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

Answer:

1) $y(t) = \frac{Ce^{4t} +3}{4}$

2) $5 = \frac{C e^0 +3}{4}$

$20 = C +3$

$C= 20-3=17$

So then the model would be given by:

$y(t) = \frac{20e^{4t} +3}{4}$

Step-by-step explanation:

For this case we have the following differential equation:

$y' = \frac{dy}{dt} = 4y-3$

We can do this using algebra:

$\frac{dy}{4y-3} = dt$

Part 1

And now we can integrate both sides like this:

$\int \frac{dy}{4y-3} = \int dt$

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

$\frac{1}{4} \int\frac{du}{u} = t+ c$

We can multiply both sides by 4 and we got:

$\int\frac{du}{4} = 4t+ 4k = 4t+ c$, 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:

$y(t) = \frac{Ce^{4t} +3}{4}$

Part 2

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

$5 = \frac{C e^0 +3}{4}$

$20 = C +3$

$C= 20-3=17$

So then the model would be given by:

$y(t) =\frac{17te^{4t} +3}{4}$