Solving Doomsday Equations: Get Help Here!

[Beez]

Hi, I found the same problem I needed a help to solve somewhere in this forum. However, I could not reach the answers only with the help provided on that page. I would really appreciate it if someone could offer me some help.

P: 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 equation in the expression ky^(1+c) is larger than that for natural growth (that is, ky).

(a) Determine the solution that satisfies the initial condition y(0)=y(subzero)

What I did was

dy/y^-(1+c)=k dt Integrate both sides I got
y(t)=1/[ck(T-t)]^(1/c) for some constant T

Is this correct?

(ba) Show that there is a finite time t = ta (doomsday) such that lim(t->T-) wy(t) = infinity

For the equation, infinity = (limt->T-)1/[ck(T-t)]^1/c, when t approaches to T, T=t or T-t=0, which makes the denominator 0, hence the value of the equation becomes infinity.

Is this what I need to say, or should I get the exact value of t (can I?)?

(c) 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 three months, then when is doomsday?

Since y^(1.01), c=0.1. and Y(3)=16, By substituting those numbers to the equation to obtain the value of k.

16^0.1=1/[0.1*k(3)]^(1/0.1)
[0.3k]^10=1/1.32
0.3k=(1/1.32)^(1/10)
k=0.9726

This time use wy(0)=2 to get the value of T

2=1/[0.1*0.9726*T]^(1/10)
[0.1*0.9726*T]^(1/10)=1/2
[0.1*0.9726*T]=(1/2)^(1/10)
T=0.9330/0.09726=(approx)9.6months

How does that sound?
I have no confident with these solutions, especially (c).

Someone, please help me!

 

[arildno]

Hi, I found the same problem I needed a help to solve somewhere in this forum. However, I could not reach the answers only with the help provided on that page. I would really appreciate it if someone could offer me some help.

P: 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 equation in the expression ky^(1+c) is larger than that for natural growth (that is, ky).

(a) Determine the solution that satisfies the initial condition y(0)=y(subzero)

What I did was

dy/y^-(1+c)=k dt Integrate both sides I got
y(t)=1/[ck(T-t)]^(1/c) for some constant T

Is this correct?

Almost correct, but do you agree that T must fulfill:
y_{0}=\frac{1}{(ckT)^{\frac{1}{c}}}

(ba) Show that there is a finite time t = ta (doomsday) such that lim(t->T-) wy(t) = infinity

For the equation, infinity = (limt->T-)1/[ck(T-t)]^1/c, when t approaches to T, T=t or T-t=0, which makes the denominator 0, hence the value of the equation becomes infinity.

Is this what I need to say, or should I get the exact value of t (can I?)?

True, by the above I've mentioned, you know the value of doomsday time T as expressed in c,k,y_{0}

(c) 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 three months, then when is doomsday?

Since y^(1.01), c=0.1. and Y(3)=16, By substituting those numbers to the equation to obtain the value of k.

16^0.1=1/[0.1*k(3)]^(1/0.1)
[0.3k]^10=1/1.32
0.3k=(1/1.32)^(1/10)
k=0.9726

This time use wy(0)=2 to get the value of T

2=1/[0.1*0.9726*T]^(1/10)
[0.1*0.9726*T]^(1/10)=1/2
[0.1*0.9726*T]=(1/2)^(1/10)
T=0.9330/0.09726=(approx)9.6months

How does that sound?
I have no confident with these solutions, especially (c).

Someone, please help me!

You have instead:
2^{0.1}=y(0)^{0.1}=\frac{1}{(0.1kT)^{\frac{1}{0.1}}}
16^{0.1}=y(3)^{0.1}=\frac{1}{(0.1k(T-3))^{\frac{1}{0.1}}}
These equations determines k and T. It's easiest to first solve for kT from the first equation, then k from the second equation, and then determine T.
Welcome to Pf.

 

[Beez]

Thank you

Thank you so much for you help. Your explanation was very clear and helped me understand the problems in great extent! I'm taking an independent class, so I could not get any help from my instructor (it would take a long time to get answers). I should have registered on this forum long time ago!
Thank you again.