my text book for differential equations has a nice applied 'project'/investigation that I have been working through over the weekend (this is not a homework question I just thought it may be interesting because my grandfather was a pilot.)
Have I perhaps made a mistake along the way?
I don't think you've made a mistake, but maybe you've complicated your life a bit more than you wanted with all these differentiations. Let's start with $$\begin{pmatrix}\dot x\\ \dot y \end{pmatrix} = \frac{v_2}{\sqrt{x^2+(v_1t-y)^2}} \begin{pmatrix}-x\\(v_1 t-y)\end{pmatrix}$$ When tackling something like this, you'll want to "mix up" both sides. Note that $\|(\dot x,\dot y)\|^2=(v_2)^2$, which means that you can take the dot product of your expression above, with $(\dot x,\dot y)$. On the left you will then have $(v_2)^2$ and on the right you will have $\frac{v_2}{\sqrt{\dots}}[-x\dot x+(v_1t-y)\dot y] $. You should move the square root left and square everything, expand it etc., and you should get something simpler. You will also need to eliminate $t$ everywhere if you want a curve! Notice that $\frac{d y}{d x} = \frac{dy/dt}{dx/dt}=\frac{\dot y}{\dot x}$, and that $v_2 t$ is the arc length of the curve that the chasing plane traces, ie $v_2t=\int dx \sqrt{1+(dy/dx)^2}$.
Also I'd like to perhaps present you with the derivation from Simmons' book Differential Equations with Applications and Historical Nots (I really love that book, and seeing that you want to know more about the subject, I suggest you look it up!). Simply draw the curve and note that $$\frac{dy}{dx}=\frac{y-v_1t}{x}\Rightarrow x y' - y = -v_1t.$$
where the prime ' denotes differentiation with respect to $x$. Now differentiate with respect to $x$ again to get $$x y''=-v_1\frac{dt}{dx}.$$ Write $\frac{dt}{dx}=\frac{dt}{ds}\frac{ds}{dx}=-\frac{1}{v_2}\sqrt{1+(y')^2}$, where the minus comes from the fact that the arc length $s$ increases as $x$ goes to 0, ie as it approaches the line through the origin where the first plane is flying. You can probably solve this using the techniques you know already! Good luck and I hope this was helpful!