Logistic differential equation:

\(\frac{dP}{dt}=kP(1-\frac{P}{M})\)

where M is carrying capacity

Hence, to obtain M and k we need to rewrite the given equation:

\(\frac{dP}{dt}=0.05P-0.0005P^2\)

\(=0.05P(1-0.01P)\)

\(=0.05P(1-\frac{P}{100})\)

Therefore we have:

\(\frac{dP}{dt}=0.05P(1-\frac{P}{100})\)

And from this equation we get:

\(k=0.05\quad M=100\)

\(\frac{dP}{dt}=kP(1-\frac{P}{M})\)

where M is carrying capacity

Hence, to obtain M and k we need to rewrite the given equation:

\(\frac{dP}{dt}=0.05P-0.0005P^2\)

\(=0.05P(1-0.01P)\)

\(=0.05P(1-\frac{P}{100})\)

Therefore we have:

\(\frac{dP}{dt}=0.05P(1-\frac{P}{100})\)

And from this equation we get:

\(k=0.05\quad M=100\)