Respuesta :
The differential equation that describes the situation is given by:
[tex]\frac{dN}{dt} = k(350 - N)[/tex]
The solution is given by:
[tex]N(t) = -250e^{-0.4581t} + 350[/tex]
What is the differential equation that describes the situation?
Considering that the rate of change of bears, N, in a population is directly proportional to 350-N, it is given by:
[tex]\frac{dN}{dt} = k(350 - N)[/tex]
In which k is the rate of change.
How do we solve the differential equation?
Applying separation of variables, which means that every term with N goes to one side of the equality, while every term without N goes to the other side, and then both sides are integrated. Hence:
[tex]\frac{dN}{dt} = k(350 - N)[/tex]
[tex]\frac{dN}{350 - N} = k dt[/tex]
[tex]\int \frac{dN}{350 - N} = \int k dt[/tex]
[tex]-\ln{|350-N|} = kt + K[/tex]
[tex]\ln{|350-N|} = -kt + K[/tex]
[tex]e^\ln{|350-N|}^ = -Ke^{-kt}[/tex]
[tex]350 - N = -Ke^{-kt}[/tex]
[tex]N(t) = Ke^{kt} + 350[/tex]
When t=0, populations is 100, hence:
[tex]100 = K + 350[/tex]
[tex]K = -250[/tex]
Then:
[tex]N(t) = -250e^{kt} + 350[/tex]
When t=2, the population has increased to 250, hence:
[tex]N(t) = -250e^{kt} + 350[/tex]
[tex]250 = -250e^{2k} + 350[/tex]
[tex]e^{2k} = \frac{100}{250}[/tex]
[tex]\ln{e^{2k}} = \ln{\frac{2}{5}}[/tex]
[tex]2k = \ln{\frac{2}{5}}[/tex]
[tex]k = \frac{\ln{\frac{2}{5}}}{2}[/tex]
[tex]k = -0.4581[/tex]
Hence, the solution to the differential equation is:
[tex]N(t) = -250e^{-0.4581t} + 350[/tex]
More can be learned about differential equations at https://brainly.com/question/14318343
