Application of Differential Equation. The rate of increase in the number of a species of fish in a lake is described by the differential equation $$\frac{dp}{dt}=(a-b)p$$ where $p$ is the number of fish at time $t$ week, $a$, the rate of reproduction, and $b$, the mortaliry rate, with $a$ and $b$ as constants. i)Assuming that $p=p_0$ at time $t=0$ and $a>b$, solve the differential equation and sketch its solution curve. ii)At a certain instant, there is an outbreak of an epidemic of a disease. The epidemic results in no more offspring of the fish being produced and the fish die at a rate directly proportional to $\sqrt{\frac{1}{p}}$. There are 900 fish before the outbreak of the epidemic and only 400 fish are alive after 6 weeks. Determine the length of time from the outbreak of the epidemic until all the fish of that species die. I solved i. But I've no idea for ii. Anyone can solve for me? thanks.

After the epidemic $$\frac{dp}{dt}=-c\left(\sqrt{\frac{1}{p}}\right)p=-c\sqrt{p}$$ Equivalently $$\frac{dp}{p^{1/2}}=-cdt$$ Integrating in the interval $[t_1,t]$ we obtain $$2\sqrt{p(t)}=2\sqrt{p(t_1)}-c(t-t_1)$$ which means that the population extincts ($p(T)=0$) at $$T=t_1+\frac{2}{c}\sqrt{p(t_{1})}$$ You can also calculate the rate $c$ since you have $p(t_1)=900$ and within six weeks $t-t_1=6$ the population is $p(t)=400$ i.e. $$2\sqrt{400}=2\sqrt{900}-6c$$ that results in $c=10/3$. So the population will be extinct within 18 weeks since $$T-t_1=\frac{6}{10}\sqrt{900}=18$$