Math 431 Section 4 Solution of Midterm 2 Fall 1998
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 
.
, then 
and .
The mass stretches the spring L feet where L satisfies the equation
and k is the spring constant. We can read k from
the equation (1) as the coefficient of u(t).
Hence, k=64 and 
.
are
.
Hence, 
. The initial condition determines that
and 
.
It follows that the position function is
Above, 
.
Hence, 
, where n is an integer.
The first time occurs when 
.
All times are given by
,
where n is a non-negative integer.
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.
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
.
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:
is
a solution of the differential equation
by plugging in. Indeed,
So, 
.
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
