Non-Linear Differential Equation and Phase Plane Help

[SilkLemur]

Consider the system:
$$\frac{dx}{dt} = y$$

and

$$\frac{dy}{dt}= x + 2 x^3$$

a) Show that the equilibrium solution x=0, y=0 of the linearized system $$\dot{x} = y$$, $$\dot{y}= x$$ is a saddle, and draw the phase portrait of the linearized system.
b) Find the orbits of the system and draw its phase portrait.
c) Show that there are exactly two orbits of the system (one for x>0 and one for x<0) on which $$x \rightarrow 0$$, $$y \rightarrow 0$$ as $$t \rightarrow\infty$$. Similarly, there are exactly two orbits of the system (one for x>0 and one for x<0) on which $$x \rightarrow 0$$, $$y \rightarrow 0$$ as $$t \rightarrow -\infty$$. Thus, observe that the phase portraits of the system and the linearized system look the same near the origin.

2. Attempt at a solution
a) $$\dot{x} = \left[\begin{array}{cc}0&1\\1&0\end{array}\right] x$$
Thus,
$$det( \left[\begin{array}{cc}0&1\\1&0\end{array}\right] - \lambda I) = ( \lambda +1)( \lambda -1)$$
When $$\lambda_1$$<0<$$\lambda_2$$ we know that there will be a saddle and the equilibrium point.

b) The orbits of the system are the solution curves of the scalar equation:
$$\frac{dy}{dx}= \frac{x+2x^3}{y}$$
which is separable and every solution is of the form $$y^2 = x^2 + x^4 +c$$.

c) I am not quite sure what this part of the question is asking.

 

[mighty2000]

However, based on the given information, I can observe that there are two distinct orbits on either side of the y-axis where the solutions approach the origin as t approaches positive or negative infinity. This is similar to the linearized system, where the phase portrait also shows two distinct orbits approaching the origin from either side of the y-axis. Therefore, it can be observed that the phase portraits of the system and the linearized system look the same near the origin.