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\centerline{Mathematics 304 -- Ordinary Differential Equations}
\centerline{Problem Set 5 -- Selected Solutions}
\centerline{November 13, 2000}
\bigskip
\noindent
Section 4.8
\bigskip
56. a) We want to find the general solution of the ODE:
$$mx'' + bx' + kx = \sin(\beta t)\leqno(1)$$
The homogeneous equation $mx'' + bx' + kx = 0$ has auxiliary
equation $mr^2 + br + k = 0$, so 
$$r = {-b \pm \sqrt{b^2 - 4mk}\over 2m}.$$
Since we are given $b^2 < 4mk$, the two values for $r$
are complex conjugates, and we can write the general solution
of the homogeneous equation as 
$$c_1 e^{-bt\over 2m} \cos(\omega t) + 
c_2 e^{-bt\over 2m} \sin(\omega t)$$
where $\omega = {\sqrt{4mk - b^2}\over 2m}$.  

To solve the inhomogeneous equation, we use the undetermined
coefficients method.  By form III in the table on page 208 of
the text, we try a particular solution of the form
$$x_p = A\cos(\beta t) + B\sin(\beta t)$$
(Note that neither term here can solve the homogeneous
equation -- no exponential terms -- hence we take $s = 0$
in the undetermined coefficients formula.)  Substituting
$x_p$ into (1), we get:
$$\eqalign{
(k -m\beta^2) A + b\beta B &= 0\cr
-b\beta A + (k - m\beta^2)B &=1\cr}$$
The simplest way to write down the solution of a system like
this is to use Cramer's Rule:
$$\eqalign{
A &= {\det\pmatrix{0&b\beta\cr
                    1&k-m\beta^2\cr}\over
       \det\pmatrix{k-m\beta^2&b\beta\cr
                    -b\beta&k - m\beta^2\cr}}\cr
&= {-b\beta\over (k-m\beta^2)^2 + (b\beta)^2}\cr}$$
and
$$\eqalign{
B &= {\det\pmatrix{k-m\beta^2&0\cr
                   -b\beta&1\cr}\over
       \det\pmatrix{k-m\beta^2&b\beta\cr
                    -b\beta&k - m\beta^2\cr}}\cr
&= {k - m\beta^2\over (k-m\beta^2)^2 + (b\beta)^2}\cr}$$
Then the general solution of the equation (1) is 
$$c_1 e^{-bt\over 2m} \cos(\omega t) + 
c_2 e^{-bt\over 2m} \sin(\omega t) + A \cos(\beta t) + B \sin(\beta t),$$
where $A,B$ are given by the formulas above.

b)  Because $b/2m > 0$, the exponentials in the general
solution tend to zero as $t\to \infty$.  The solution tends
to $A \cos(\beta t) + B \sin(\beta t)$.  This is a sinusoidal
function of the form $C = \cos(\beta t + \phi)$ for suitable
$C,\phi$.

\bigskip
\noindent
57.  The idea here is to follow the same method used in problem
55.  We first solve the IVP:
$$\cases{y'' + y = \cos(Vt) &\cr
         y(-\pi/(2V)) = 0&\cr
         y'(-\pi/(2V)) = 0&\cr}\leqno(2)$$
(Note: it is given that $y(t) = 0$ for all $t \le -\pi/(2V)$ --
the car is not moving vertically at all before it hits the
speed bump.  So this gives the initial conditions at
$t=-\pi/(2V)$.)  Call the solution of (2) $y_a(t)$.
{\it Then}, at $t = \pi/(2V)$, we solve a second IVP:
$$\cases{y'' + y = 0 &\cr
         y(\pi/(2V)) = y_a(\pi/(2V))&\cr
         y'(\pi/(2V)) = y_a'(\pi/(2V))&\cr}\leqno(3)$$
to get the continuation of the solution to $t > \pi/(2V)$.

The first step is to find the function $y_a(t)$ solving
(2).  The general solution of the homogeneous equation
$y''+y' = 0$ is $y = c_1\cos(t) + c_2\sin(t)$.  By the
form of the right hand side of the ODE in (2), for $V \ne 1$,
we expect a particular solution of the form
$$y_p = A\cos(Vt) + B\sin(Vt).$$
Substituting into (2), we find that 
$$A(1 - V^2)\cos(Vt) + B(1-V^2)\sin(Vt) = \cos(Vt)$$
so  $A={1\over 1-V^2}$ and 
$B = 0$.  

Now, we determine the constants $c_1,c_2$ in the general solution
$$y = c_1\cos(t) + c_2\sin(t) + {1\over 1 - V^2}\cos(Vt)$$
to get $y(-\pi/(2V)) = 0$ and $y'(-\pi/(2V)) = 0$.
These initial conditions say:
$$\eqalign{
c_1\cos(-\pi/(2V)) + c_2\sin(-\pi/(2V)) + 0 &= 0\cr
-c_1\sin(-\pi/(2V))+ c_2\cos(-\pi/(2V)) &= {V\over V^2-1}\cr}$$
By Cramer's Rule again,
$$\eqalign{
c_1 &= {\det\pmatrix{0&\sin(-\pi/(2V))\cr
                     {V\over V^2-1} &\cos(-\pi/(2V))\cr}\over 
        \det\pmatrix{\cos(-\pi/(2V))&\sin(-\pi/(2V))\cr
                    -\sin(-\pi/(2V)) &\cos(-\pi/(2V))\cr}}\cr
    &= {V \sin(\pi/(2V))\over V^2-1}\cr
c_2 &= {\det\pmatrix{\cos(-\pi/(2V))&0\cr
                     -\sin(-\pi/(2V))&{V\over V^2-1} \cr}\over 
        \det\pmatrix{\cos(-\pi/(2V))&\sin(-\pi/(2V))\cr
                    -\sin(-\pi/(2V)) &\cos(-\pi/(2V))\cr}}\cr
    &= {V \cos(\pi/(2V))\over V^2-1}\cr }$$ 
This gives the coefficients in $y_a(t)$, the formula for
the solution on $-\pi/(2V) < t < \pi/(2V)$.

Now, we want to solve the second IVP (3).  If we call the solution
$y_b(t)$, then from the ODE in (3) -- the homogeneous
equation $y'' + y = 0$, we know that 
$y_b = d_1\cos(t) + d_2\sin(t)$, and we must determine
$d_1,d_2$ so that 
$$y_b(\pi/(2V)) = y_a(\pi/(2V)) 
= c_1\cos(\pi/(2V)) + c_2\sin(\pi/(2V)) = {2V\sin(\pi/(2V))\cos(\pi/(2V))\over V^2-1}$$ 
and
$$y_b'(\pi/(2V)) = y_a'(\pi/(2V)) = {2V\cos^2(\pi/(2V))\over V^2-1}.$$
So,
$$\eqalign{
\cos(\pi/(2V)) d_1 + \sin(\pi/(2V)) d_2 
&= {2V\sin(\pi/(2V))\cos(\pi/(2V))\over V^2-1}\cr
-\sin(\pi/(2V)) d_1 + \cos(\pi/(2V)) d_2 
&= {2V\cos^2(\pi/(2V))\over V^2-1}\cr}$$
Solving by Cramer's Rule, we find $d_1 = 0$, and $d_2 = {2V\cos(\pi/(2V))\over V^2-1}$.
Thus the solution on $t \ge \pi/(2V)$ is
$$y_b(t) = {2V\cos(\pi/(2V))\over V^2-1}\sin(t)$$
Plotting the amplitude as a function of $V$, we find
$A(V) =  {2V\cos(\pi/(2V))\over V^2-1}$ has 
a maximum at roughly $V \doteq .73$.
\bye
