 ## Solving Systems

(continued from last page...)

### Exercises

Each of the exercises that you are going to do will have essentially the same instructions:

1. Use NDSolve to solve the initial value problem on the interval given.
2. Extract the individual components of the solution into separate variables using the "[[1,#,2]] trick" that you saw being used in the preliminary example.
3. Plot the graphs of the each of the components of the solution on separate coordinate systems. There is no need to use color.

### Exercise 1

#### A Simple First Order System in Two Dependent Variables

The following initial value problem is extremely similar to the one you saw me do in the preliminary example.

System Initial Conditions
x' = - x + 2y
y' = 2x - 3y
x(0) = 1
y(0) = 2

Solve it numerically, and plot the solution's components on the interval -1 ≤ t ≤ 2. I'm not going to give you any further instructions, but you can check the graphs of the solution's components if you want to see if you are on the right track.

### Exercise 2

#### A Simple First Order System in Three Dependent Variables

The following initial value problem involves three independent variables. The method of solution is just a simple extension to what you have already been doing. Note the fact that the extra variable brings along an additional initial condition also. You will, of course, be providing three graphs as your answer to this problem, the usual two, and one of the z-component.

System Initial Conditions
x' = 3x + 2y - z
y' = x - y + 4z
z' = 4x + y
x(0) = 1
y(0) = 2
z(0) = 0

Solve it numerically, and plot the solution's components on the interval -1 ≤ t ≤ 1. I'm not going to give you any further instructions, but you can check the graphs of the solution's components if you want to see if you are on the right track.

### Exercise 3

#### A Second Order System in Two Dependent Variables

The following initial value problem involves second derivatives. As a result, each variable requires two initial conditions. The method of solution is once again just a simple extension of the problems you have already done. We'll be back to making just two graphs for the solution of this problem. Don't forget that to form the "double prime" in Mathematica you need to hit the single quote key twice, not the double quote key.

System Initial Conditions
x'' = x' + 2y'
y'' = 2x' - 3y
x(3) = 1
x'(3) = 0
y(3) = -2
y'(3) = 1

Solve it numerically, and plot the solution's components on the interval 0 ≤ t ≤ 6. I'm not going to give you any further instructions, but you can check the graphs of the solution's components if you want to see if you are on the right track.

You have now completed this laboratory. You may now return to its beginning, go to the Table of Contents of all the Differential Equations Laboratories, or simply quit altogether. If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.

ODE Laboratories: A Sabbatical Project by Christopher A. Barker

©2017 San Joaquin Delta College, 5151 Pacific Ave., Stockton, CA 95207, USA
e-mail:
cbarker@deltacollege.edu