# Question on Logistic Equations?

A rabbit population satisfies the logistic equation:

dy/dt = 2 x 10^-7 y (10^6 – y)

where t is the time measured in months. The population is suddenly reduced to 20% of its steady state size by myxamatosis. If the myxamatosis then has no further effect, how large is the population 6 months later? How long will it take for the population to build up again to 90% of its steady state size?

Relevance
• 9 years ago

To solve this separate the variables onto different sides then integrate.

dy / dt = 2 x 10ˉ⁷ x y(10⁶ - y)

dy / dt = 2y(10⁶ - y) / 10⁷

dy / y(10⁶ - y) = 2dt / 10⁷

∫ 1 / y(10⁶ - y) dy = 2 ∫ 1 dt / 10⁷

To integrate the first part we need to use partial fractions:

1 / y(10⁶ - y) = A / y + B / (10⁶ - y)

1 = A(10⁶ - y) + By

1 = 10⁶A - Ay + By

1 = (B - A)y + 10⁶A

10⁶A = 1

A = 1 / 10⁶

B - A = 0

B = A

B = 1 / 10⁶

1 / y(10⁶ - y) = 1 / 10⁶y + 1 / 10⁶(10⁶ - y)

Now we can integrate both sides of the equation:

∫ 1 / y(10⁶ - y) dy = 2 ∫ 1 dt / 10⁷

∫ 1 / 10⁶y + ∫ 1 / 10⁶(10⁶ - y) = 2 ∫ 1 dt / 10⁷

∫ 1 / 10⁶y - ∫ 1 / 10⁶(y - 10⁶) = 2 ∫ 1 dt / 10⁷

ln|y| / 10⁶ - ln|y - 10⁶| / 10⁶ = 2t / 10⁷ + C

Finally solve the equation for y:

(ln|y| - ln|y - 10⁶|) / 10⁶ = 2t / 10⁷ + C

ln|y / (y - 10⁶)| / 10⁶ = 2t / 10⁷ + C

ln|y / (y - 10⁶)| = t / 5 + C

y / (y - 10⁶) = e ^ (t / 5 + C)

y / (y - 10⁶) = e ^ C * e ^ (t / 5)

y / (y - 10⁶) = ke ^ (t / 5)

(y - 10⁶ + 10⁶) / (y - 10⁶) = ke ^ (t / 5)

(y - 10⁶) / (y - 10⁶) + 10⁶ / (y - 10⁶) = ke ^ (t / 5)

1 + 10⁶ / (y - 10⁶) = ke ^ (t / 5)

10⁶ / (y - 10⁶) = ke ^ (t / 5) - 1

(y - 10⁶) / 10⁶ = 1 / [1 - ke ^ (t / 5)]

y - 10⁶ = 10⁶ / [1- ke ^ (t / 5)]

y = 10⁶ + 10⁶ / [1 - ke ^ (t / 5)]

From here you would need some condition to find the constant k and continue further.