Is there a more specific name for this PDE and how do I solve it?

I have reduced a problem I am working on down to finding a solution $u: \mathbb{R}^2 \to \mathbb{R}$ for a PDE of the form

$$u_t(t, x) + (a x f(t) + b) u(t, x) + (c x + d)u_x(t, x) = 0$$

where $a, b, c, d \in \mathbb{R}$ and $f: \mathbb{R} \to \mathbb{R}$ satisfies some "niceness" properties.

I am not a PDE expert, so I searched online to see if this is a standard form for a PDE. I have found that this is a "First-order partial differential equation" of dimension 2. This section on the Wikipedia describes how to solves these PDEs in general, but I can't make sense of this section or figure out how to apply it to my PDE. I believe that this section may be too general for my needs, and that the problem I am solving may be fall into a more specific class of PDEs.

I believe that PDEs of this form will continue to show up frequently in the work I am doing, so: does a PDE of this form fall into a more specific category than "first-order partial differential equation of dimension 2," and how do I solve it?

$$u_t(t, x) + (c x + d)u_x(t, x) = - (a x f(t) + b) u(t, x) $$ This is a quasi-linear first order PDE.

Charpit-Lagrange characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{cx+d}=\frac{du}{- (a x f(t) + b) u}$$ A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{cx+d}$ : $$\left(x-\frac{d}{c}\right)e^{-ct}=C_1 $$ A second characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{- (a x f(t) + b) u}$

$\frac{du}{u}+\left(a \left(C_1 e^{ct}+\frac{d}{c}\right) f(t) + b\right)dt=0$ $$u\:\exp\left(\int_0^t \left(a \left(C_1 e^{c\,\tau}+\frac{d}{c}\right) f(\tau) + b\right) d\tau \right)=C_2$$

$$u\:\exp\left(b\,t+a \,C_1\int_0^t e^{c\,\tau} f(\tau) d\tau +\frac{a\,d}{c}\int_0^t f(\tau) d\tau \right) =C_2$$ The general solution of the PDE on the form of implicit equation $C_2=\Phi(C_1)$ is : $$u\:\exp\left(b\,t+a \,C_1\int_0^t e^{c\,\tau} f(\tau) d\tau +\frac{a\,d}{c}\int_0^t f(\tau) d\tau \right) =\Phi\left(e^{-ct}\left(x-\frac{d}{c}\right)\right)$$ $\Phi$ is an arbitrary function. $$u(t,x)=\exp\left(-b\,t-a \,C_1\int_0^t e^{c\,\tau} f(\tau) d\tau -\frac{a\,d}{c}\int_0^t f(\tau) d\tau \right) \:\Phi\left(e^{-ct}\left(x-\frac{d}{c}\right)\right)$$

$$\boxed{u(t,x)=\exp\left(-b\,t-a \,e^{-ct}\left(x-\frac{d}{c}\right)\int_0^t e^{c\,\tau} f(\tau) d\tau -\frac{a\,d}{c}\int_0^t f(\tau) d\tau \right) \:\Phi\left(e^{-ct}\left(x-\frac{d}{c}\right)\right)}$$ The function $\Phi$ has to be determined if some initial condition is specified.