Question

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 the exponent 1 for natural growth. An especially prolific breed of rabbits has the growth term My^1.01. If 2 such rabbits breed initially and the warren has 16 rabbits after three months, then when is doomsday?

Answer 1

The doomsday is 146 days

Explanation:

Given

`(dy)/(dt) = ky^(1 +c)`

First, we calculate the solution that satisfies the initial solution

Multiply both sides by

`(dt)/(y^(1+c))`

`(dt)/(y^(1+c)) * (dy)/(dt) = ky^(1 +c) * (dt)/(y^(1+c))`

`(dy)/(y^(1+c)) = k\ dt`

Take integral of both sides

`\int (dy)/(y^(1+c)) = \int k\ dt`

`\int y^(-1-c)\ dy = \int k\ dt`

`\int y^(-1-c)\ dy = k\int\ dt`

Integrate

`(y^(-1-c+1))/(-1-c+1) = kt+C`

`-(y^(-c))/(c) = kt+C`

To find c; let t= 0

`-(y_0^(-c))/(c) = k*0+C`

`-(y_0^(-c))/(c) = C`

`C =-(y_0^(-c))/(c)`

Substitute
`C =-(y_0^(-c))/(c)` in
`-(y^(-c))/(c) = kt+C`

`-(y^(-c))/(c) = kt-(y_0^(-c))/(c)`

Multiply through by -c

`y^(-c) = -ckt+y_0^(-c)`

Take exponents of
`-c^{-1`

`y^{-c*-c^(-1)} = [-ckt+y_0^(-c)]^{-c^(-1)`

`y = [-ckt+y_0^(-c)]^{-c^(-1)`

`y = [-ckt+y_0^(-c)]^{-(1)/(c)}`

i.e.

`y(t) = [-ckt+y_0^(-c)]^{-(1)/(c)}`

Next:

`t= 3` i.e. 3 months

`y_0 = 2` --- initial number of breeds

So, we have:

`y(3) = [-ck * 3+2^(-c)]^{-(1)/(c)}`

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We have the growth term to be:
`ky^(1.01)`

This implies that:

`ky^(1.01) = ky^(1+c)`

By comparison:

`1.01 = 1 + c`

`c = 1.01 - 1 = 0.01`

`y(3) = 16` --- 16 rabbits after 3 months:

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`y(3) = [-ck * 3+2^(-c)]^{-(1)/(c)}`

`16 = [-0.01 * 3 * k + 2^(-0.01)]^{(-1)/(0.01)}`

`16 = [-0.03 * k + 2^(-0.01)]^(-100)`

`16 = [-0.03 k + 0.9931]^(-100)`

Take -1/100th root of both sides

`16^(-1/100) = -0.03k + 0.9931`

`0.9727 = -0.03k + 0.9931`

`0.03k= - 0.9727 + 0.9931`

`0.03k= 0.0204`

`k= (0.0204)/(0.03)`

`k= 0.68`

Recall that:

`-(y^(-c))/(c) = kt+C`

This implies that:

`(y_0^(-c))/(c) = kT`

Make T the subject

`T = (y_0^(-c))/(kc)`

Substitute:
`k= 0.68`,
`c = 0.01` and
`y_0 = 2`

`T = (2^(-0.01))/(0.68 * 0.01)`

`T = (2^(-0.01))/(0.0068)`

`T = (0.9931)/(0.0068)`

`T = 146.04`

The doomsday is 146 days