Solving a linear PDE by the method of characteristics without using a parameter

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.

• I get that $$u = \frac{x^{2}}{4} + g(x-2t)$$ and hence \begin{align} u(x,0) &= \frac{x^{2}}{4} + g(x) \\ &= f(x) \\ \implies g(x) &= f(x)-\frac{x^{2}}{4} \\ \implies g(x-2t) &= f(x-2t)-\frac{(x-2t)^{2}}{4} \end{align} Nov 11 '19 at 3:43

You correctly found two equations of characteristic curves : $$x - 2t = c$$ $$u = \frac14 x^2 + k$$ Then you correctly wrote $$k = g(c)$$ but you made a mistake at next step. You wrote $$k= g(x - ct)$$ which is not correct since $$c=x-2t$$. The correct writting is : $$k=g(x-2t)$$ $$u=\frac14 x^2+g(x-2t)$$ Then the condition $$u(x,0)=f(x)=\frac14 x^2+g(x)$$ gives $$g(x)=f(x)-\frac14 x^2$$
Now the function $$g$$ is determined, any variable $$X$$ : $$g(X)=f(X)-\frac14 X^2$$ We put this function into the general solution $$u=\frac14 x^2+g(x-2t)$$ where $$X=x-2t$$ : $$u=\frac14 x^2+f(x-2t)-\frac14 (x-2t)^2$$