Question

Last question ( calculus )

Answer 1

The growth rate of the population of bears during a given year P(t) varies with logistic model with k=0.2 and a carrying capacity of 300. So it has the following differential equation.

\(\displaystyle\large \frac{dP}{dt}=\frac{P}{5}\left(1-\frac{P}{300}\right) \)

If initially there were 30 bears, approximately how many will there be in 10 years?

Answer 2

im thinkg a bernoiulli approach

Answer 3

I am so stupid. Haven't heard of this name.

Answer 4

its a change of variable technique

p' = p - p^2

p ^(-1) p' = 1 - p

let p = z^-n

p^(-1) = z^n

p' = -n z^-(n-1) z'
----------------------

-n z^n z^-(n-1) z' = 1 - z^-n

-n z z' = 1 - z^-n

let n=-1

-n z z' = 1 - z^-n

z z' = 1 - z

z z' + z= 1

might have to do it again, but thats the basic idea

Answer 5

http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx

Answer 6

yeah, sorry for wasting your time. It is not something that our teacher taught us.
I know about logistic equation P(t)=e^(kt), separation of variables, Euler's method of stepsize, and separation of variables in DE.

Answer 7

said separation of variables twice -:()

Answer 8

well we arent seperable as is ..

Answer 9

I tried to separate them.

\[\frac{5}{P}\left(\frac{300}{300-P}\right)\frac{dP}{dt}=1\]

Answer 10

then I would integrate with respect to t on both sides.

Answer 11

then I get
5log(P)-5log(300-P)=t+C

Answer 12

then perhaps I can use the given that P=30 when t=0

or am I off somewhere ?

Answer 13

\[\frac{dP}{dt}=\frac{P}{5}\left(1-\frac{P}{300}\right)\]
\[\frac{5}{P}\frac{dP}{dt}=1-\frac{P}{300}\]
\[\frac{5}{P}\frac{dP}{dt}=\frac{300-P}{300}\]
\[\frac{1500}{P(300-P)}dP=dt\]

Answer 14

i spose we can work a decomp now

Answer 15

1500 = A(300-P) + B(P)

1500 = A(300-300) + B(300); B=500, A = 500

500(1/P + 1/(300-P)) dP = dt

500(ln(P) - ln(P-300)) = t + C

ln(P) - ln(P-300) = t/500 + C

ln(P) - ln(P-300) = C e^(t/500)

P/(P-300) = C e^(t/500)

Answer 16

5 not 500

Answer 17

yes, I got the integral.
after integrating both sides I have

5log(P)-5log(300-P)=t+C

log(P)-log(300-P)=(1/5)t+C

log[P/(300-P)]=(1/5)t+C

yes, 5, I am working on it now

Answer 18

P/(P-300) = C e^(t/5)
| K |

P = K(P-300)

P = KP-300K

P-KP = -300K

P(1-K) = -300K

P = 300K/(1-K)

Answer 19

sorry I glitched for real

Answer 20

the first line is prolly best if we have aninitial condition to solve for C

Answer 21

C e^(t/5) is the k ?

Answer 22

yeah, easier to keep track of while typing

Answer 23

30/(30-300) = C

-1/9 = C