A population of animals in an ecological niche is growing in time so that it's rate of growth dp/dt is related to its current size by the differential equation dp/dt = 900/p^2. If time is measured in years and initially there are P(0) = 10 animals present, find the population function P(t) giving the size of the population after t years.
If dp/dt = 900/p^2, you can solve that differential equation using separation of variables
Integral of p^2 dp = Integral or 900 dt
p(0) to p(t) ........ = 0 to t
900 t = p(t)^3/3 - p(0)^3/3
p(t)^3 = p(0)^3 + 2700 t = 1000 + 2700 t
Solve for p(t)
Note the growth is not exponential in this case. It is much slower than that, approaching t^(1/3) growth as t goes to infinity
To find the population function P(t), we need to solve the given differential equation.
The differential equation is dp/dt = 900/p^2.
To solve this differential equation, we can rearrange it by multiplying both sides by p^2 and dt:
p^2 dp = 900 dt.
Integrating both sides, we get:
∫p^2 dp = ∫900 dt.
The integral of p^2 with respect to p is (1/3) p^3, and the integral of 900 with respect to t is 900t plus a constant C.
So the equation becomes:
(1/3) p^3 = 900t + C.
Now, we can substitute p(0) = 10 and t = 0 into the equation to find the value of the constant C.
(1/3) (10)^3 = 900(0) + C.
(1/3) (1000) = C.
C = 1000/3.
Substituting this value of C back into the equation, we have:
(1/3) p^3 = 900t + 1000/3.
Now, we can solve for p:
p^3 = 2700t + 1000.
Taking the cube root of both sides gives us:
p = (2700t + 1000)^(1/3).
This is the population function P(t) that gives the size of the population after t years.