Math Question?

Question

The populatin p(t) of a certain group of rabbits satisfies the initial value problem:
dp/dt = kp^2 , p(0) = p(not)
where k is a positive constant, derive this solution

The answer is (p) = p(not) /( 1 - kp(not)t)
Can someone explain how I can get to the final equation

Answer 1

dp/dt = kp^2
dp/p^2 = kdt
-1/p = kt +c
-1/p(not) = c
1/p = 1/p(not) - kt = (1-kp(not))/p(not)
p = p(not) /( 1 - kp(not)t)

Answer 2

First you need to rearrange the equation
dp/p^2 = k dt
integrate 1/p^2 with respect to p will give – 1/p,
similarly for the right hand side, integrate k with respect to t will give kt + C, where C is a constant.
Therefore, – 1/p = kt + C.
p = – 1/ ( kt + C ), when t = 0, p = p (not)
p (not) = – 1/ ( k(0) + C )
C = – 1/ p (not).
Therefore p = – 1/ ( kt + (– 1/ p (not)) )
p = – p (not) / (kp (not)t – 1)
p = p (not) / ( 1 – kp (not)t )

Answer 3

dp/dt = kp^2 , p(0) = p(not)

dp/p^2=kdt
take integral
-1/p=kt+C
C=-1/p-kt

for t=0
C=-1/p(not)

plug this value in p