dp/dt=kp(L-P), where L is the upper limit of P and k is a constant. Prove that L and k are positive constant of p=L/1+Ce^(-klt)

Question

anonymous · Answer

$$\frac{ dp }{ dt }=kP(L-P)       P=\frac{ L }{ 1+Ce ^{-klt} }$$

anonymous · Answer

thanks for ur help

anonymous · Answer

sadly thats only one of the question i find really hard

anonymous · Answer

ill post the other question some other times

loser66 · Answer

@chris00 @oldrin.bataku @eliassaab @tkhunny @druminjosh

anonymous · Answer

Can you separate the variables and set up the antiderivative you need to solve?

loser66 · Answer

Can you do something for her/him? I have no clue.

anonymous · Answer

the only way to prove is to solve  the equation. its a separable ODE (the logistic equation)

anonymous · Answer

@Loser66

loser66 · Answer

I got this part and posted it but the asker and me couldn't see the link between the problem and the solution. The question is Proving and we Solve. Ha!!

anonymous · Answer

http://books.google.com.au/books?id=Z4hwkWDh5NIC&pg=PA388&lpg=PA388&dq=dp/dt%3Dkp(1000-p)&source=bl&ots=pmKEXtV9O6&sig=vHWZI4nqdYJmzgVY0uXfHl2Vcaw&hl=en&sa=X&ei=5bj1UeX7Ho-akgXH9oGQAw&ved=0CGMQ6AEwCA#v=onepage&q=dp%2Fdt%3Dkp(1000-p)&f=false

anonymous · Answer

they have similar equations

anonymous · Answer

yes but when you integrate and solve you will see that L has to be positive