How do I solve this ode: $1+(y')^2=cy^{-1}$

Question

I have an ODE, which is the solution to the Brachistochrone problem: $$1+(y')^2=cy^{-1}$$

How does one solve this?

Can you just give me a general direction/hint, rather than a specific solution?

Answer 1

you can write $$\frac{dy}{dx}=\pm\sqrt{\frac{c-y}{y}}$$ then you will get $$\sqrt{\frac{y}{c-y}}dy=\pm dx$$ after integrating we have $$\frac{\sqrt{\frac{y}{c-y}} \left(\sqrt{y} (y-c)+c \sqrt{c-y} \tan ^{-1}\left(\frac{\sqrt{y}}{\sqrt{c-y}}\right)\right) }{\sqrt{y}}=\pm x+K$$

Answer 2

Use $y=c\sin^2(t)$ in $$ \sqrt{\frac{c-y}y}=\frac{\mathrm{d}y}{\mathrm{d}x} $$ to get $$ \frac{\cos(t)}{\sin(t)}=2c\sin(t)\cos(t)\frac{\mathrm{d}t}{\mathrm{d}x} $$ which simplifies to $$ 1=c(1-\cos(2t))\frac{\mathrm{d}t}{\mathrm{d}x} $$ giving us the parametric equation of the curve: $$ \begin{align} x&=c\left(t-\tfrac12\sin(2t)\right)\\ y&=c\sin^2(t) \end{align} $$