(a) $P=50\left(2^{0.1 t}\right) $

(b) $P=\frac{500\left(2.25^{0.1 t}\right)}{9+2.25^{0.1 t}} ; 500$

(a)
$\frac{d P}{d t}=k P$, where $k$ is a positive constant.
$\int \frac{1}{P} d P=\int k d t$
$P=e^{k t+c}$
$P=A e^{k t}$, where $A=e^c$ is an arbitrary constant.
When $t=0, P=50$, then $A=50$
When $t=10, P=100$, then $e^k=2^{0.1}$
Thus, $P=50\left(2^{0.1 t}\right)$
(b)
$\frac{d P}{d t}=\lambda P(500-P)$
$\int \frac{1}{P(500-P)} d P=\int \lambda d t$
$\frac{1}{500} \int \frac{1}{P}+\frac{1}{500-P} d P=\int \lambda d t$
$\ln P-\ln |500-P|=500 \lambda t+c$
$\frac{P}{500-P}=B e^{500 \lambda t}$, where $B=\pm e^c$ is an arbitrary constant.
When $t=0, P=50$, then $B=\frac{1}{9}$
When $t=10, P=100$, then $e^{5000 \lambda}=2.25 \Rightarrow e^{500 \lambda=2.25^{0.1}}$
$\frac{P}{500-P}=\frac{1}{9} 2.25^{0.1 t}$
$P=\frac{500\left(2.25^{0.1 t}\right)}{9+2.25^{0.1 t}}$
(c) Using $P=\frac{500\left(2.25^{0.1 t}\right)}{9+2.25^{0.1 t}}=\frac{500}{9\left(2.25^{-0.1 t}\right)+1}$, we we that as $t \longrightarrow \infty, e^{-0.1 t} \longrightarrow 0$, then
$P \longrightarrow 500$. Thus, the population will increase and tend to 500 .
Refined model is more realistic as it suggests a limiting value for the population while the first model suggests the population will go to indefinitely.