Solving Doomsday Equations: Initial Conditions and Finite Time Limits

[hangainlover]

Homework Statement

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)

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

(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?

Homework Equations

integration

The Attempt at a Solution

First i wanted to find solution to the dy/dx (meaning i integrated it)

I got y^c = -c(kx+T)

but i could not define it as function y because of the negative sign in front of C

What should i do?

 

[Mark44]

Homework Statement

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)

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

(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?

Homework Equations

integration

The Attempt at a Solution

First i wanted to find solution to the dy/dx (meaning i integrated it)

I'm not sure what you mean by "solution to the dy/dx." There is no x in this problem, so it doesn't make sense to talk about dy/dx. And maybe you are using "dy/dx" as shorthand for differential equation, which is needlessly confusing.

I got y^c = -c(kx+T)

Your solution above is incorrect, and not only because you have x instead of t.

The differential equation dy/dt = ky^1+c is separable. Your next equation should be dy/(y^{1 + c)} = k dt. Integrate both sides. What do you get?

but i could not define it as function y because of the negative sign in front of C

What should i do?

 

[hangainlover]

sorry what i meant by dx or x were dt and t
i apologize for that confusion.
so starting from dy/(y1 + c) = k dt
If i integrate that, 1/(1-1-c)y^(-1-c+1) =kt + T (upper case T is the constant)
-c^(-1)*y^(-c)=kt+T
I got y^(-c)=-c(kt+T)

what am i doing wrong?

 

[hangainlover]

Not knowing how to isolate Y by getting rid of that the exponent, -c, i decided to plug in the initial condition to define the constant T.

at t=0, y^(-c)=-c*T
A(initial)^(-c) =-cT
Therefore, T= ((A(initial)^(-c))/(-c)

 

[Mark44]

sorry what i meant by dx or x were dt and t
i apologize for that confusion.
so starting from dy/(y1 + c) = k dt
If i integrate that, 1/(1-1-c)y^(-1-c+1) =kt + T (upper case T is the constant)
-c^(-1)*y^(-c)=kt+T
I got y^(-c)=-c(kt+T)

what am i doing wrong?

Looks fine so far. You can simplify it some more, though.
$$y^{-c} = -c(kt + T)$$
$$\Rightarrow \frac{1}{y^c}= -(ckt + cT)$$
$$\Rightarrow y^c= \frac{-1}{ckt + cT}$$
$$\Rightarrow y = y(t) = \left(\frac{-1}{ckt + cT}\right)^{1/c}$$

Now use your initial condition that y(0) = y₀ and continue from there to parts b and c of your problem.

 

[hangainlover]

so at t=0, ((-1)/(cT))^(1/c) =A initial

-1/(cT) = A (initial)^c
T = -1 /(cA^c)

after substituting that for the constant T, i get, y(t) = (-A^c)/((A^c)*(kt)-1)

you can see that at t=1/k, as t approaches T from the negative side, the y value approahces infinity
does this answer part b?