This is a two-part question:
1) Suppose we have the eigenvalues $λ_1 = a + bi$, $λ_2 = a – bi$. Then, is it true that:
For $a>0,$ the phase portrait is a spiral out
For $a<0$, the phase portrait is a spiral in
For $a=0$, the phase portrait is a closed loop (stable center, like a circle, an ellipse etc.)
2) By solving a system, I arrived at the parametric equations $x(t) = 10\sin(2t),
y(t) = 4\cos(2t) + 2\sin(2t)$.
The solution shows that this figure is a skewed ellipse, but can someone explain how I could derive this from the parametric equations (and maybe even find the equation of the ellipse in terms of $x,y$), given that I have not learned the general equation for a skewed ellipse?
Thanks in advance.
Best Answer
To find the Cartesian equation, notice $\frac{1}{4}\left(y-\frac{x}{5}\right)=\cos(2t)$ and $\frac{x}{10}=\sin(2t)$ so $$\left(\frac{x}{10}\right)^2+\frac{1}{16}\left(y-\frac{x}{5}\right)^2=1.$$ Making a change of variables $X=x$ and $Y=y-\frac{x}{5}$ we have $$\frac{X^2}{100}+\frac{Y^2}{16}=1,$$ which is the equation of an ellipse.
It is a skewed ellipse since one of its axes is $Y=0$, i.e. $y=\frac{x}{5}$.