I have the following system of equations: $\def\b{\begin{pmatrix}}\def\e{\end{pmatrix}}$ $\b\dot{y}_1 \\ \dot{y}_2\e=\b2&0\\3&-1\e\b y_1\\ y_2 \e$
and I need to find the equation of straight line trajectories where the tangent to the trajectories is horizontal or vertical.
Now I have solved this and drawn the phase diagram and see it:
1) An unstable sadle centred at the origin.
2) Has eigenvectors for repulsive $\lambda=2$, $(1,1)^T$ and attactive $\lambda=-1$, $(0,1)^T$
But I got a different answer for the $0$ and $\pm\infty$ tangents.
I would have thought this occurs at $\frac{\dot{y}_2}{\dot{y}_1}=\frac{3y_1-y_2}{2y_1}$
where we can see that $$\frac{\dot{y}_2}{\dot{y}_1}=\frac{3y_1-y_2}{2y_1}=\left\{ \begin{array}00,\quad y_2=3y_1\\\pm\infty,\quad y_1=0 \end{array}\right.$$
Apparently the answer is:
$$\left\{ \begin{array}0\frac32,\quad y_2=0\\\pm\infty,\quad y_1=0 \end{array}\right.$$
Why is this the solution? I can't see why for the $0$ tangent, they didn't set $y_2=3y_1$ to get the numerator to $0$, but rather set $y_2=0$ to get $\frac32$, what good does $\frac32$ even do? We want the horizontal line, e.g. $\frac{\dot{y}_2}{\dot{y}_1}=0$ as far as I understand?