Solving $u_t + cu_x = k$ by method of characteristics

Given the 1st order linear PDE $$u_t + cu_x = k$$ with initial condition $$u(x,0)=\mathrm{cosh}2x$$, I am required to find a solution using the method of characteristics.

Characteristic equations are $$\frac{\mathrm{d}t}{\mathrm{d}s}= 1$$ $$\frac{\mathrm{d}x}{\mathrm{d}s} = c$$ $$\frac{\mathrm{d}u}{\mathrm{d}s} = k$$

I have calculated, $$t=s+c_1, x=cs+c_2, u=ks+c_3$$ I'm unsure on how to proceed.

Well, you need to use the initial data, i.e. the values of $$u(s)|_{s=0}$$: $$u(x(s),t(s))|_{s=0}=u(x_0,0)=\cosh(2x_0),$$ which gives $$t(s)=s$$, $$x(s)=cs+x_0$$ and $$u(s)=ks+\cosh(2x_0)$$. The latter rewrites as $$u(x,t)=kt+\cosh(2(x-ct))$$ by using $$s=t$$ and $$x_0=x-ct$$.