This is a note for the Kolmogorov differential equations from the MIT OpenCourseWare Discrete Stochastic Processes. For the example 6.3.1 a 2-state Markov process where \(q_{01}=\lambda\) and \(q_{10}=\mu\)

Use the Kolmogorov forward equations for \(P_{01}(t)\), we get the following equation:

\[\dfrac{dP_{01}(t)}{dt} = \lambda -P_{01}(t)(\lambda+\mu) \tag{1}\]

Using the boundary condtion \(P_{01}(0)=0\), the solution is

\[P_{01}(t) = \dfrac{\lambda}{\lambda+\mu}\left[ 1-e^{-(\lambda+\mu)t}\right] \]

We show how to get this solution

From \((1)\) we have

\[\dfrac{dP_{01}(t)}{dt} +(\lambda+\mu) P_{01}(t)= \lambda \tag{2}\]

This is just a first order linear differential equation.

Let \[u(t)=e^{(\lambda+\mu)t} \tag{3}\]

Multiply \((3)\) at both side of the equation \((2)\)

\[\dfrac{dP_{01}(t)}{dt} e^{(\lambda+\mu)t}+(\lambda+\mu) P_{01}(t)e^{(\lambda+\mu)t}= \lambda e^{(\lambda+\mu)t} \tag{4}\]

From \((4)\) we can see

\[(e^{(\lambda+\mu)t}P_{01})'=\lambda e^{(\lambda+\mu)t} \tag{5}\]

Integrate both side of \((5)\) in term of \(dt\), we get

\[\int (e^{(\lambda+\mu)t}P_{01})'dt=\int \lambda e^{(\lambda+\mu)t}dt\] Then we have

\[e^{(\lambda+\mu)t}P_{01}(t)+k=\frac{\lambda}{\lambda+\mu}e^{(\lambda+\mu)t}+c \tag{6}\] Rearrange \((6)\) we have

\[P_{01}(t)=\frac{\lambda}{\lambda+\mu}+c'e^{-(\lambda+\mu)t}\] Now we use the boundary condition \(P_{01}(0)=0\) to solve \(c'\) and we get \[c'=-\frac{\lambda}{\lambda+\mu}\]

Therefore \[P_{01}(t) = \dfrac{\lambda}{\lambda+\mu}\left[ 1-e^{-(\lambda+\mu)t}\right]\], we got the solution for the Kolmogorov forward equation.