TITLE: What the orbits of the following DS look like?
QUESTION [0 upvotes]: Let us consider the following DS:
$$x'(t) = y(t)$$
$$y'(t) = −ax(t) + bx^2(t)$$
I have to describe the orbits of the system when $a>0$ and $b=0$. The eigenvalues of the matrix associated to the system are $λ = ±i√a$. Two eigenvectors are $u = (1, 0)$ e $v = (0,√a)$. I know that because the real part of the eigenvalue is zero, then the orbits are circles, centred in the origin. But what about the orbits in the original basis? I think they are ellipses because the second component are somohow multiplied by a scaling factor, but I am not sure whether the scaling factor for the second component is $√a$ or $1/√a$. Thank you.

REPLY [1 votes]: $$\begin{cases}
\frac{dx}{dt} = y(t)\\
\frac{dy}{dt} = −ax(t) + bx^2(t)
\end{cases} \quad\implies\quad \frac{dy}{dx}=\frac{−ax + bx^2}{y}$$
$$ydy=(-ax+bx^2)dx$$
$$\frac12 y^2=-\frac{a}{2}x^2+\frac{b}{3}x^3+c$$
The equation of the trajectory is :
$$y^2+ax^2-\frac{2b}{3}x^3+C=0$$ 
Or explicitly :
$$y(x)=\pm\sqrt{-\frac{2b}{3}x^3+ax^2-C}$$
The trajectories are particular cubic curves.
If $b<0$ :
The trajectory is not closed. For $x\to+\infty\quad\implies\quad y\to\pm\infty$ depending on the initial conditions (not defined in the wording of the question).
If $b>0$ :
The trajectory is not closed. For $x\to-\infty\quad\implies\quad y\to\pm\infty$ depending on the initial conditions.
if $b=0$ :
$$y^2+ax^2+C=0$$
The trajectory is hyperbolic if $a<0$.
The trajectory is elliptic if $a>0$.
For more consistent answer the initial condition should be specified.
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