Question: The velocity of light above a hot surface decreases with the height from that surface. The velocity is given by $v=v_0(\frac{1-y}{\alpha})$ where $y$ is the vertical distance above the starting position. Show that the path taken by light from $(0,0)$ to $(2X,0)$ is the arc of a circle centered at $(X,\alpha)$. Note that the origin is chosen to be well above the ground, so that $y<0$ is allowable.
Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE: $\dfrac 1{v_0 (\frac{1+y}{\alpha})\sqrt{1+y ^2}}=-C$ where $C$ is a constant.
We're trying to enter it into Mathematica to hopefully come up with the solution, and show this resembles something similar to a circle.
The question concerns the light path in a vertically inhomogeneous medium with the velocity $v(y)$ supposed linearly dependent on the height $y(x)$.
Here $$v=v_0\left(1-\frac y\alpha \right)$$From Fermat's principle one obtains the ode $$\frac 1{v\sqrt{1+y'^2}}=\text{const}$$Here one solves the "boundary value problem" $$\begin {cases}y'=\dfrac{\sqrt {\alpha^2-C^2v_0^2\,(y-\alpha)^2}}{Cv_0\,(y-\alpha)} \\ \\ \\y(0)=0 \\ \\ y(2X)=0 \end {cases}$$The ode is solved by separation of variables remembering that $(\sqrt u)'=u'/(2\sqrt u)$.
One obtains $$\sqrt {\alpha^2-C^2v_0^2\,(y-\alpha)^2}=Cv_0(x+B)$$ with $$\begin {cases} B=-X \\ \\ C=\dfrac \alpha{v_0\sqrt{X^2+\alpha^2}}\end {cases}$$Finally the equation of the circle $$(x-X)^2+(y-\alpha)^2=X^2+\alpha^2$$ I used a CAS to check the steps but not an ode solver.