The rate of growth of a colony of ants is proportional to the square root of the size of the population......
Show that if $P$ is the size of the population and $t$ is the time in days then $\dfrac{dP}{dt} = k\sqrt{P}$, where $k$ is a constant.
When $t=0$, $P=100$ and when $t=5$, $P=625$. Find an expression for $P$ as a function of $t$ and the size of the population when $t=10$. I have integrated the equation, but I can't seem to find a single value for $k$.
If
$$P(t)=c(t-A)^2$$
then $$\frac{ dP}{dt}=2\sqrt c \sqrt c (t-A)=2\sqrt c\sqrt{P(t)}. $$
So, $k=2\sqrt c$ and we have two equaions:
$$cA^2=100 \ \text{ and } \ c(5-A)^2=625.$$
From these equations we get $A=\frac{10}7$ and $c=49$.
Using approach to solving (one kind of) differential equations I explained here, you get $P(t)=\frac{1}{4}(k^2t^2+2ktc+c^2)$. The two initial conditions give you two equations allowing you to pin down $k$ and $c$.
Let $P$ be the size of the population ants, since the rate of growth of a colony of ants $\Delta P$ is proportional to the square root of the size of the population $P$, so $\dfrac{\Delta P}{\Delta t}\propto\sqrt{P}$ which $t$ is the time in days. We convert this proportion with a constant $k$: $\dfrac{\Delta P}{\Delta t}=k\sqrt{P}$
($k$ determines with experiments).
When varieties are calculated in small time $\Delta t\to0$, this proportion induced a limit
$$\lim_{\Delta t\to0}\dfrac{\Delta P}{\Delta t}\propto\sqrt{P}$$
Calculus rules make this limit as a differential
$$\dfrac{dP}{dt}=k\sqrt{P}$$
When $t=0$, $P=100$ and when $t=5$, $P=625$ thus integrating shows $$\int_{100}^{625}\dfrac{dP}{\sqrt{P}}=\int_0^5kdt$$ so $$2\sqrt{P}\Big|_{100}^{625}=kt\Big|_0^5$$ or $50-20=5k$ and $k=6$. Furthermore $\displaystyle\int\dfrac{dP}{\sqrt{P}}=\int6dt$ that is $2\sqrt{P}=6t+C$ or $$\color{blue}{P(t)=(3t+10)^2}$$
when $t=10$, population $P(10)=1600$.
By definition, $\frac{dP}{dt}=k\sqrt{P}$. Then we have, $$\frac{dP}{\sqrt{P}}=kt\rightarrow2\sqrt{P}=kt+c\rightarrow P=\frac{(kt+c)^2}{4}.$$ Using the first initial condition, we have, $$100=\frac{c^2}{4}\rightarrow c=20.$$ We can use this $c$ to solve for $k$ then using the second initial condition, after which you can simply plug in the value of $t$ at which you want $P$.