Question

First order differential equations--
Suppose there's a lake. A population of fish P(t) is attacked by a disease in time t=0; resulting in fishes stopping their reproduction (birth rate = 0), and a death rate proportional to 1/(sqrt(P)). If there were initially 900 fish in the tank and 441 were lost after 6 weeks, how long will it take for all fishes to die?

Answer 1

According to the book, the answer is 20 weeks, or t=20, i'm thinking I need to integrate there?

Answer 2

you want to put the p term with the dp term and the dt on the other side. then integrate

Answer 3

Having only one constant, how can I evaluate for two different points of time? 1st should be t(0) = 900, and the second t(6) = 459?

Answer 4

Solving for C gives me 900 for the initial population in time = 0, but not for the 459 in t (6). Also not for t = 20 weeks.

Answer 5

I will post something if I can make sense of it...

Answer 6

The death rate is proportional to 1/sqrt(P). The number of fish dying is
P times the death rate or
\[ \frac{dP}{dt} = -C_0P\cdot P^{-\frac{1}{2}}\]
\[ \frac{dP}{dt} = P^{\frac{1}{2}}\]
separate the variables:
\[ P^{-\frac{1}{2}} dP= -C_0 dt\]

integrate to get
\[ 2\sqrt{P}= -C_0 t + C_1 \]
as these are unknown constants, we can divide by 2, and use different constants
\[\sqrt{P}= -At+B \]
at t=0, P=900 and we find B=30

now, if we use 441 as the number still alive (because 441 is a perfect square), and solve for A at t=6:
\[ \sqrt{441}= -6A+30 \]
\[ A= (21-30)/-6= \frac{3}{2} \]
\[ \sqrt{P}= -1.5t+30 \]
setting P = 0 (no more fish!)
\[ t= 30\cdot \frac{2}{3}= 20 \]

Answer 7

Thank you SO MUCH!, I was seriously stuck at the two constants part. I knew there had to be two, I just didn't know where I could stick them in into the problem. Thanks lots!