Part (b) has already been done by Yue Du above, so I will provide an explanation of the equations.

The first equation follows from a consequence of Faraday's law of induction: the emf induced in a inductor is related to the rate of change in current by $\mathcal{E}=-L i_t$ where $L$ is the self inductance. Consider a segment of wire of length $dx$ with the left endpoint at $x$. Since $L$ is defined as the inductance per length here, the inductance of this segment is $L dx$. The difference in potential between the two ends is $v(x+dx,t)-v(x,t)$. Plugging in to the law of induction we have

\begin{equation}

v(x+dx,t)-v(x,t)=-L i_t\,dx

\end{equation}

Moving the $dx$ to the LHS gives the result.

Now for the second equation. I am going to assume here that "capacitance" means "self capacitance of a conductor," which is the amount of charge needed to raise an isolated conductor's potential by 1 volt. In other words, $C=dq/dv$ where $q$ is the charge. Consider once again a segment of wire of length $dx$ with its left endpoint located at position $x$. Between times $t$ and $t+dt$, the change in potential of this segment is $dv=v(x,t+dt)-v(x,t)$. In this amount of time, an amount of charge $dq=(i(x,t)-i(x+dx,t))dt$ has flown into the region. Meanwhile, the capacitance of this length of wire is $C\,dx$. Putting it all together we have

\begin{equation}

C\,dx=\frac{(i(x,t)-i(x+dx,t))dt}{v(x,t+dt)-v(x,t)}\implies -Cv_t=i_x

\end{equation}