Eliminating $P_2=1-P_1$ gives $\dot P_1=\beta-(\alpha+\beta)P_1$. Therefore
$$
P_1(t)=\frac\beta{\alpha+\beta}
+\left(p-\frac\beta{\alpha+\beta}\right)e^{-(\alpha+\beta)t}.
$$
With $r=\alpha p-\beta(1-p)$,
$$
\boxed{P_1=\frac{\beta+re^{-(\alpha+\beta)t}}{\alpha+\beta},
\qquad P_2=\frac{\alpha-re^{-(\alpha+\beta)t}}{\alpha+\beta}.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 356 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-356/1/c/solution