Solve the IVP $$ \begin{cases} u_t + cu_x = 1 & c \in \mathbb{R} \\ u(x,0) = \sin x \end{cases}$$
To solve this, I have used characteristics as follows:
Note that $$\frac{\partial u}{\partial t} = u_x \frac{dx}{dt} + u_t$$ So this equation is equal to the one in the IVP if $\frac{\partial u}{\partial t} = 1$ and $\frac{dx}{dt} = c$
$$\frac{\partial u}{\partial t} = 1 \Rightarrow u(x,t) = t + A(x)$$ Using the initial condition, we find that $A(x) = \sin x$ so that $u(x,t) = t + \sin x$.
But this is not the correct solution and I have not used the fact that $\frac{dx}{dt} = c$ either.
Can someone explain where I am wrong and perhaps give a rigorous answer to the problem using my method so I can also see just how to solve the problem properly as I am currently doing it without really understanding anything...