Background. We wish to study the system of D. E.'s
x' = a x + b y
y' = c x + d y

Computer Lab Work.

Example 1. Find the general solution to the system of D. E.'s

Plot the solution curves where the starting points are (1,1), (-1,1), (-1,-1), (1,-1)
and the parameter t is in the interval 0 < = t < = 12.566

Put the D.E.'s in operator form and eliminate y to obtain a higher order D.E. for x, and find the roots of its characteristic equation.

The roots are complex, , so the general solution is formed as follows.

Now construct the vector form of the solution R[t,c1,c2] and look at the initial value R[0,c1,c2].

Plot the solution curves where the starting points are (1,1), (-1,1), (-1,-1), (1,-1).

Example 2. Find the general solution to the system of D. E.'s

Plot the solution curves where the starting points are (1,0), (0,1), (0.5,1), (1,0.5),
and the parameter t is in the interval 0 < = t < = 2.

Put the D.E.'s in operator form and eliminate y to obtain a higher order D.E. for x, and find the roots of its characteristic equation.

The roots are real and distinct, , so the general solution is formed as follows.

Now construct the vector form of the solution R[t,c1,c2] and look at the initial value R[0,c1,c2].

It is desirable to have the general solution in a form where (c1,c2) is an initial point that the solution curve passes through. So we adjust the above solution to fit this requirement

We will construct the vector function P so that the I.C.'s are easier to input.

Plot the solution curves where the starting points are (1,0), (0,1), (0.5,1), (1,0.5).

Example 3. Find the general solution to the system of D. E.'s

Plot the solution curves where the starting points are (1,1), (2,2), (3,3), (4,4).
and the parameter t is in the interval 0 < = t < = 5.9

Put the D.E.'s in operator form and eliminate y to obtain a higher order D.E. for x, and find the roots of its characteristic equation.

The roots are pure complex, , so the general solution is formed as follows.

Now construct the vector form of the solution R[t,c1,c2] and look at the initial value R[0,c1,c2].

It is desirable to have the general solution in a form where (c1,c2) is an initial point that the solution curve passes through. So we adjust the above solution to fit this requirement

We will construct the vector function P so that the I.C.'s are easier to input.

Plot the solution curves where the starting points are (1,1), (2,2), (3,3), (4,4).

Example 4. Find the general solution to the system of D. E.'s

Plot the solution curves where the starting points are (1,0), (0,0.6), (0.6,0), (0,0.3).
and the parameter t is in the interval -0.2 < = t < = 1

Put the D.E.'s in operator form and eliminate y to obtain a higher order D.E. for x, and find the roots of its characteristic equation.

The roots are real and distinct, , so the general solution is formed as follows.

Now construct the vector form of the solution R[t,c1,c2] and look at the initial value R[0,c1,c2].

It is desirable to have the general solution in a form where (c1,c2) is an initial point that the solution curve passes through. So we adjust the above solution to fit this requirement

We will construct the vector function P so that the I.C.'s are easier to input.

Plot the solution curves where the starting points are (1,0), (0,0.6), (0.6,0), (0,0.3).

Example 5. Find the general solution to the system of D. E.'s

Plot the solution curves where the starting points are (0.2,0), (0.6,0), (0,0.4), (0,1),
and the parameter t lies in the interval 2 < = t < = 0.8

Put the D.E.'s in operator form and eliminate y to obtain a higher order D.E. for x, and find the roots of its characteristic equation.

The roots are real and equal, , so the general solution is formed as follows.

Now construct the vector form of the solution R[t,c1,c2] and look at the initial value R[0,c1,c2].

It is desirable to have the general solution in a form where (c1,c2) is an initial point that the solution curve passes through. So we adjust the above solution to fit this requirement

We will construct the vector function P so that the I.C.'s are easier to input.

Plot the solution curves where the starting points are (0.2,0), (0.6,0), (0,0.4), (0,1).

(c) John H. Mathews, 1998