Answer:
![\frac{dP}{\sqrt{P}} = k dt](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7B%5Csqrt%7BP%7D%7D%20%3D%20k%20dt)
And if we integrate both sides we got:
![2 \sqrt{P} = kt +C](https://tex.z-dn.net/?f=%202%20%5Csqrt%7BP%7D%20%3D%20kt%20%2BC)
Where C is a constant., we can rewrite the expression like this:
![\sqrt{P} = \frac{1}{2} (kt +C)](https://tex.z-dn.net/?f=%20%5Csqrt%7BP%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28kt%20%2BC%29)
If we square both sides we got:
![P = \frac{1}{4} (kt +C)^2](https://tex.z-dn.net/?f=%20P%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%28kt%20%2BC%29%5E2%20)
If we use the initial condition we have that:
![P(0) = 100 = \frac{1}{4} (k*0 +C)^2](https://tex.z-dn.net/?f=%20P%280%29%20%3D%20100%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%28k%2A0%20%2BC%29%5E2%20)
And we can solve for C like this:
![400 = C^2](https://tex.z-dn.net/?f=%20400%20%3D%20C%5E2)
![C = 20](https://tex.z-dn.net/?f=%20C%20%3D%2020)
And now we can find the derivate of the function and we got:
![P'(t) = 2* \frac{1}{4} (kt + 20) * k](https://tex.z-dn.net/?f=%20P%27%28t%29%20%3D%202%2A%20%5Cfrac%7B1%7D%7B4%7D%20%28kt%20%2B%2020%29%20%2A%20k)
Using the condition
we got:
![10 = \frac{1}{2} k (k*0 +20)](https://tex.z-dn.net/?f=%2010%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20%28k%2A0%20%2B20%29)
![20 = 20 k](https://tex.z-dn.net/?f=%2020%20%3D%2020%20k)
k= 1
And then the model is defined as:
![P = \frac{1}{4} (t +20)^2](https://tex.z-dn.net/?f=%20P%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%28t%20%2B20%29%5E2%20)
And for t =12 months we have:
![P(12) = \frac{1}{4} (12 +20)^2 = 256](https://tex.z-dn.net/?f=%20P%2812%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2812%20%2B20%29%5E2%20%3D%20256%20)
Step-by-step explanation:
For this case we cna use the proportional model given by:
![\frac{dP}{dt} = k \sqrt{P}](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7Bdt%7D%20%3D%20k%20%5Csqrt%7BP%7D)
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
and ![P'(0) = 10](https://tex.z-dn.net/?f=%20P%27%280%29%20%3D%2010%20)
we can rewrite the differential equation like this:
![\frac{dP}{\sqrt{P}} = k dt](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7B%5Csqrt%7BP%7D%7D%20%3D%20k%20dt)
And if we integrate both sides we got:
![2 \sqrt{P} = kt +C](https://tex.z-dn.net/?f=%202%20%5Csqrt%7BP%7D%20%3D%20kt%20%2BC)
Where C is a constant., we can rewrite the expression like this:
![\sqrt{P} = \frac{1}{2} (kt +C)](https://tex.z-dn.net/?f=%20%5Csqrt%7BP%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28kt%20%2BC%29)
If we square both sides we got:
![P = \frac{1}{4} (kt +C)^2](https://tex.z-dn.net/?f=%20P%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%28kt%20%2BC%29%5E2%20)
If we use the initial condition we have that:
![P(0) = 100 = \frac{1}{4} (k*0 +C)^2](https://tex.z-dn.net/?f=%20P%280%29%20%3D%20100%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%28k%2A0%20%2BC%29%5E2%20)
And we can solve for C like this:
![400 = C^2](https://tex.z-dn.net/?f=%20400%20%3D%20C%5E2)
![C = 20](https://tex.z-dn.net/?f=%20C%20%3D%2020)
And now we can find the derivate of the function and we got:
![P'(t) = 2* \frac{1}{4} (kt + 20) * k](https://tex.z-dn.net/?f=%20P%27%28t%29%20%3D%202%2A%20%5Cfrac%7B1%7D%7B4%7D%20%28kt%20%2B%2020%29%20%2A%20k)
Using the condition
we got:
![10 = \frac{1}{2} k (k*0 +20)](https://tex.z-dn.net/?f=%2010%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20%28k%2A0%20%2B20%29)
![20 = 20 k](https://tex.z-dn.net/?f=%2020%20%3D%2020%20k)
k= 1
And then the model is defined as:
![P = \frac{1}{4} (t +20)^2](https://tex.z-dn.net/?f=%20P%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%28t%20%2B20%29%5E2%20)
And for t =12 months we have:
![P(12) = \frac{1}{4} (12 +20)^2 = 256](https://tex.z-dn.net/?f=%20P%2812%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2812%20%2B20%29%5E2%20%3D%20256%20)