a) By plugging $u=\phi(x-vt)$ into equations (16) and (17) on the problems page, we obtain:
\begin{equation}
v^{2}\phi''-c^{2}\phi''+m^{2}\phi=0
\end{equation}\begin{equation}
v^{2}\phi''-c^{2}\phi''-m^{2}\phi=0
\end{equation}
for (16) and (17), respectively.
Proceeding to solve (1) using methods of linear ODEs yields:
\begin{equation}
\phi''=-\frac{m^{2}}{v^{2}-c^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)
\end{equation}
Where $a,b$ depend on initial conditions. This is assuming that $v^{2} > c^{2}$, which gives us a periodic solution that is bounded, as desired. Proceeding the same way with (2):
\begin{equation}
\phi''=\frac{m^{2}}{v^{2}-c^{2}}\phi
\end{equation}
If we assume again that $v^{2} > c^{2}$, we will get:
\begin{equation}\large
\phi(z)=ae^{\frac{m}{\sqrt{v^{2}-c^{2}}}z}+be^{-\frac{m}{\sqrt{v^{2}-c^{2}}}z}
\end{equation}
But this solution does not tend to $0$ as $|z|\rightarrow\infty$ unless $a=b=0$, and we do not want the trivial solution. Therefore $v^{2}$ must be less than $c^{2}$. Then:
\begin{equation}
\phi''=-\frac{m^{2}}{c^{2}-v^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)
\end{equation}
Now $u=\phi(x-vt)$ using $\phi(z)$ in (3) and (6) solve the original problems (16) and (17), respectively.
b) Plugging $u=\phi(x-vt)$ into problem (18):
\begin{equation}
-v\phi'-K\phi'''=0
\end{equation}\begin{equation}
\phi'''+\frac{v}{K}\phi'=0
\end{equation}
If $v$ and $K$ have the same sign, then the characteristic equation will have imaginary roots and give a bounded periodic solution:
\begin{equation}
\phi(z)=a\sin \left(\sqrt{\frac{v}{K}}z\right)+b\cos \left(\sqrt{\frac{v}{K}}z\right)+c
\end{equation}
Same with problem (19):
\begin{equation}
-v\phi'-iK\phi''=0
\end{equation}\begin{equation}
\phi''-\frac{iv}{K}\phi'=0
\end{equation}
In this case, the solution will be suitable regardless of the value of $v$, and the solution will be complex:
\begin{equation}\large
\phi(z)=ae^{\frac{iv}{K}z}+b=a\left(\cos \left(\frac{vz}{K}\right)+i\sin \left(\frac{vz}{K}\right)\right)+b
\end{equation}
Same with problem (20):
\begin{equation}
v^{2}\phi''+K\phi''''=0\Rightarrow\phi''''+\frac{v^{2}}{K}\phi''=0
\end{equation}
Here, since $v$ only appears in terms of its square, we do not need to worry about its value. As long as $K>0$, solving the above ODE gives us a suitable solution:
\begin{equation}
\phi(z)=a\sin \left(\frac{vz}{\sqrt{K}}\right)+b\cos \left(\frac{vz}{\sqrt{K}}\right)+c+d
\end{equation}
Again, $u=\phi(x-vt)$ using $\phi(z)$ in (9), (12), and (14) solve the original problems (18), (19), and (20), respectively.
