Logistic differential equation?

Question

dy/dt=k(1-y/L)y

L= max. carrying capacity

prove y= y0L / [ y0+(L-y0)e^(-kt)

Please show me the prove steps, thank you very much!

Answer 1

So dy / (y(1-y/L)) = kt dt
or dy / (y(y/L-1)) = -kt dt
Expand the left side with partial fractions:
A/y + B/(y/L - 1) = 1/(y(y/L-1))

To get the constants, first multiply through by y and set it equal to 0:
A = -1. Now you can read off B (or get it in the same way):
B = 1/L

Thus, we have:
dy/(y-L) - dy/y = -kt dt

Now we integrate both sides. On the left, we integrate from y0 to y, on the right we integrate from t0 to t:
ln(y-L) - ln(y0 - L) - ln(y) + ln(y0) = -k(t-t0)
or
ln((y-L)/(y0-L)) - ln(y/y0) = -k(t-to)
Exponentiate both sides:
(y-L)/(y0-L) / (y/y0) = e^(-k(t-t0))
Clear the denominators:
y0(y-L) = y(y0-L) e^(-k(t-t0))
Collecting y's we get:
y*y0 + y(L-y0) e^(-k(t-t0)) = L * y0
and now we divide through:
y = L * y0 / (y0 + (L-y0) e^(-k(t-t0)))

Clearly, you are assuming that t0 = 0

Answer 2

dy/dt =(k/L)y(L - y)

dy/[y(L-y)] = (k/L)dt

by partial fractions:

1/[y(L-y)] = (1/L)(1/y) +(1/L)(1/(L-y))

so we have:

dy/y + dy/(L-y) = kdt
(note that (1/L) divides out)

the LHS integrates as ln:

ln[y/yo] - ln[(L - y)/(L - yo)] = kt

by log rules:

ln[ (y/yo)(L-yo)/(L-y)] = kt

(y/yo)(L-yo)/(L-y) = e^kt

y/(L-y) = [yo/(L-yo)]e^kt

y = (L - y) [yo/(L-yo)]e^kt

y = [Lyo/(L-yo)]e^kt - y [yo/(L-yo)]e^kt

y(1 + [yo/(L-yo)]e^kt] = [Lyo/(L-yo)]e^kt
(multiply both sides by (L-yo) )

y( L - yo + yoe^kt) = Lyoe^kt

y = Lyoe^kt/ (L - yo + yoe^kt)

y = Lyo/ ([L - yo]e^-kt + yo)