In the process of solving a continuous-time Markov chain problem, I derived the following PDE for the probability distribution $P_0(t,T)$, $t\in[0, T], T\in \mathbb{R}$: \begin{align} \frac{\partial^2 P_0}{\partial T^2} + \frac{\partial^2 P_0}{\partial t \, \partial T} + (k_{\text{on}} + k_{\text{off}}) \frac{\partial P_0}{\partial T} + k_{\text{on}} \frac{\partial P_0}{\partial t} = 0. \end{align} The rates $k_{\text{on}}$ and $k_{\text{off}}$ are positive numbers. This equation is augmented by the condition \begin{align} P_0(0,T) = p_0^{\text{ss}} e^{-k_{\text{on}}T}. \end{align} I don't have much experience solving PDEs, so I looked up some worked out examples and following their approach reduced the equation into its canonical form given by \begin{align} \frac{\partial^2 P_0}{\partial \eta \partial t} + (k_{\text{on}} + k_{\text{off}}) \frac{\partial P_0}{\partial \eta} + k_{\text{on}} \frac{\partial P_0}{\partial t} = 0, \end{align} where $\eta = T-t$. Then, I used the separation of variables approach ($P_0(t, \eta) = Q(t)E(\eta)$), which is what I would normally do to solve, say, the diffusion equation. Unfortunately, the solution I get leads to unphysical results when I impose the 'initial' condition. Skipping some of the steps, I find that \begin{align} Q(t)E(\eta) = C \exp\left( -\frac{k_{\text{on}} + k_{\text{off}}}{1+c_e}t + \frac{k_{\text{on}}}{c_e} (T-t) \right), \end{align} where $C$ and $c_e$ are constants. Imposing the condition leads to $C=p_0^{\text{ss}}$ and $c_e=-1$. This second number is what blows the solution. It suggests that $1+c_e=0$ and the denominator in the exponent makes the whole expression diverge.
Any suggestions on how to approach this problem? Can it be that the separation of variables does not apply in this case? What other methods can I try? Thanks in advance!
[Edit] There is also a second condition that can be helpful in solving this: $P_0(T,T)=0$.
Simplifying your notation to make the typing a bit easier you have $$P_{TT} + P_{tT} + (k_1 + k_2)P_T + k_1P_t = 0$$ subject to $P(0, T) = p_0 e^{-k_1 T}$. You're on the right track separating the variables, $P(t, T) = A(t)B(T)$.
Doing this I get $B' + cB = 0$, where $c$ is some constant, and $ A'' + (k_1 + k_2 - c)A' - ck_1 A = 0$. Using the initial condition, the former tells us $c = k_1$, so the latter becomes $A'' + k_2 A' - k_1^2 = 0$.
Solve the characteristic equation associated with this ordinary differential equation. You'll see it has roots $-\tfrac{1}{2}k_2 \pm \tfrac{1}{2}\sqrt{k_2^2 + 4k_1^2}$. Since your problem seems to deal with probablility distributions I imagine you'll want to pick the negative root so that $A \rightarrow 0$ as $t \rightarrow \infty$, so with $\kappa = \tfrac{1}{2}k_2 + \tfrac{1}{2}\sqrt{k_2^2 + 4k_1^2}$ your solution is $$ P(t, T) = p_0 e^{-k_1T} e^{-\kappa t}. $$