**a.** This is a quasilinear equation.

\begin{equation}

\frac{d t}{1} = \frac{d x}{u} = \frac{d u}{0};

\end{equation}

Then we get $ut-x= C$ for some arbitrary constant C.

Consider the initial condition $u|_{t=0}=x$, take an initial point $(0, x_0)$ such that

\begin{equation}

u(x, 0) = x

\end{equation}

Therefore we have $u = f(x_0) = f(x-ut)$ along characteristics, so

\begin{equation}

u(x,y) = f(x-ut) = x-ut\\

u(x, y) = \frac{x}{1+t}

\end{equation}

**b.**

When $t>-1$, the solution is clearly hold.

However when $t<-1$ the solution breaks because the characteristics lines cannot be interpreted if they are intersected or undefined.