a) This is a 1-D wave PDE general solution of which is discussed in the class.

\begin{equation*}

u(t,x)=\phi(x+3t) + \phi(x-3t)

\end{equation*}

b) Using D'Alemblert formula we get:

\begin{equation*}

u(t,x)=\frac{1}{2} \bigl[ (x+3t)^2-(x-3t)^2 \bigr] + \frac{1}{6}\int_{x-3t}^{x+3t} s\,ds \\

=\frac{1}{2}\bigl[ (x+3t)^2-(x-3t)^2 \bigr]+\frac{1}{12}(x+3t)^2-\frac{1}{12}(x-3t)^2 \\

=\frac{7}{6}(x+3t)^2+\frac{5}{6}(x-3t)^2

\end{equation*}

c) We impose Goursat problem boundary conditions to general solution and get:

$$

\phi(6t)+\psi(0)=t \\

\phi(0)+\psi(-6t)=2t

$$

Letting $t=0$ in first equation and subtracting it from the second we get $\phi(0)=\psi(0)=0$. Therefore $$\phi(t)=\frac{t}{6} \\

\psi(t)=\frac{-t}{3}

$$

Final solution for Goursat problem is $$ u(t,x)=\frac{-1}{6}x+\frac{3}{2}t $$