You are working with the rate of change (dN/dt) rather than the population itself (N). You need to integrate this equation - this one is separable so it's not too hard as far as DEs go.
I suggest using partial fractions to integrate the LHS
Hi, need help answering a question in a past exam paper. Below is the question:
5 mice in a stable population of 500 are intentionally infected with a disease. The rate of change in the infected population is proportional to the product of the number of mice who have the disease with the number that are disease free. If N(t) denotes the number of mice with the disease at time, t, the theory predicts that:
dN/dt = kN(500 - N)
where k is a constant. Show that the time taken for half the population to contract the disease is
t= 0.00919/k
I have substitued N=5 to give dN/dt = k x 5 x (495) but i dont understand what to do next.
You are working with the rate of change (dN/dt) rather than the population itself (N). You need to integrate this equation - this one is separable so it's not too hard as far as DEs go.
I suggest using partial fractions to integrate the LHS
Can someone solve this step by step so i can see how its done?
Using MSUMathStdnt method i got K=0.00202020...what do i do from here?
Sorry to sound like a novice but I have no examples to look at or too learn from which is why i am finding it difficult.
You're not asked to solve for k so why bother? What don't you understand about integrating by partial fractions?
Spoiler:
Using partial fractions we arrive at
These are much simpler to integrate. It seems surprising that you're doing differential equations without covering integration by partial fractions
You've only integrated the left hand side. Now, integrate on the right hand side. Remeber RHS will have a constant, C.
Use (t,N)=(0,5) in the integrated equation to determine the value of C. Then you can substitute N=250 to find t when half the population is infected.
If my math is correct t=11.02729384/k. Maybe e^{i*pi) (should we call you "One") can confirm (better yet, see if you can confirm).
again it does not work out...
>>1/500 (ln5 + ln495) = k(0) + C
>>0.01562799 = C
so now i sub in 250 this time
>>1/500 (ln250 + ln250) = kt + 0.01562799
>>0.02200858 = kt + 0.01562799
>>t = 0.00645785/K
this is not the answer i should be getting as in the question it says i should get t=0.00919/K