# Math Help - doomsday equation..

1. ## doomsday equation..

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?

so....
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
does it always have to be function y of c ? can we just leave it at that?

What should i do?

2. Originally Posted by hangainlover
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?

so....
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
does it always have to be function y of c ? can we just leave it at that?

What should i do?

$dy\slash dx=ky^{1+c}\Longleftrightarrow \int\frac{dy}{y^{1+c}}=\int k\,dx\Longrightarrow y^{-c}=kx+T\,,\,\,T=$ a constant $\Longrightarrow y=(kx+T)^{-1\slash c}$ $\Longrightarrow y_0=y(0)=T^{-1\slash c} \Longrightarrow T=y_0^{-c}\Longrightarrow y=\left(kx+y_0^{-c}\right)^{-1\slash c}$

Tonio