Question

How did they go from
\[\frac{dp}{dt} = k_1p-k3\frac{p(p-1)}{2} \] to
\[\frac{dp}{dt} = -Ap(p-p_1), p(0)=p_0\] where
\[A=\frac{k_3}{2}, p_1 = \left( \frac{2k_1}{k_3} \right)+1\]