Answer
a)
$\frac{d^2P}{dt^2}=k^2P(1-\frac{P}{M})(1-\frac{2P}{M})$
b)
$P=\frac{M}{2}$
Work Step by Step
a)
As we know $\frac{dP}{dt}=kP(1-\frac{P}{M})$
$\frac{dP}{dt}=kP(1-\frac{P}{M})$
$\frac{d^2P}{dt^2}=k[P(-\frac{1}{M}\frac{dP}{dt})+(1-\frac{P}{M})\frac{dP}{dt}]$
$\frac{d^2P}{dt^2}=k\frac{dP}{dt}(-\frac{P}{M}+1-\frac{P}{M})$
$\frac{d^2P}{dt^2}=k[kP(1-\frac{P}{M})](1-\frac{2P}{M})$
$\frac{d^2P}{dt^2}=k^2P(1-\frac{P}{M})(1-\frac{2P}{M})$
b)
$P$ grows faster when $P'$ has a maximum, that is when $P''=0$
from (a) $P''=0$
$P=0,~ P=M$ or $P=\frac{M}{2}$
As we know $0
