Answer:
a) ![t = 2 *ln(\frac{82}{5}) =5.595](https://tex.z-dn.net/?f=%20t%20%3D%202%20%2Aln%28%5Cfrac%7B82%7D%7B5%7D%29%20%3D5.595)
b) ![t = 2 *ln(-\frac{820}{p_0 -820})](https://tex.z-dn.net/?f=%20t%20%3D%202%20%2Aln%28-%5Cfrac%7B820%7D%7Bp_0%20-820%7D%29%20)
c) ![p_0 = 820-\frac{820}{e^6}](https://tex.z-dn.net/?f=%20p_0%20%3D%20820-%5Cfrac%7B820%7D%7Be%5E6%7D)
Step-by-step explanation:
For this case we have the following differential equation:
![\frac{dp}{dt}=\frac{1}{2} (p-820)](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdp%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%28p-820%29)
And if we rewrite the expression we got:
![\frac{dp}{p-820}= \frac{1}{2} dt](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdp%7D%7Bp-820%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%20dt)
If we integrate both sides we have:
![ln|P-820|= \frac{1}{2}t +c](https://tex.z-dn.net/?f=ln%7CP-820%7C%3D%20%5Cfrac%7B1%7D%7B2%7Dt%20%2Bc)
Using exponential on both sides we got:
![P= 820 + P_o e^{1/2t}](https://tex.z-dn.net/?f=%20P%3D%20820%20%2B%20P_o%20e%5E%7B1%2F2t%7D)
Part a
For this case we know that p(0) = 770 so we have this:
![770 = 820 + P_o e^0](https://tex.z-dn.net/?f=%20770%20%3D%20820%20%2B%20P_o%20e%5E0)
![P_o = -50](https://tex.z-dn.net/?f=%20P_o%20%3D%20-50)
So then our model would be given by:
![P(t) = -50e^{1/2t} +820](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%20-50e%5E%7B1%2F2t%7D%20%2B820)
And if we want to find at which time the population would be extinct we have:
![0=-50 e^{1/2 t} +820](https://tex.z-dn.net/?f=%200%3D-50%20e%5E%7B1%2F2%20t%7D%20%2B820)
![\frac{820}{50} = e^{1/2 t}](https://tex.z-dn.net/?f=%20%5Cfrac%7B820%7D%7B50%7D%20%3D%20e%5E%7B1%2F2%20t%7D)
Using natural log on both sides we got:
![ln(\frac{82}{5}) = \frac{1}{2}t](https://tex.z-dn.net/?f=%20ln%28%5Cfrac%7B82%7D%7B5%7D%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dt)
And solving for t we got:
![t = 2 *ln(\frac{82}{5}) =5.595](https://tex.z-dn.net/?f=%20t%20%3D%202%20%2Aln%28%5Cfrac%7B82%7D%7B5%7D%29%20%3D5.595)
Part b
For this case we know that p(0) = p0 so we have this:
![p_0 = 820 + P_o e^0](https://tex.z-dn.net/?f=%20p_0%20%3D%20820%20%2B%20P_o%20e%5E0)
![P_o = p_0 -820](https://tex.z-dn.net/?f=%20P_o%20%3D%20p_0%20-820)
So then our model would be given by:
![P(t) = (p_o -820)e^{1/2t} +820](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%20%28p_o%20-820%29e%5E%7B1%2F2t%7D%20%2B820)
And if we want to find at which time the population would be extinct we have:
![0=(p_o -820)e^{1/2 t} +820](https://tex.z-dn.net/?f=%200%3D%28p_o%20-820%29e%5E%7B1%2F2%20t%7D%20%2B820)
![-\frac{820}{p_0 -820} = e^{1/2 t}](https://tex.z-dn.net/?f=%20-%5Cfrac%7B820%7D%7Bp_0%20-820%7D%20%3D%20e%5E%7B1%2F2%20t%7D)
Using natural log on both sides we got:
![ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t](https://tex.z-dn.net/?f=%20ln%28-%5Cfrac%7B820%7D%7Bp_0%20-820%7D%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dt)
And solving for t we got:
![t = 2 *ln(-\frac{820}{p_0 -820})](https://tex.z-dn.net/?f=%20t%20%3D%202%20%2Aln%28-%5Cfrac%7B820%7D%7Bp_0%20-820%7D%29%20)
Part c
For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:
![12 = 2 *ln(\frac{820}{820-p_0})](https://tex.z-dn.net/?f=%2012%20%3D%202%20%2Aln%28%5Cfrac%7B820%7D%7B820-p_0%7D%29%20)
![6 = ln (\frac{820}{820-p_0})](https://tex.z-dn.net/?f=%206%20%3D%20ln%20%28%5Cfrac%7B820%7D%7B820-p_0%7D%29%20)
Using exponentials we got:
![e^6 = \frac{820}{820-p_0}](https://tex.z-dn.net/?f=%20e%5E6%20%3D%20%5Cfrac%7B820%7D%7B820-p_0%7D)
![(820-p_0) e^6 = 820](https://tex.z-dn.net/?f=%20%28820-p_0%29%20e%5E6%20%3D%20820)
![820-p_0 = \frac{820}{e^6}](https://tex.z-dn.net/?f=%20820-p_0%20%3D%20%5Cfrac%7B820%7D%7Be%5E6%7D)
![p_0 = 820-\frac{820}{e^6}](https://tex.z-dn.net/?f=%20p_0%20%3D%20820-%5Cfrac%7B820%7D%7Be%5E6%7D)