Non-Linear Differential Equation 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.

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]

It seems to be pointing out that the phase portraits of the system and the linearized system look the same near the origin, but I am not sure how to show this mathematically. Can you please provide more clarification on what is being asked in this part? Thank you.