Brachistochrone Differential Equation

[Bennigan88]

The first part of this problem asks me to solve the following for y' :
[tex] \left( 1 + {y'}^2 \right)y = k^2 [/tex]
So I have:
[tex] 1 + {y'}^2 = \frac{k^2}{y} [/tex]
[tex] {y'}^2 = \frac{k^2}{y} - 1 [/tex]
[tex] y' = \sqrt{{\frac{k^2}{y} - 1}} [/tex]

Then I am asked to show that if I introduce the following:
[tex] y = k^2 \sin^2t [/tex]

Then the equation for y' found above takes the form:
[tex] 2k^2\sin^2t dt = dx [/tex]

My attempt looks like this:
[tex] y' = \sqrt{ \frac{k^2}{y} - 1} [/tex]
[tex] y' = \sqrt{ \frac{k^2}{k^2 \sin^2t}-1} [/tex]
[tex] y' = \sqrt{ \csc^2t - 1 } [/tex]
[tex] y' = \sqrt{ \cot^2t } [/tex]
[tex] \frac{dy}{dx} = \cot t [/tex]

At this point I feel like I have gotten off track or that I am following the wrong line of argumentation. Either that or I'm out of steam and I can't see how keep this going and get what I'm being asked for. Any insight would be greatly appreciated.

 

[haruspex]

[tex] \frac{dy}{dx} = \cot t [/tex]

You need to use the substitution for y in dy/dx also. What does that give you?

 

[Bennigan88]

Thank you! I get:

[tex] \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} [/tex]
[tex] \frac{dy}{dx} = 2k^2 \sin t \cos t \frac{dt}{dx} [/tex]
[tex] \cot t = 2k^2 \sin t \cos t \frac{dt}{dx} [/tex]
[tex] 1 = 2k^2 \sin^2 t \frac{dt}{dx} [/tex]
[tex] dx = 2k^2 \sin^2 t dt [/tex]

Your genius rivals that of Gauss himself!