I have formulated the following PDE during my research: $$ \rho(W - K + \delta R)\partial_W U + \mu R\, \partial_R U - \gamma = 0\ , $$ where $U = U(W,R)$, and all other symbols in the above are constants. The $\delta R$ term is what is driving me crazy: if it were absent, a solution of the form $a \log{(W-K)} + b \log{(R)}$ would work just fine.
I would be grateful for any ideas on how to go about solving for, or approximating $U$.
EDIT. Following the suggestion in the comments, I attempted to apply the method of characteristics as follows.
I first of all introduce the initial condition $U(0, R) = \log{R}$. I then have the following IVPs: \begin{alignat}{3} \partial_\tau W &= \rho(W - K + \delta R) \qquad &&W(0, s) &&&= 0\\ \partial_\tau R &= \mu R &&R(0, s) &&&= s\\ \partial_\tau U &= \gamma &&U(0, s) &&&= \log{s} \end{alignat} Solving the IVP for $U$ yields $$ U = \gamma \tau + \log{s}\ , $$ whereas the IVP for $R$ yields $$ R = s \exp{\mu \tau}\ . $$ Plugging this expression into the IVP for $W$, Mathematica solves it as $$ W = \frac{K (\rho -\mu ) \left(e^{\rho \tau }-1\right)+\delta \rho s \left(e^{\mu \tau }-e^{\rho \tau }\right)}{\mu -\rho }\ . $$ Have I proceeded correctly thus far? Is the only remaining step to solve for $\tau$ and $s$ in terms of $R$ and $W$ and plug this back into the expression for $U(\tau, s)$? If so, what does one do if the system cannot be inverted, as is the case here?