Please, use

MathJax for proper displaying equations. Also you need either repeat a problem here, or to provide a clickable link, like

thisSo, we have equation

\begin{equation}

u_t+3u_x-2u_y=x

\label{eqn-1}

\end{equation}

with the IVP

\begin{equation}

u |_{t=0}=0.

\label{eqn-2}

\end{equation}

Writing characteristics

\begin{equation}

\frac{dt}{1}=\frac{dx}{3}=\frac{dy}{-2}=\frac{du}{x}.

\label{eqn-3}

\end{equation}

Solving the first equality: $x-3t=c_1$, second $y+2t =c_2$ and the last one $u-\frac{x^2}{6}=C$, with $c_1, c_2, C$ constants along characteristics, which are marked by $c_1,c_2$. Then $C=\varphi(c_1,c_2)$ and finally

\begin{equation}

\boxed{u = \frac{x^2}{6} + \varphi (x-3t, y+2t)}

\label{eqn-4}

\end{equation}

is the general solution to (\ref{eqn-1}).

Plugging (\ref{eqn-4}) into (\ref{eqn-2}) we get $ \frac{x^2}{6} + \varphi (x, y) =0\implies \varphi(x,y)= -\frac{x^2}{6}$ and plugging into (\ref{eqn-4}) we get

\begin{equation}

\boxed{u = \frac{x^2}{6} - \frac{(x-3t)^2}{6} = xt - \frac{3}{2}t^2.}

\label{eqn-5}

\end{equation}