Thread: Doomsday Equation Example (Can you check my work?)

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,

"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."

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?

Originally Posted by Coop

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,

"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."

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

Yes, that's good.

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

Okay

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

No. y= 1/(-.00681t+ .993)^(1/.01)= 1/(-.00681t+ .993)^(100).

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?

Except for the ".01" power where it should be 100, yes. And that does not affect the final answer, the number of months.

Thanks!

P.S. Can anyone tell me the units for k and A?[/QUOTE]

"Except for the ".01" power where it should be 100, yes. And that does not affect the final answer, the number of months."

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Ah of course, thanks!