what is the solution for dy/dt= kY(L-Y)

Question

anonymous · Answer

$$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$$

anonymous · Answer

Then?

anonymous · Answer

Use the formula there..
u = KY

v = (L - Y)

x = t

anonymous · Answer

i dont get it

anonymous · Answer

This is a logistic equation. A sketch of the work needed follows.
a) dy/dt = k y [ L-y]....L = carrying capacity and solution is {1/L] [ ln y - ln(L-y)] = kt + C or y = [CL / ( e^[-Lkt] + C)]
b) letting y(0) be initial condition we get y = Ly(0) / [ L e^(-Lkt) + y(0) { 1- e^[-Lkt ] } ] or Lc / [ L e^(-Lkt) + c {1 - e^(-Lkt)} ]
c) L= 2400, c = 120, y(4) = 1200..use this to find k ..{ 0.73391} .....now find t so that y(t) = 2260......about 3 PM...{t=6.9939}

anonymous · Answer

how to derive the answer?

anonymous · Answer

By the way, you don't really need to solve for k. Since
 e^{kt}= (e^k)^t
 You really only need [itex]e^k[/tex].

anonymous · Answer

derive please how do it arrive to that answer?

anonymous · Answer

It is based on Newtons law dear.  it complicated.  It is calculas problem and somethig that you will need to figure out on your own.  I am not trying to be mean but it is just to hard to explain over the computer.