I am instructed to solve the following: $$\frac{\partial U(x,t)}{\partial t}+c\frac{\partial U}{\partial x}=2(x-ct)$$ $$u(x,0)=f(x)$$
I said $\frac{dx}{dt}=C \implies x=ct+x_0$ and $\frac{dU}{dt}=g(t)=2(x-ct)$. then $U(x,t)=2xt-ct^2+c_2.$ Using $U(x,0)=f(x)=2x(0)-c(0)+c_2$, then $f(x)=c_2$. $$U(x,t)=2xt-ct^2+f(x).$$ Then for some reason, I feel that I must sub in $x-ct$ in $f(x)$, and I have $u(x,t)=2xt-ct^2+f(x-ct)$. $U(x,0)=f(x)$ confirming the initial condition. $$U_t = 2x-2ct+f'(x-ct)(-c)$$ and $$U_x = 2t+f'(x-ct).$$ $$U_t+cU_x=2x-2ct-cf'(x-ct)+2ct+cf'(x-ct)$$ $$=2x,$$ not $2(x-ct)$ as required. I have puzzled over this problem many times. Can anyone help?
You forgot factor $2$ between steps $\frac{dU}{dt}=2(x-ct)$ and $U(x,t)=2xt-2ct^2+c_2(x)$. that makes the problem till the end. the final solution would be : $$ U(x,t)=2xt-2ct^2 + f(x-ct) = 2t(x-ct)+f(x-ct) $$ you can see this satisfy the PDE.
But it seems your method has some problems (for example you don't know how correctly add $f(x-ct)$) so here's a more detailed solution:
This is non-homogeneous first order wave equation, you can solve this similar to D'Alembert solution. $$ \left\{ \begin{array}{l} \xi =x-ct \\ \eta=x +ct \end{array}\right. \longrightarrow \left\{ \begin{array}{l} \frac{\partial U}{\partial t} =-c \frac{\partial U}{\partial \xi} + c\frac{\partial U}{\partial \eta} \\ \frac{\partial U}{\partial x} = \frac{\partial U}{\partial \xi} + \frac{\partial U}{\partial \eta} \end{array}\right. $$ put these back into equation and you get : $$ 2c \frac{\partial U}{\partial \eta} = 2(x-ct) = 2\xi \rightarrow \\ \frac{\partial U}{\partial \eta}= \frac{\xi}{c} $$ you can easily integrate this : $$ U(\xi,\eta) = \frac{1}{c}\xi \eta + g(\xi) \\ \Longrightarrow U(x,t) = \frac{1}{c} (x^2-c^2t^2)+g(x-ct) $$ now use initial condition $t=0$ : $$ U(x,0) = \frac{x^2}{c}+g(x)=f(x) \Rightarrow g(x) = f(x)-\frac{x^2}{c} $$ if you substitute this result into final $U(x,t)$ ,with a little work you get the final solution: $$ U(x,t) = 2t(x-ct)+f(x-ct) $$