Using separation of variables for the differential equation, it is found that:
a) The differential equation is: [tex]\frac{dP}{dt} = kP[/tex]
b) The solution is: [tex]P(t) = 5000e^{kt}[/tex]
c) In 6 months, the population is: [tex]P(6) = 5000e^{6k}[/tex]
What is the differential equation?
Population of fishes in a pond is directly proportional to amount of fishes present in the bond, hence the differential equation is:
[tex]\frac{dP}{dt} = kP[/tex]
In which k is the growth rate.
What is separation of variables?
In separation of variables, we place all the factors of y on one side of the equation with dy, all the factors of x on the other side with dx, and integrate both sides.
Hence, for the differential equation in this problem:
[tex]\frac{dP}{dt} = kP[/tex]
[tex]\frac{dP}{P} = k dt[/tex]
[tex]\int \frac{dP}{P} = \int k dt[/tex]
[tex]\ln{P} = kt + P(0)[/tex]
[tex]P(t) = P(0)e^{kt}[/tex]
In which P(0) = 5000 is the initial population, hence:
[tex]P(t) = 5000e^{kt}[/tex]
In 6 months, the population is:
[tex]P(6) = 5000e^{6k}[/tex]
More can be learned about separation of variables in differential equations at https://brainly.com/question/14318343
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