Respuesta :
Answer:
[tex] \frac{dP}{\sqrt{P}} = k dt[/tex]
And if we integrate both sides we got:
[tex] 2 \sqrt{P} = kt +C[/tex]
Where C is a constant., we can rewrite the expression like this:
[tex] \sqrt{P} = \frac{1}{2} (kt +C)[/tex]
If we square both sides we got:
[tex] P = \frac{1}{4} (kt +C)^2 [/tex]
If we use the initial condition we have that:
[tex] P(0) = 100 = \frac{1}{4} (k*0 +C)^2 [/tex]
And we can solve for C like this:
[tex] 400 = C^2[/tex]
[tex] C = 20[/tex]
And now we can find the derivate of the function and we got:
[tex] P'(t) = 2* \frac{1}{4} (kt + 20) * k[/tex]
Using the condition [tex] P'(0) = 10 [/tex] we got:
[tex] 10 = \frac{1}{2} k (k*0 +20)[/tex]
[tex] 20 = 20 k[/tex]
k= 1
And then the model is defined as:
[tex] P = \frac{1}{4} (t +20)^2 [/tex]
And for t =12 months we have:
[tex] P(12) = \frac{1}{4} (12 +20)^2 = 256 [/tex]
Step-by-step explanation:
For this case we cna use the proportional model given by:
[tex] \frac{dP}{dt} = k \sqrt{P}[/tex]
Where k is a proportional constant, P the population and the represent the number of months
For this case we know the following initial condition [tex] P(0) =100[/tex] and [tex] P'(0) = 10 [/tex]
we can rewrite the differential equation like this:
[tex] \frac{dP}{\sqrt{P}} = k dt[/tex]
And if we integrate both sides we got:
[tex] 2 \sqrt{P} = kt +C[/tex]
Where C is a constant., we can rewrite the expression like this:
[tex] \sqrt{P} = \frac{1}{2} (kt +C)[/tex]
If we square both sides we got:
[tex] P = \frac{1}{4} (kt +C)^2 [/tex]
If we use the initial condition we have that:
[tex] P(0) = 100 = \frac{1}{4} (k*0 +C)^2 [/tex]
And we can solve for C like this:
[tex] 400 = C^2[/tex]
[tex] C = 20[/tex]
And now we can find the derivate of the function and we got:
[tex] P'(t) = 2* \frac{1}{4} (kt + 20) * k[/tex]
Using the condition [tex] P'(0) = 10 [/tex] we got:
[tex] 10 = \frac{1}{2} k (k*0 +20)[/tex]
[tex] 20 = 20 k[/tex]
k= 1
And then the model is defined as:
[tex] P = \frac{1}{4} (t +20)^2 [/tex]
And for t =12 months we have:
[tex] P(12) = \frac{1}{4} (12 +20)^2 = 256 [/tex]
