Solving PDE like ODE

Question

Is it possible to treat $$\frac{\partial}{\partial\tau}g(\tau,s)+g(\tau,s)=e^s e^{3\tau}$$ like an ODE and solve for $g(\tau,s)$? Thank you.

Answer 1

Yes, consider $s$ as a "parameter" and solve the linear ODE with respect to $\tau$: $$\frac{\partial g}{\partial \tau}+g=e^s e^{3\tau}\implies g(\tau,s)=\underbrace{C(s)e^{-\tau}}_{\text{hom. sol.}}+\underbrace{\frac{e^se^{3\tau}}{4}}_{\text{part. sol.}}.$$ where $C(s)$ is a function in $s$.