Birth and death rates of animal populations typically are...

Birth and death rates of animal populations typically are not constant; instead, they vary periodically with the passage of seasons. Find $P(t)$ if the population $P$ satisfies the differential equation $$ \frac{d P}{d t}=(k+b \cos 2 \pi t) P $$ where $t$ is in years and $k$ and $b$ are positive constants. Thus the growth-rate function $r(t)=k+b \cos 2 \pi t$ varies periodically about its mean value $k$. Construct a graph that contrasts the growth of this population with one that has the same initial value $P_{0}$ but satisfies the natural growth equation $P^{\prime}=k P$ (same constant $k$ ). How would the two populations compare after the passage of many years?

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that the birth and death rates of animal populations typically are not constant, and instead they arrive periodically with the passage of season.

00:13 So we need to find the function b of d if the population b satisfies the differential equation.

00:21 And the differential equation is the following.

00:23 The derivative of the population with respect to time is equal to k, which is a constant plus b, which is another constant, cosine of two times pi times the time, and all of this times the population p.

00:46 Where t is in years and k and b are positive constants.

00:51 Now, the growth rate function, the one that is in between the break in here, varies periodically about this mean value k.

01:02 So what we need to do is to construct a graph that construct the growth of this population with the one that has the same initial value, p -0.

01:13 So we are also given the initial value at the time equals to zero.

01:17 The population is b -0.

01:19 But satisfies the natural growth equation.

01:23 And that natural growth equation, we need to compare this to p -prime is equal to k times p, where we know that that is going to have an exponential solution.

01:37 So for that, we just going to first, we're going to solve this equation, this one right here.

01:50 So to do that, we just separate the variables.

01:53 We pass everything that depends on the population to the left and everything that depends on the time to the right.

02:00 So in this case, it's quite easy because only there is this p on the right side that depends on the population.

02:08 And all of this, as you can see, well, it's going to depend on the time.

02:13 So we just put that in here.

02:16 K plus b, cosine of two times pi times the time t.

02:24 And this times the differential in time.

02:27 Now, we integrate both sides of this expression.

02:31 Then the left side is quite easy to obtain because we know that the integral of 1 over something is then a perial logarithm in this case of b and the other side will give us well in this case we will have the integral of a constant so we will obtain k times the time t and in the other way we will have the we can take b out of the integral because it is a constant and we divide this by 2 times pi because that is a turn that multiplies the parameter and the parameter at which we are integrating...

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