has the form
where 
is the
general solution of the associated homogeneous equation.
, 
,
are particular solutions of 
found via the
method of undetermined coefficients. For 
we look for a solution of
the form 
. Plugging in we find that
. Equating coefficients we get
. For 
we look for a solution of
the form 
. Plugging in we find that 
. For
we look for a solution of
the form 
because 2 is a root
of the characteristic equation with multiplicity 1. Plugging in we get:
Equating coefficients we get the two equations
and 
. Hence, 
and
. The general solution is
using the method of variation of parameters. The characteristic equation
factors 
.
A fundamental set of the
associated homogeneous equation is 
and 
.
Hence, the general solution has the form
with initial condition 
and 
.
- (2 points)
If the mass weighs
, then 
.
Hence, the mass stretches the spring L feet where L satisfies the equation
and k is the spring constant.
Since the equation (1) is normalized,
. Hence, 
-
(8 points)
If
, the roots of the characteristic equation
are
.
Hence, 
. The initial condition determines that
and 
.
It follows that the position function is
Above,
. - (6 points)
-
(3 points)
The points in time at which the mass returns to its equilibrium
are solutions of the equation u(t)=0. These must be solutions of
Hence, 
, where n is an integer.
The first time occurs when 
.
All times are given by
,
where n is a non-negative integer. - (6 points)
The system returns to equilibrium infinitely many times
if
because in that case the characteristic equation
has two complex roots and the system vibrates.
The system never come back to its
equilibrium if the system is over-damped 
or
critically damped 
.
In other words, when 
. To see that it suffices to check
the case 
(since, if 
, the system will move even more
slowly having a larger damping force). When
the solution to the initial value problem is
. We see that
u(t) is always positive for t;SPMgt;0.
Note: (the following was not required to get full credit)
If , the two roots 
, 
are negative.
Say 
. Then
and the initial condition implies
that 
,
,
and 
.
Hence, 
. It follows that
and u(t) is positive for all t;SPMgt;0.
- (15 points)
The general solution of the equation
has the form 
where
a particular solution 
can be found
using the method of undetermined coefficients. Plugging in we get
the equation
. Hence,
8A-6B=0 and 6A+8B=1 and
- (reduced to 5 points due to poor performance of entire class)
If u(0)=1, then
.
The behavior of y(t) in (2)
is dominated by the term 
if 
and by
the term 
if 
. Hence,
if 
, then 
is negative.
The slope of the graph of the solution
at t=0 satisfies in that case the inequality:
- (2 point)
We verify that
is
a solution of the differential equation
by plugging in. Indeed,
So,
. - (13 points, a generous partial credit scheme)
We first find the general solution using the method of reduction of order.
Given one solution
of a second order D.E. y''+p(x)y'+q(x)y=0
we look for a solution of the form 
.
Then v(x) satisfies the differential equation
In our case, p(x)=0. So v(x) satisfies the differential equation
If we let u(x)=v'(x) and integrate both sides of (3) we get the equation
.
Hence, 
. We may choose C to be 1. Integrating once
more we get
Hence,
.
We can now find the particular solution satisfying the initial condition
(It was not necessary to determine
and in order to get the full
credit). We get the equations
and 
.
The solution is
