Here is a 3 (really 4) part problem that I am a little unsure about. I give the context to the problem, then each question/part, followed by what I have done so far for each part.
I would really appreciate any help or guidance as to if I am doing this right, or where I am mistaken.
Given only a bucket with a hole at the bottom, we can not determine how long ago the bucket might have been filled with water to a height $h_0$.
Applying Torricelli's Law gives us an equation modeling the changing height of the water in the bucket:
$\frac{dh}{dt} = -k\sqrt{h}$
with $k$ being a positive constant, dependent upon the size of the hole, and the size and shape of the bucket.
1.) If the condition for uniqueness of an equation of the form $\frac{dh}{dt} = f(t,h)$, with $h(0)=h_0$, is that $\frac{\partial f}{\partial h}$ be bounded. Show that the differential equation does not satisfy this condition. What conclusion can you make about the uniqueness of the solutions?
A.1.)
$\frac{\partial f}{\partial h}=-\frac{k}{2\sqrt{h}}$, we know that as time $t$ goes on, the height $h$ of water in the bucket approaches $0$.
Taking $\lim_{h\to0}$$|\frac{\partial{f}}{\partial{h}}|$ then $\lim_{h\to0}$$(\frac{k}{2\sqrt{h}})$ (from the right-hand side) gives us $\frac{k}{2}\infty$, thus $\frac{\partial f}{\partial h}$ is not bounded, and therefore the condition is not satisfied. We can conclude that there is not a unique solution.
2.) [a.] Make up a reasonable value for $h(0)=h_0$ and solve the differential equation. [b.] Why is it impossible to say when the bucket was full?
A.2.) a.) let $h_0=h(0)=9$
Solving the differential equation leaves us with the general solutions:
$h(t)= \frac{k^2t^2}{4} + (c_1)^2 - ktc_1$
I.V.P. $h(0)=9=0+(c_1)^2 +0$; $c_1 = +3$ or $-3$
I do not know if both are proper, or if one is the proper answer so I did both.
if $c_1 = +3$ then $h(t)=\frac{1}{4}(kt-6)^2$
if $c_1 = -3$ then $h(t)=\frac{1}{4}(kt+6)^2$
I am also unsure as to how to answer part B of this part...
3.) Show that the time it takes to empty the bucket is $\frac{2\sqrt{h_0}}{k}$.
If we are given a full bucket at $t=0$ then we can say $h(0)=h_0$
From part 2.a. we found $h(t)= \frac{k^2t^2}{4} + (c_1)^2 - ktc_1$
then $h(0)=0+(c_1)^2 +0=h_0$ therefore $c_1 = \sqrt{h}$ or $-\sqrt{h}$
Similarly unsure which result to take here, I assume it will be the positive one but do not know the intuition behind this choice...
Taking $c_1 = \sqrt{h}$ gives us $h(t)=\frac{1}{4}(kt-2\sqrt{h_0})^2$
When the bucket empties, $h(t)=0=\frac{1}{4}(kt-2\sqrt{h_0})^2$
$kt-2\sqrt{h_0}=0$; $\to$ $t=\frac{2\sqrt{h_0}}{k}$