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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.

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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 $x_0=x-ct$.

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