Differential equation (Brachistochrone problem)
I'm really only supposed to solve the differential equation $(1+(y')^2)y=k^2.$
I haven't encountered any problem with $(y')^2$. How do you start with a problem like this, I did try googling it but all I got was basic differential equation, should I substitute for something?
Grateful for any help at this point!
$\endgroup$ 32 Answers
$\begingroup$$$\left(1+\left(\frac{dy}{dx}\right)^2\right)y(x)=k^2$$ $$\frac{dy}{dx}=\pm \sqrt{\frac{k^2}{y}-1}$$ ODE of the separable kind : $$dx=\pm \sqrt{\frac{k^2-y^2}{y}}\:dy \quad\to\quad x=\pm\int \sqrt{\frac{k^2-y^2}{y}}\:dy $$ $$\pm x=k^2\tan^{-1}\left(\sqrt{\frac{y}{k^2-y}}\right)-\sqrt{(k^2-y)y}\:+c$$ The solution is on the form of $x$ as a function of $y$. There is no simple closed form for the inverse function $y(x)$.
The result can be presented on parametric form , with $\quad\tan(\theta)=\sqrt{\frac{y}{k^2-y}}$ :
$$\begin{cases} x=k^2\left(\theta -\sin(\theta)\cos(\theta)\right)+c \\ y=k^2\sin^2(\theta) \end{cases}$$ which is a parametric form of equation of cycloid.
$\endgroup$ 0 $\begingroup$There is an algebraic mistake in the above answer: In the third line,
dx should be sqrt[ y/(k^2 -y^2) ] dy
(The argument of the sqrt has the numerator and denominator interchanged).
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