Backward and forward Kolmogorov equation for pure birth

Question

I am dealing with a pure birth process and trying to solve the forward and backward kolmogorov equations. I am stuck solving the backward equation.

$\begin{bmatrix}P'_{0,0}(t)&P'_{0,1}(t)\\P'_{1,0}(t)&P'_{1,1}(t)\end{bmatrix}= \begin{bmatrix}- \lambda& \lambda\\0&0\end{bmatrix} \begin{bmatrix}P_{0,0}(t)&P_{0,1}(t)\\P_{1,0}(t)&P_{1,1}(t)\end{bmatrix} $

Where, $P'_{i,j}$ is the derivative.

I did try to attempt this and got the following

$P'_{00}(t) = \lambda - 2\lambda P_{00}(t)$

I am not sure how to proceed further, and would love if you could show 1) that the back ward and forward Kolmogorov equations are equal 2) How to solve the backward kolmogorov equations

Answer 1

Note that $P_{1,0}(t)$ is actually a constant since the derivative is $0$.

$$P'_{0,0}(t) = -\lambda P_{0,0}(t) + \lambda P_{1,0}$$

$$P'_{0,0}(t) +\lambda P_{0,0}(t) = \lambda P_{1,0}$$

$$e^{\lambda t}P'_{0,0}(t) +\lambda e^{\lambda t}P_{0,0}(t) = \lambda e^{\lambda t}P_{1,0}$$

$$\frac{d}{dt}\left( e^{\lambda t}P_{0,0}(t) \right)= \lambda e^{\lambda t}P_{1,0}$$

$$e^{\lambda t}P_{0,0}(t) = e^{\lambda t} P_{1,0} + A $$

$$P_{0,0}(t) = P_{1,0}+Ae^{-\lambda t}$$

Also by symmetry,

$$P_{0,1}(t) = P_{1,1}+Be^{-\lambda t}$$

Try to think of what relationship can you say about $P_{0,0}$ and $P_{0,1}$ to relate $A$ and $B$ together.