Chapter 6

Visualizing second-order linear ODE’s, continued.

Second-order linear ODE’s with constant coefficients, continued.

Here we continue to visualize the solutions to the second-order linear ODE

\begin{align} y'' + py' + qy = 0, \end{align}

with constants $p,q$.

To play with the visulization yourself, pick values for the constants $p,q$, specify the initial conditions $y(0)$ and $y'(0)$, and choose your prefered display options. You can also manipulate the figures yourself; move your mouse to any figure and you will see various viewing options on the top right corner of the figure. Click on the lengend to hide or show a curve. When you hover your mouse over the 2d figures, you can see the data plotted; the black number you see at the bottom is the coordinate value of the horonzonal axis, and the colored number on each curve is the corresponding value of the vertical coordinate. Click on the legend to hide or show a curve. For the 3d figure, press and rotate it from top right to bottom left to get the same view as the top left figure, or press and rotate it from bottom left to top right to get the same view as the top right figure.

Constants

$p$:

$q$:

Initial Condtions

$y(0)$:

$y'(0)$:

Display Options Independent solutions:
hide
show

Output

As we have learned, the associated characteristic polynomial is $\lambda^2 + p \lambda + q = \lambda^2$+$\lambda$-1.5.

The roots are $\lambda_1 =$ , $\lambda_2 =$ .

Two independent solutions are $y_1(t) = e^{\lambda_1 t}$ and $y_2(t) = e^{\lambda_2 t}$.

Two independent solutions are $y_1(t) = e^{\lambda_1 t}$ and $y_2(t) = te^{\lambda_1 t}$.

Let $\lambda_1, \lambda_2 = r_0 \pm i s_0$. Then two independent solutions are $y_1(t) = e^{r_0t}\cos(s_0t)$ and $y_2(t) = e^{r_0t}\sin(s_0t)$.

$y(t) = c_1y_1(t) + c_2y_2(t) = $ $y_1(t)$$+$ $-$ $y_2(t)$.

Graph