So I was presented a problem with 3 components; the first 2 I believe I've done right; but I need a bit of push for the 3rd component:
dP/dt = (r/K)(P PU )(PS P)
where P(t) is the population size, r is the intrinsic reproduction rate, K is the carrying capacity, PU is the unstable equilibrium and PS is the stable equilibrium.
1) Use separation of variables and partial fractions to determine the general solution to the above differential equation. Leave your solution in implicit form, you do not need to make P the subject.
dP/dt = (r/k) (P - Pu)(Ps - P)
dP/(P - Pu)(Ps - P) = (r/k) dt
Integrating both sides:
∫dP/(P - Pu)(Ps - P) = rt/k + C
1/(Ps - Pu) * ∫[(P - Pu) + (Ps - P)]/(P - Pu)(Ps - P) dP = rt/k + C
1/(Ps - Pu) * ∫1/(Ps - P) + 1/(P - Pu) dP = rt/k + C
1/(Ps - Pu) * [-ln|Ps - P| + ln|P - Pu|] = rt/k + C
1/(Ps - Pu) * ln|(P - Pu)/(Ps - P)| = rt/k + C
2)Now assume that PU < P < PS and show that the general solution can be written in the forma ln(P-PU)/(PS-P) = (r/K)t + C
where C is an arbitrary constant. Give an expression for the coecient a.
1/(Ps - Pu) * [-ln|Ps - P| + ln|P - Pu|] = rt/k + C
1/(Ps - Pu) * ln|(P - Pu)/(Ps - P)| = rt/k + C
If Pv < P < Ps, then you can take out the absolute values and use the log property: ln(a) - ln(b) = ln(a/b) to factor the logs.
Therefore a represents 1/(Ps - Pu)
a ln|(P - Pu)/(Ps - P)| = rt/k + C
3)The constant harvest model has r = 0:3 per year, K = 3000 kilotons, PS = 1600 kilotons and PU = 1400 kilotons. A disaster has reduced the population close to the unstable equilibrium. Assuming P = 1410 when t = 0 calculate the time taken for the population to reach P = 1500.
After attempts; I know I'm not getting near the right answer.. Well, more or less, I'm unsure how to tackle the problem..
dP/dt = (r/K)(P PU )(PS P)
where P(t) is the population size, r is the intrinsic reproduction rate, K is the carrying capacity, PU is the unstable equilibrium and PS is the stable equilibrium.
1) Use separation of variables and partial fractions to determine the general solution to the above differential equation. Leave your solution in implicit form, you do not need to make P the subject.
dP/dt = (r/k) (P - Pu)(Ps - P)
dP/(P - Pu)(Ps - P) = (r/k) dt
Integrating both sides:
∫dP/(P - Pu)(Ps - P) = rt/k + C
1/(Ps - Pu) * ∫[(P - Pu) + (Ps - P)]/(P - Pu)(Ps - P) dP = rt/k + C
1/(Ps - Pu) * ∫1/(Ps - P) + 1/(P - Pu) dP = rt/k + C
1/(Ps - Pu) * [-ln|Ps - P| + ln|P - Pu|] = rt/k + C
1/(Ps - Pu) * ln|(P - Pu)/(Ps - P)| = rt/k + C
2)Now assume that PU < P < PS and show that the general solution can be written in the forma ln(P-PU)/(PS-P) = (r/K)t + C
where C is an arbitrary constant. Give an expression for the coecient a.
1/(Ps - Pu) * [-ln|Ps - P| + ln|P - Pu|] = rt/k + C
1/(Ps - Pu) * ln|(P - Pu)/(Ps - P)| = rt/k + C
If Pv < P < Ps, then you can take out the absolute values and use the log property: ln(a) - ln(b) = ln(a/b) to factor the logs.
Therefore a represents 1/(Ps - Pu)
a ln|(P - Pu)/(Ps - P)| = rt/k + C
3)The constant harvest model has r = 0:3 per year, K = 3000 kilotons, PS = 1600 kilotons and PU = 1400 kilotons. A disaster has reduced the population close to the unstable equilibrium. Assuming P = 1410 when t = 0 calculate the time taken for the population to reach P = 1500.
After attempts; I know I'm not getting near the right answer.. Well, more or less, I'm unsure how to tackle the problem..
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You've done well on 1.) and 2.). :)
3.) So we have (1/(Ps - Pu))*ln[(P - Pu)/(Ps - P)] = (r/k)*t + C.
We are given values for r, k, Pu and Ps, and that P(0) = 1410. So we have
[1/(1600 - 1400)]*ln[(1410 - 1400)/(1600 - 1410)] = (0.3/3000)*0 + C --->
(1/200)*ln(10/190) = C ---> C = ln(1/19) /200 = -ln(19) /200.
So to find t so that P(t) = 1500 we have
(1/200)*ln(100/100) = (0.3/3000)*t - (ln(19) /200) --->
0 = (0.3/3000)*t - (ln(19) /200) --->
(0.3/3000)*t = ln(19) /200 --->
t = (3000/0.3) * (ln(19) /200) = 147.2 years.
3.) So we have (1/(Ps - Pu))*ln[(P - Pu)/(Ps - P)] = (r/k)*t + C.
We are given values for r, k, Pu and Ps, and that P(0) = 1410. So we have
[1/(1600 - 1400)]*ln[(1410 - 1400)/(1600 - 1410)] = (0.3/3000)*0 + C --->
(1/200)*ln(10/190) = C ---> C = ln(1/19) /200 = -ln(19) /200.
So to find t so that P(t) = 1500 we have
(1/200)*ln(100/100) = (0.3/3000)*t - (ln(19) /200) --->
0 = (0.3/3000)*t - (ln(19) /200) --->
(0.3/3000)*t = ln(19) /200 --->
t = (3000/0.3) * (ln(19) /200) = 147.2 years.