If I have the differential equation:
$$\begin{cases} \pi(2\,y\,R-y^2)\frac{dy}{dt}=-\pi\,r^2\sqrt{2\,g\,y} \\ y(0) = R \\ y'(0) < 0 \end{cases}$$
Where $r, R,$ and $g$ are all constants.
If I want to find when $y(t) = 0$, is there any easy method for doing this? NB: The solution for this differential equation is very tedious; using Wolfram and plugging in values for $r, R,$ and $g$ gives a numerical solution which is very long.
I found one solution at this site (see details); however, I don't understand how he integrates the equation (would love a step-by-step) , and also how it is true that the $t$ I'm seeking equals:
$\dfrac{14/15\,R^{5/2}}{r^2\sqrt{2\,g}}$
I do know that a regular function $y(x) = -3x+6$ means that
$$y(t)=0 \iff 0 = 3x + 6 \iff x = -2$$
How does this translate to differential equations? His last step in DETAILS just seems like a big statement without anything to back it up.
How did he find the solution? I don't know. But you can verify that it IS the solution by differentiating and seeing that it satisfies the equation, and then also checking that at time $0$ the value for $y$ is indeed $R$.
Actually, I think that I do know. He divided both sides by $\sqrt{y}$ to get
\begin{align} \pi(2y^\frac{1}{2}R-y^\frac{3}{2}) dy=-\pi r^2\sqrt{2g} dt \end{align} and then integrated both sides to get \begin{align} \pi(2\frac{2}{3}y^\frac{3}{2}R-\frac{2}{5}y^\frac{5}{2}) =-\pi r^2\sqrt{2g} t + C \text{, i.e.}\\ \pi(\frac{4}{3}y^\frac{3}{2}R-\frac{2}{5}y^\frac{5}{2}) =-\pi r^2\sqrt{2g} t + C \text{, i.e.}\\ \frac{4}{3}y^\frac{3}{2}R-\frac{2}{5}y^\frac{5}{2} =- r^2\sqrt{2g} t + C' \end{align}
As for the second part, once you've found an equation relating $y$ to $t$ (but not involving $y'$!) you can set $y$ to zero and see what this implies about $t$.
In this case, the formula given is \begin{align} \frac{4}{3} R y^\frac{3}{2} - \frac{2}{5} y^\frac{5}{2} = -r^2 \sqrt{2g} t + \frac{14}{15} R^\frac{5}{2} \end{align} Setting $y = 0$ gives \begin{align} \frac{4}{3} R 0^\frac{3}{2} - \frac{2}{5} 0^\frac{5}{2} = -r^2 \sqrt{2g} t + \frac{14}{15} R^\frac{5}{2}\\ 0 = -r^2 \sqrt{2g} t + \frac{14}{15} R^\frac{5}{2}\\ r^2 \sqrt{2g} t = \frac{14}{15} R^\frac{5}{2}\\ t = \frac{1}{r^2 \sqrt{2g}}\frac{14}{15} R^\frac{5}{2}\\ \end{align}