how to solve this differiential equation? dy/dt=ky(1-y)?

Answer 1

dy/dt = ky(1-y)
seperation of variables
(1/y*(1-y)) dy = k dt
integrate
-ln ((y-1)/y) + c = kt +c
move c's to right side and apply both sides as power of e
(y/(y-1)) = e ^ (kt +c)
drop down c since e ^ (kt +c) = c*e ^ (kt)
(y/(y-1)) = c*e ^ (kt)
Uh, well, not sure where to go with solving here but it ends up as
y = (c*e ^ (kt)) / (c*e ^ (kt) - 1)

Good ol' TI-89...Sorry I wasn't really sure where to go with the solving for y. Maybe you can see it... Also, did it have an inital condition?

Haha, never mind, ok, here we go, figured it out:

y / (y - 1) = e ^ (kt + c)
multiply by y-1
y = (y - 1)*e ^ (kt + c)
distribute
y = y*e^(kt + c) - e ^ (kt + c)
move y*e^(kt + c) to left
y - y*e ^ (kt + c) = -e ^ (kt + c)
factor out y on left side
y (e ^ (kt + c) - 1) = e ^ (kt + c)
divide (e ^ (kt + c) - 1) to move to right side
y = e ^ (kt + c) / (e ^ (kt + c) - 1)

Yay!

Answer 2

This is separable. Dividing through, we get dy/y(1-y) = kdt. Integrating both sides, Log[y/y-1] = kt+C. So, solving for y, we get
y = Ce^kt / (Ce^kt-1). Note that y=0 and y=1 are also solutions.

Steve

Answer 3

dy/dt=ky(1-y)
by solving the eqn, (1-y)dy=kydt
divide through by y (1/y-1)dy=kdt

Intergrate both sides with respect to y and t
==> lny-y=kt
==> lny-y-kt=0 Is the general solution

Answer 4

Definition: y=t^a dy/dt=at^(a-1)

Step 1. a=k
Step 2. (a-1)=(1-y)
Step 3. t=y

Substitute 1 & 2, (k-1)=(1-y)
Answer, y=t