How to prove $\frac{dh}{dt}= \frac{5 }{h^2} - \frac{1}{20}$ and a couple other related questions (complete information inside)?

Question

This is a differential equation question. Since I might not be able to explain is well, I will attach a link to the question as well as a screenshot of the mark scheme. Question: Mark Scheme:

Here's what I've done so far:

$\frac {dh}{dt} = kh^2$

when $h = 1, \frac {dh}{dt} = 4.95$

$k = 4.95$

$\frac {dh}{dt} = 4.95h^2$

As you can see, I'm not on the right track. I would really appreciate it if someone would guide me on where I'm going wrong. I'm not THAT bad at differential equations, but the ones with a scenario end up being too complex for me at times. I will add that I have knowledge of product and quotient rules and the chain rule (not too good at this which explains my lack of skills in this topic). I am also comfortable with partial fractions (looks like part $(ii)$ of the question is about partial fractions. Thanks in advance! :D

Answer 1

It is pretty much summed up in the solution you provided. $$ \text{net flow of fluid in the system} = \text{flow into system} -\text{flow out of system} $$ Now we get the equivalent $$ \text{rate of net flow of fluid in the system} = \text{rate of flow into system} -\text{rate of flow out of system} $$ Or in symbols/variables $$ \dfrac{dV}{dt} = \dfrac{d}{dt}V_{in} - \dfrac{d}{dt}V_{out} $$ So we know what $V$ is its the volume of the container. We know what the rate in will be (its the constant) and you also know what the form of the out flow is the proportional to $h^2$ part.

$\textbf{Edit:}$

It seems that the link has disappeared but i will continue. $$ \begin{align} \dfrac{d}{dt}V_{in} &=20&\\ \dfrac{d}{dt}V_{out}&=kh^2&\\ \dfrac{dV}{dt}&=\dfrac{d}{dt}\frac{4}{3}h^3& \end{align} $$ now using chain rule $$ \dfrac{d}{dt}f(h) = \frac{df(h)}{dh}\frac{dh}{dt} $$ setting $f(h) = \frac{4}{3}h^3$ you can compute the last equation. bring it all together and re-arrange for $\frac{dh}{dt}$.