Figure 4.26 contains graphs of solutions to the Duffing, Airy, and van der Pol equations. Try to match the solution to the equation.

The Energy Integral Lemma:

Let y(t) be a solution to the differential equation $\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t. Let F(y) is an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$, that is, $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$is constant; i.e.$\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$ .

Change of angular momentum:

$\mathrm{m}{\ell }^2\mathrm{\theta }=-\ell \mathrm{mgsin\theta }$…… (1)

Newton’s rotational law: The rate of change of angular momentum is equal to torque.

The mass–spring oscillator equation:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{ext}}=\left[\mathrm{inertia}\right]\mathrm{y}+\left[\mathrm{damping}\right]\mathrm{y}\text{'}+\left[\mathrm{stiffness}\right]\mathrm{y} \\ =\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky} \end{gathered}$$…… (2)

Given that, the graph of the solutions of equations Duffing, Airy, and van der Pol.

Part (a):

It is the same as Figure 4.13 so this is the Airy function and look at the graph there is negative time and for positive time amplitude increases without any bound. So, this is the Airy equation.

Part (b):

It is the same as Figure 4.16 so this is Duffing function and look at the graph there is no amplitude change means no damping and the stiffness increases as t increases. So, it is a Duffing equation.

Part (c):

It is the same as Figure 4.17 so this is the van der Pol function and looking at the graph there are more oscillations and the amplitude is not constant for every cycle. So it is a van der Pol equation.