I am looking to solve a linear first-order PDE of the form
$$(e^{\beta x} - \alpha x -1 ) f_x(x,t) - f_t(x,t) + \gamma x f(x,t) =0$$
with boundary conditions $f(0,t)=1$ and $f(x,0)=e^{x\lambda}$.
I tried the methods of characteristics, but did not get any meaningful conclusion. Does this one has an explicit solution? And if yes, how can I get it constructively?
$$(e^{\beta x} - \alpha x -1 ) f_x(x,t) - f_t(x,t) =- \gamma x f(x,t) $$ The Charpit-Lagrange characteristic ODEs are : $$\frac{dx}{e^{\beta x} - \alpha x -1}=\frac{dt}{-1}=\frac{df}{-\gamma x\:f }$$ A first characteristic equation comes from solving $\frac{dx}{e^{\beta x} - \alpha x -1}=\frac{dt}{-1}$ : $$t+\int_0^x \frac{d\xi}{e^{\beta \xi} - \alpha \xi -1}=c_1$$ A second characteristic equation comes from solving $\frac{dx}{e^{\beta x} - \alpha x -1}=\frac{f}{-\gamma x }$ : $$f+\int_0^x \frac{\gamma\: \xi\:d\xi}{e^{\beta \xi} - \alpha \xi -1}=c_2$$ The general solution of the PDE on implicit form $c_2=F(c_1)$ leads to : $$\boxed{f(x,t)=-\gamma \int_0^x \frac{\xi\:d\xi}{e^{\beta \xi} - \alpha \xi -1}+F\left(t+\int_0^x \frac{d\xi}{e^{\beta \xi} - \alpha \xi -1}\right)}$$ $F$ is an arbitrary function.
BOUNDARY CONDITION : $f(0,t)=1$
$f(0,t)=1=-\gamma \int_0^0 \frac{\xi\:d\xi}{e^{\beta \xi} - \alpha \xi -1}+F\left(t+\int_0^0 \frac{d\xi}{e^{\beta \xi} - \alpha \xi -1}\right)=F(t)$
The function $F$ is determined as a constant : $F=1$. We put it into the above general solution. $$f_1(x,t)=1-\gamma \int_0^x \frac{\xi\:d\xi}{e^{\beta \xi} - \alpha \xi -1} \tag1$$
INITIAL CONDITION : $f(x,0)=e^{x\lambda}$
$$e^{\lambda x}=-\gamma \int_0^x \frac{\xi\:d\xi}{e^{\beta \xi} - \alpha \xi -1}+\Phi\left(0+\int_0^x \frac{d\xi}{e^{\beta \xi} - \alpha \xi -1}\right)$$ The notation $F$ is changed to $\Phi$ to make clear that this is not the same function than above for the boundary condition. $$\Phi\left(\int_0^x \frac{d\xi}{e^{\beta \xi} - \alpha \xi -1}\right)=e^{\lambda x}+\gamma \int_0^x \frac{\xi\:d\xi}{e^{\beta \xi} - \alpha \xi -1}$$
$$\text{Let}\quad X(x)=\int_0^x \frac{d\xi}{e^{\beta \xi} - \alpha \xi -1}\quad \text{and let}\quad x(X)\quad \text{the inverse function.}$$ $$\Phi\left(X\right)=e^{\lambda x(X)}+\gamma \int_0^{x(X)} \frac{\xi\:d\xi}{e^{\beta \xi} - \alpha \xi -1}$$ This determines the function $\Phi$. One put it into the above general solution where the argument is $\left(t+\int_0^x \frac{d\xi}{e^{\beta \xi} - \alpha \xi -1}\right)$. This gives a second solution $f_2(x,t)$.
Of course this is purely theoretical. One cannot find a closed form for the integral and even more one cannot find a closed form for the inverse function $x(X)$.
So the conclusion is that the solution of the problem probably cannot be expressed on a closed form. The problem will require numerical methods of solving.
Note : Two solutions $f_1(x,t)$ and $f_2(x,t)$ wre mentioned. Of course they are not both valid together everywhere. The solution is a picewise function made of $f_1(x,t)$ and $f_2(x,t)$ each one valid in a limited region limited by the respective conditions.