Question

Consider a population of rabbits in a region with unlimited food resources. Suppose that the growth rate is proportional to the population with a proportionality factor of 4 per month, and every month we collect 3 rabbits. Then, the rabbit population is described by the differential equation

y′=4y−3.

1. Find an explicit expression of all solutions y of the differential equation above. Denote by c any arbitrary integration constant.

2. Suppose the initial rabbit population is 5. From all the solutions above, find the only solution that satisfies the initial condition y(0)=5

Answer 1

1)
`y(t) = (Ce^(4t) +3)/(4)`

2)
`5 = (C e^0 +3)/(4)`

`20 = C +3`

`C= 20-3=17`

So then the model would be given by:

`y(t) = (20e^(4t) +3)/(4)`

Explanation:

For this case we have the following differential equation:

`y' = (dy)/(dt) = 4y-3`

We can do this using algebra:

`(dy)/(4y-3) = dt`

Part 1

And now we can integrate both sides like this:

`\int (dy)/(4y-3) = \int dt`

We can use the u substituton for the left part
`u = 4y-3`
`du = 4 dy`

`(1)/(4) \int(du)/(u) = t+ c`

We can multiply both sides by 4 and we got:

`\int(du)/(4) = 4t+ 4k = 4t+ c`, where
`c=4k`

and after integrate the left part we got:

`ln |u| = 4t+c`

And if we apply exponential on both sides we got:

`u = e^(4t) e^c`

Now we can replace u and we got:

`4y-3 = Ce^(4t)` where
`C=e^c`

And then finally we have:

`y(t) = (Ce^(4t) +3)/(4)`

Part 2

For this case we have the initial condition y(0) =5 and if we use it we got:

`5 = (C e^0 +3)/(4)`

`20 = C +3`

`C= 20-3=17`

So then the model would be given by:

`y(t) =(17te^(4t) +3)/(4)`