MA2051  Ordinary Differential Equations
Review For Exam II  Solutions
REVIEW QUESTIONS
1.
This problem analyzes .

(a)

The characteristic equation is
and so the roots are and . This gives two
solutions and . To check
independence, compute the Wronskian and show that it is never zero.

(b)

To find a particular solution, use the method of undetermined
coefficients and look for a solution of the form
. Plug this into the
original equation and solve for A and B to obtain
and B=0. Hence, a particular solution is
.

(c)

The general solution is
. The initial
data give you equations for and . Solving these
equations give the solution
.

(d)

The transient component is (because this is
the part that converges to zero as t goes to infinity).
This is less than for .
(Just set and solve for t;
it is a quadratic equation for .)

(e)

The limiting periodic solution (obtained in part (c)) is
. The period is and the amplitude is .

(f)

Go back to the general solution in part (c) and choose the initial
data so that . The right choice is
and .
2.
The characteristic equation has roots , so that
. Square both sides ()
and simplify to obtain the
characteristic equation . It follows that the
original differential equation was .
For part (b), .
Similarly, with a little differentiation,
.
3.
This problem analyzes a springmass system.

(a)

The system is undamped and there is no forcing term, so
the mathematical model that you want is
where m = 6 and k = 98.

(b)

The general solution is
where
is the natural frequency
and the period is seconds.
To find the amplitude of the motion, use the initial data to solve for
and to obtain .
The amplitude is meters.

(c)

The period is . If you increase the mass
by a factor of 4, you will double the period.
The amplitude will not change with a change in the mass. (It depends
only on the initial position in this problem.)

(d)

The new model is
and the system will exhibit oscillations as long as the characteristic
equation has complex roots. This occurs for
. Plug in the numbers to find that you have oscillations
for .
4. Solve a system of firstorder equations.

(a)

The characteristic equation is
The roots are (where ).
The corresponding solution pairs are
(You obtain from , for example, by substituting
back into the first differential equation in the system.)
To verify independence, compute the Wronskian
The Wronskian is never zero, so the solutions are linearly independent.
The general solution pair is
where and are arbitrary constants.

(b)

The initial data give you and in the formula
for the general solution.

(c)

Plot a few points to obtain a spiral heading away from the origin.

(d)

Differentiate the quantity to obtain
This last quantity is always positive and so, as soon as the
solution curve is away from the stationary point ,
its distance from the origin is strictly increasing.
5.
The first step is to write the secondorder equation as a firstorder
system:
Euler's method starts with and (the initial data).
Because you want the solution at t=1 and you are using
two steps, .
The next two steps are
Your approximation for is . From Problem #1,
the exact value is .
The absolute error is . The relative
error is , or about 237%. (Bigerror.)
© 1996 by Will Brother.
All rights Reserved. File last modified on April 25, 1996.