Question

dN/dt=r(1-N/K) solve for N(t) using Partial Fractions.

Answer 1

\[\frac{dN}{dt}=r(1-\frac{N}{K})=r-\frac{rN}{K}\]
If that's your problem, then it's a differential equation. I feel like that's not what you are trying to solve. Could you clarify if the quest is as such?

Answer 2

how can i use partial fractions in this case to solve for N(t)

Answer 3

This web sit does a good job of explaining what partial fractions is.
http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/partial.html

You'll find out that this equation doesn't have a partial fraction.

Answer 4

its an Allee Effect problem

if you know how to do that one

Answer 5

Yes, you solve it using a first order linear differntial equation. Have you taken differntial equation yet?

Answer 6

not yet, need to do it for a biology question

Answer 7

i know what the individual parts represent like what K is , what r is

Answer 8

how do i solve it

Answer 9

\[\frac{dN}{dt}=r(1-\frac{N}{K})=r-\frac{rN}{K}\]

\[\frac{dN}{dt}+\frac{r}{K}N=r\]
\[\mu*\frac{dN}{dt}+N\frac{d\mu}{dt}=\mu*r\]
where
\[\frac{r}{K}=\frac{1}{\mu}\frac{d\mu}{dt}\]
Solve for that using separation of variables and integrating, you get.
\[\mu=e^{\frac{r}{K}t}\]

Essentially,
\[\mu*N(t) = Ke^{\frac{r}{K}t}+c\]
therefore
\[N(t)=K+c*e^{\frac{r}{K}t}\]
where c is a constant. If you don't follow, then I don't think this problem was properly assigned to you. Or this problem has more of a theoretical understanding rather than mathematical.

Answer 10

If you look at the equation at the end, you'll see that it's a exponential rate. The population of N(t) grows at a rate of r/K as time increases.

Answer 11

so if K=1 and r=1 and at N(0)=0.01 i can find c?

Answer 12

yes

Answer 13

i get C=.99

Answer 14

Did they really ask you to solve that problem in biology? That seems very farfetched. You must either have a hardcore teacher or misunderstood the problem.
You C has the wrong sign.
.01-1=-0.99

Answer 15

thanks man

Answer 16

when i graphed my problem from 0 to 10 for t i got a negative exponetial graph

Answer 17

yes because you are subtracting the exponent from your K value. Your initial condition made c negative.
If your initial problem was dN/dt=r+rN/K (notice its positive instead of negative) then your solution would be N(t) = K + .99e^(rt/K)

Answer 18

i tried to also find the equilibrium dN/dt=0. i got 2 but the question says 3 equilibrium points

Answer 19

when dN/dt = 0 you can just solve the initial equation dN/dt = r - rN/K =0 where N=K

Answer 20

I found the book where this guy got the question from its rediculous. he used a question from a diff eq book