
## Section6.2Using CalcPlot3D

### Subsection6.2.1Visualizing Systems of Differential Equations

When we study systems of 1st-order differential equations, it is helpful to be able to visualize the phase plots associated with these systems, and to see that no matter what values we use for the two (or three) parameters we obtain in the solution, the general solution (expressed parametrically) will always pass through the vectors in the (two or three dimensional) phase plot just as the solution curves to 1st-order differential equations pass nicely through the corresponding direction fields for any given value of the single parameter C.

Let's do a 2 x 2 system of differential equations example!

###### Exploration6.2.1

a. First, clear the view by clicking on the clear plot button or by deleting the function object by clicking the small x in the upper-right corner of the object.

b. Then select Vector Field on the Add to graph menu. Note that a parameter slider for the parameter $t$ is also automatically added to the plot. This can be used for animating motion along a solution curve (or flowline) as described below in step g.

c. Click the checkbox labeled, “Show system of DEs notation”, so that it looks like the system we are entering.

d. Now enter the equations of the following system of 1st-order differential equations to see its phase plot.

\begin{align*} \frac{dx}{dt} &= -2x + y\\ \frac{dy}{dt} &= -0.5x - y\\ \frac{dz}{dt} &= 0 \end{align*}

e. Since this is a 2 x 2 system, let's Restrict the view to 2D by checking this checkbox.

f. Clicking on a point on the phase plot generates a solution curve. As you drag this point around the plot, the curve dynamically adjusts to fit the field.

The CalcPlot3D app should look something like this:

g. Clicking on the Animate button by the $t$- parameter animates a point along this path. This shows not only the direction of travel of an object traveling through this phase portrait from the chosen starting point, but also the speeds at which it travels as time passes.

h. Now add a Space Curve representing the general solution of this system (in parametric form) to the plot, using parameters a and b. We can use it to visually verify this general solution.

The general solution for this system is:

\begin{equation*} \vec{X} = c_1 e^{-3t/2}\left[\begin{pmatrix}1\\1\end{pmatrix}\cos{\scriptsize\frac{t}{2}}-\begin{pmatrix}-1\\0\end{pmatrix}\sin{\scriptsize\frac{t}{2}}\right] \end{equation*}
\begin{equation*} + \quad c_2 e^{-3t/2}\left[\begin{pmatrix}-1\\0\end{pmatrix}\cos{\scriptsize\frac{t}{2}}+\begin{pmatrix}1\\1\end{pmatrix}\sin{\scriptsize\frac{t}{2}}\right] \end{equation*}

This can then be written in parametric form as:

\begin{align*} x(t) &= e^{-3t/2}\left[ c_1\left(\cos{\scriptsize{\frac{t}{2}}} + \sin{\scriptsize{\frac{t}{2}}} \right) + c_2\left(\sin{\scriptsize{\frac{t}{2}}} - \cos{\scriptsize\frac{t}{2}}\right)\right]\\ y(t) &= e^{-3t/2}\left(c_1\cos{\scriptsize\frac{t}{2}} + c_2 \sin{\scriptsize\frac{t}{2}} \right)\\ z(t) &= 0 \end{align*}

Now replacing $c_1$ with a and $c_2$ with b, we can enter the following equations in the Space Curve object and graph it.

\begin{align*} x(t) &= e^{-3t/2}\left[\textbf{a}\left(\cos{\scriptsize{\frac{t}{2}}} + \sin{\scriptsize{\frac{t}{2}}} \right) + \textbf{b}\left(\sin{\scriptsize{\frac{t}{2}}} - \cos{\scriptsize\frac{t}{2}}\right)\right]\\ y(t) &= e^{-3t/2}\left(\textbf{a}\cos{\scriptsize\frac{t}{2}} + \textbf{b} \sin{\scriptsize\frac{t}{2}} \right)\\ z(t) &= 0 \end{align*}

The parameters a and b can then be varied using their respective sliders to see how the solution curves given by this general solution always fit the system's phase plot.

Animate the motion along this curve to see that the orientation is also correct. The motion should go in the direction of the vectors in the plot.