MA2051 - Ordinary Differential Equations
Review For Exam II - Solutions


1. This problem analyzes .
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.

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 .

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

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 .)

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

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 spring-mass system.
The system is undamped and there is no forcing term, so the mathematical model that you want is

where m = 6 and k = 98.

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.

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.)

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 first-order equations.
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.

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

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

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 second-order equation as a first-order 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.