Hi,

I just wanted to make sure that I completed this problem correctly, because I had a little bit of a rough time with it.

The question reads,

"Letcbe a positive number. A differential equation of the formk

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

whereis a positive constant, is called adoomsday equationbecause the exponent in the expression ky^(1+c) is larger than the exponent 1 for natural growth."

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

Here's what I did...

(dy/dt) = ky^(1.01); c = .01

Separating the DE I get...

(integral) dy/(y^1.01) = k integral (dt)

Integrating...

(y^-.01)/-.01 = kt + A <- "A" is my constant

Now I plug in the initial condition y(0) = 2

(2^-.01)/-.01 = k(0) + A

A = (2^-.01)/-.01 = -99.3

Now I plug in y(3) = 16 and A = -99.3

(16^-.01)/-.01 = k(3) + -99.3

Solving for k I get...

[ (16^-.01)/-.01 + 99.3] / 3 = k = .681

Now I plug k and A back into the original equation...

(y^-.01)/-.01 = .681(t) + -99.3

The question is asking me when the lim(t -> T) = infinity, right? So that means there is an asymptote at t = T. This means I need to find a t that y(t) does not occupy.

Solving for y...

y^-.01 = (.681(t) + -99.3)*-.01

y^-.01 = -.00681t + .993

y = 1/(-.00681t + .993)^.01

So if I solve -.00681t + .993 = 0 for t, then that should be my answer, right? Because that would mean there is a 0 in the denominator.

t = -.993 / -.00681 = (approx) 146 months

Is what I did correct?

Thanks!

P.S. Can anyone tell me the units for k and A?