**a.** This is a first order linear PDE with constant coefficient.

\begin{equation}

\frac{d x}{4} = \frac{d y}{3} = \frac{d u}{0};

\end{equation}

Then we get $u(x, y) = f(3x+4y)$ for some arbitrary $f$.

**b.** Use IVP $u|_{x=0}=\sin (y)$, we get

\begin{equation}

f(4y) = \sin(y)\\f(w) = \sin(\frac{y}{4})

\end{equation}

Hence solution is

\begin{equation}

u(x,y) = f(3x+4y) = \sin(\frac{3}{4}x+y)

\end{equation}

**c.** With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is

\begin{equation}

u(x, y) = \frac{3}{4}x+y.

\end{equation}

Since $\forall x>0, y>0\implies \frac{3}{4}x+y >0$.

Thus this solution is defined on the whole domain $\{x>0,y>0\}$.

**d.** With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is

\begin{equation}

u(x, y) = \frac{3}{4}x+y.

\end{equation}

Since $f$ is define when $y>0$, the solution only is defined where $\frac{3}{4}x+y >0$, then when $y> -\frac{3}{4}x$, the solution is defined. However when $y< -\frac{3}{4}x$ we need to impose condition at $y = 0$, we get

\begin{equation}

f(3x) = x \\

f(w) = \frac{w}{3}\\

\end{equation}

Then we get:

\begin{equation}

u(x,y) = x + \frac {4}{3}y, (y< -\frac{3}{4}x)

\end{equation}

Final solution would be:

\begin{equation}

u(x,y) = \left\{

\begin{array}{l l}

\frac{3}{4}x+y & \quad y>-\frac{3}{4}x\\

x+\frac{3}{4}y & \quad y<-\frac{3}{4}x

\end{array} \right.

\end{equation}