I have the PDE a(t)*(df/dx) + df/dt + b(t)f=0.
I have to make a change of variables now where g=fc(t), y=xd(t) and s=e(t). I have no idea how this works with coefficients depending on t.
It wasn't clear to me. In general, for the PDE:
$\displaystyle a(x,y)\frac{\partial u}{\partial x}+b(x,y)\frac{\partial u}{\partial y}+c(x,y)u(x,y)=f(x,y)$, we solve the characteristic DE:
$\displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}$
Assume the characteristic solution can be put into the form $\displaystyle h(x,y)=k$. Then the change of variable $\displaystyle w=h(x,y),\; z=y$ will convert the PDE to an ODE in $\displaystyle z$ for fixed $\displaystyle w$. Probably not what you want. That's why I didn't say anything. Also while I'm sayin', this is right out of "Basic Partial Differential Equations" by Bleecker and Csordas. It's a nice PDE book that's easy to read. They go over first order PDEs like this pretty good.
This is how I'd do it; maybe someone can suggest a better approach to both of us. I first write it in standard form:
$\displaystyle \frac{a(t)}{b(t)}\frac{\partial f}{\partial x}+\frac{1}{b(t)}\frac{\partial f}{\partial t}+f=0$
The characteristic equation becomes then:
$\displaystyle \frac{dt}{dx}=\frac{1}{a(t)}\Rightarrow x-\int a(t)dt=K$
So I make the change of variables:
$\displaystyle y=x-\int a(t)dt \quad\quad s=t$
Then define: $\displaystyle g(y,s)\equiv f(x,t)$
and under these transformations it can be shown:
$\displaystyle \frac{a}{b}\frac{\partial f}{\partial x}+\frac{1}{b}\frac{\partial f}{\partial t}=\frac{1}{b(s)}\frac{\partial g}{\partial y}$
Thus I have:
$\displaystyle \frac{1}{b(s)}\frac{\partial g}{\partial y}+g=0$