I'm trying to solve the partial differential equation $$2u_x + u_t = x,\qquad u(x, 0) = f(x)$$ by the method of characteristics while not retaining an extra parameter. I'm able to find the characteristic lines fairly easily, but for whatever reason, I cannot find a final solution for $u(x, t)$ that satisfies both the initial condition and the PDE itself.
Here's my work so far:
Using $dx/2 = dt/1$, I've found that the characteristic curves are $x - 2t = c$, for arbitrary $c$.
Using $du / x = dx / 2$, I've found that $u = 1/4 x^2 + k$, for arbitrary $k$.
My next logic here was to say that $k = g(c) = g(x - 2t)$, for arbitrary function $g$.
This is where it begins to break, though. Normally, I'd substitute the initial condition $u(x, 0) = f(x)$ into the equation and find how $g$ relates to $f$, and then finally give an answer for $u$. However, the initial condition here is completely violated no matter what $g$ equals, since there is no way to get $1/4x^4$ to equal $x$ by changing $g$, so clearly there's a big gap in my logic. How would I be able to solve this problem using this general method but without this flaw? My professor said it is possible to solve this without introducing a parameter.