differential equation dy/dt = ky^(1 + c), k > 0

Let c be a positive number. A differential equation of the form

dy/dt = ky^(1+c)

where k is a positive constant, is called a doomsday equation because the exponent in the expression ky^(1+c) is larger than that for natural growth (that is, ky).

It can be shown that there is a finite time t=T (doomsday) such that lim y(t)=infinity, as t approach T from the left.

An especially prolific breed of rabbits has the growth term ky^(1.01). If 2 such rabbits breed initially and the warren has 16 rabbits after four months, then when is doomsday?

Please round the answer to two decimal places.


Can someone explain this problem please?

What it means is the population will approach infinity, therefore, doomsday.

Separate the variables:

\(\displaystyle \L\\\frac{dy}{dy}=ky^{1+c}\), k>0 and c>0.

\(\displaystyle \L\\\int\frac{dy}{y^{1+c}}=\int{kdt}\)

\(\displaystyle \L\\\frac{-y^{-c}}{c}=kt+B\)

\(\displaystyle \L\\y^{-c}=-ckt-cB\)

Let \(\displaystyle t=0\):

\(\displaystyle y_{0}^{-c}=-cB\)

\(\displaystyle \L\\y^{-c}=y_{0}^{-c}-ckt\)

\(\displaystyle \L\\y^{c}=\frac{1}{y_{0}^{-c}-ckt}\)

\(\displaystyle \L\\y=\frac{1}{(y_{0}^{-c}-ckt)^{\frac{1}{c}}}\)

Doomsday comes about when \(\displaystyle y_{0}^{-c}-ckt=0\)

\(\displaystyle \L\\t=\frac{y_{0}^{-c}}{ck}=T\)

Now, can you finish the rabbit section?.

2 rabbits at t=0, 3 months there are 16.

c=1/100

To solve for k, use y(3)=16

Quite frankly, humans would be the appropriate model in this problem instead of rabbits.