It was mentioned in [13] that the equation (4.24) is chaotic, and it admits a homoclinic orbit. By means of the variables z$_{1}$ = z and z$_{2}$ = z $^{′}$, equation (4.24) can be written as a system in the following form, We perturb system (4.25) with the solutions of (4.23) to obtain the system According to Theorem 3.1, system (4.26) possesses chaos in the sense of Li-Yorke. Moreover, since the period of the function 0 . 2 sin(0 . 9 t ) and the periods 2 m, m ∈ N , of the unstable periodic motions of (4.23) are incommensurate, there are infinitely many unstable quasi-periodic motions embedded in the chaotic attractor of (4.26). In system (4.26), as the perturbation, we use the solution ( x$_{1}$ ( t ) , x$_{2}$ ( t )) of (4.23) which is represented in Figure 1, and depict in Figure 2 the solution of (4.26) with the initial data u$_{1}$ (0 . 41) = 0 . 12 and u$_{2}$ (0 . 41) = 0 . 013 . The simulations seen in Figure 2 support the result of Theorem 3.1 such that the perturbed system (4.26) exhibits Li-Yorke chaos. In order to verify that the unperturbed system (4.25) and the perturbed system (4.26) possess different chaotic attractors, we depict in Figure 3 the trajectories of (4.25) and (4.26) corresponding to the initial data z$_{1}$ (0 . 41) = 0 . 12 , z$_{2}$ (0 . 41) = 0 . 013 and u$_{1}$ (0 . 41) = 0 . 12 , u$_{2}$ (0 . 41) = 0 . 013 , respectively. Here, the trajectory of (4.25) is depicted in blue color and the trajectory of (4.26) is shown in red color. It is seen in Figure 3 that even if the same initial data is used, systems (4.25) and (4.26) generate completely different chaotic trajectories. In other words, the applied perturbation annihilates the chaotic attractor of (4.25) and causes a new one to be formed in the dynamics of (4.26). 16 Figure 2: The chaotic solution of the perturbed system (4.26) with u$_{1}$ (0 . 41) = 0 . 12 and u$_{2}$ (0 . 41) = 0 . 013 . The solution$_{(}$$_{x}$$_{1}$$_{(}$$_{t}$$_{)}$$_{, x}$$_{2}$$_{(}$$_{t}$$_{))}$ of (4.23), which is represented in Figure 1, is used as the perturbation in (4.26). z ′ 1 = z$_{2}$ z ′ 2 = − 0 . 15 z$_{2}$ + 0 . 5 z$_{1}$ (1 − z 2 $_{1}$) + 0 . 2 sin(0 . 9 t ) . (4.25) u ′ 1 = u$_{2}$ + 1 . 2( x$_{1}$ + 0 . 1 sin( x$_{1}$ )) u ′ 2 = − 0 . 15 u$_{2}$ + 0 . 5 u$_{1}$ (1 − u 2 $_{1}$) + 0 . 4 arctan( x$_{2}$ ) + 0 . 2 sin(0 . 9 t ) . (4.26) 0 50 100 150 −2 0 2 t u 1 0 50 100 150 −6 −4 −2 0 2 t u 2