A mouse starts at the origin, and at time $t=0$, runs up the $y$-axis with constant speed 1. A cat starts a distance $L$ to the right of the mouse at $(L, 0)$ at time $t=0$. At time $t=0$, it starts running with constant speed $v$ such that it is directly facing the mouse at all times $t$. What are the parametric equations of the cat's path?
Here is my attempt: Let $(x(t), y(t))$ be the path of the cat, so that $x(0)=L$, $y(0)=0$. My first step was to notice that the speed of the cat implies that
$$(x'(t))^2+(y'(t))^2=v^2.$$ Also, we know that since the cat directly faces the mouse, we have $$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{y(t)-t}{x(t)}.$$ Now I am stuck. When I try plugging the second equation into the first, we get $$\left(1+\left(\frac{y(t)-t}{x(t)}\right)^2\right)(x'(t))^2=v^2$$ with no way to remove $y(t)$. It seems no matter what I try there is no way to get two separate differential equations in $x$ and $y$. It seems I would need more equations to get rid of the $y(t)$ term but there is no real way for me to do that. Any tips?
