Question

Solve the IVP\[u_t+cu_x=1,\]\[u(x,0)=\sin(x).\]

Answer 1

\[\frac{du}{dt}=u_x\frac{dx}{dt}+u_t;\]\[\frac{dx}{dt}=c,\]\[\frac{du}{dt}=1.\]

Answer 2

\[\frac{du}{dt}=1\implies u=t+c_1;\]\[u(x,0)=\sin(x_0)=\sin(x-ct).\]

Answer 3

\[u(x,t)=t+\sin(x-ct).\]Is this correct?