Well, there are many solutions. In order to get a unique solution you need to first specify what the values of u will be over a "non-characteristic" curve.
Example. Find the solution if u(x,0) = g(x).
We use the method of characteristics and end up with a bunch of equations:
1. x = x_0 + 2bt + 4a^2g'(x_0)(e^{-t}-1), and
2. y = c^2t
You have to solve those equations for t and x_0 in terms of x and y. Then you substitute into
$$ u = 2a^2g'(x_0)^2e^{-2t} - (2bg'(x_0)+c^2h(x_0))e^{-t}.$$
Here, h(x) satisfies the equation g(x)+c^2h(x)+2bg'(x)-2a^2g'(x)^2=0.
Crystal clear, right?
Ok, I'll do a special case so you can at least take home one bona fide solution. Let's suppose the initial condition is u(x,0)= x. So g(x)=x,\, g'(x)=1,\, h(x)=(2a^2-2b-x)/c^2.
From equation 2, we find that t = y/c^2. Substituting that into equation 1, we find x-x_0 = (2b/c^2)y +4a^2(e^{-y/c^2} - 1). Therefore,
$$x_0 = x - (2b/c^2)y -4a^2(e^{-y/c^2} - 1).$$
So we can plug in our expressions for t and x_0 into the formula for u and obtain the solution.
$$ u(x,y) = (x-(2b/c^2)y-4a^2(e^{-y/c^2}-1)-2a^2)e^{-y/c^2} +2a^2e^{-2y/c^2}.$$
[EDIT] I found an error and corrected it, but there might be (probably are) other errors. You'd have to plug that back into the original equation to see.
