Question

A rabbit population satisfies the logic equation
dy/dt=2×10−7y(106−y)

where t is the time measured in months. The population is suddenly reduced to 40% of its steady state size by myxamatiosis. If the myxamatosis then has no further effect, how large is the population 8 months later? How long will it take for the population to build up again to 90% of its steady state size?

Answer 1

\[dy/dt=2\times10^{-7}y(10^6-y)\]

Answer 2

logistic...

Answer 3

its just integration...

Answer 4

the "steady state size" is the value of y when dy/dx is zero

\[\frac{dy}{dt}=(2×10^{−7})(10^6−y)y = 0\]

so\[ y = 10^6 \]is the steady state size

the initial value is 40% of this, 4 x 10^5 ,and we will let t = 0 at this point.

we need to find y now i think

separation of variables:

\[\frac{dy}{dt}=(2×10^{−7})(10^6−y)y\]

\[\frac{1}{(10^6 - y )y}\frac{dy}{dt}=(2×10^{−7})\]

integrate with respect to t :

\[\int\limits{\frac{1}{(10^6 - y )y}\frac{dy}{dt}}\text{ }dt=\int\limits{(2×10^{−7}})\text{ }dt\]

we will split the left up by partial fractions i think, unless there is an easier way..

Answer 5

thats great! Thanks guys. Really appreciate the help with this one! :)