsilon$, $\epsilon > 0$, 
then the roots of the auxiliary equation become
$$r = -4 \pm i\sqrt{\epsilon}$$
The form of the homogeneous solutions changes to
$$y = c_1 e^{-4x}\cos(\sqrt{\epsilon} x) +  c_2 e^{-4x}\sin(\sqrt{\epsilon} x)$$
and the form of the particular solution changes to
$A e^{-4x} + Bx^2 + Cx + D$ since $e^{-4x}$ is no longer a solution of 
the homogeneous equation.
\bigskip
\noindent
VI.  
\item{A)}  The system has two critical points $(0,4)$ and $(-6,-2)$.
\item{B)}  The linearization at $(0,4)$ has the matrix
$$\pmatrix{-1 & 1\cr -12 & 0}$$
with eigenvalues $\lambda = {-1\pm\sqrt{-47}\over 2}$ (complex 
conjugates with
negative real parts).  This is 
an {\it inward spiral} point.
The linearization at $(-6,-2)$ has the matrix
$$\pmatrix{-1 & 1\cr -6 & 18}$$
with eigenvalues $\lambda = {17\pm\sqrt{337}\over 2}$ (real with
opposite signs).  This is 
a {\it saddle} point.
\item{C)}  Solutions of linearized system at $(-6,-2)$:
Using the decimal approximations $\lambda \doteq -.678, 17.7$,
we get corresponding eigenvectors:
$$\pmatrix{1\cr .322\cr} \qquad \pmatrix{1\cr 18.7\cr}$$
so the solutions are
$$c_1e^{-.678 t}\pmatrix{1\cr .322} + c_2 e^{17.7t}\pmatrix{1\cr 18.7}$$
\bye





