Solution: The given differential equation can be written as \begin{equation}\label{eq:28Apr2023-1} -x\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1-u. \end{equation} We recall that Lagrange's auxiliary equation for the linear partial differential equations \[ P(x,y,u) \frac{\partial u}{\partial x} + Q(x,y,u) \frac{\partial u}{\partial u} = R(x,y,u) \] is given by \[ \frac{\mathrm{dx} }{P} = \frac{\mathrm{d} y}{Q} = \frac{\mathrm{d} u}{R}. \]

Thus, Lagrange's auxiliary equations for \eqref{eq:28Apr2023-1} will be \begin{equation}\label{eq:28Apr2023-2} -\frac{\mathrm{d} x}{x} = \frac{\mathrm{d} y}{1} = \frac{\mathrm{d} u}{1-u}. \end{equation} Considering first two terms of \eqref{eq:28Apr2023-2}, \begin{align*} -\frac{\mathrm{d} x}{x} = \frac{\mathrm{d} y}{1} & \implies -\ln x + \ln c_1 = y \\ & \implies \ln \left( \frac{c_1}{ x } \right) = y \\ & \implies c_1 = x e^y \tag{$\star_1$} \label{eq:28Apr2023-3} . \end{align*} Similarly, considering the last two terms of \eqref{eq:28Apr2023-2}, \begin{align*} \frac{\mathrm{d} y}{1} = -\frac{\mathrm{d} u}{u-1} & \implies y = -\ln (u-1) + \ln c_2 \\ & \implies y = \ln \left( \frac{c_2}{u-1} \right) \\ & \implies c_2 = (u-1) e^y \tag{$\star_2$} \label{eq:28Apr2023-4}. \end{align*}

General solution of the given linear differential equation will be \begin{align*} c_2 = \varphi\left( c_1 \right), \end{align*} where $\varphi $ is an arbitrary function. So using \eqref{eq:28Apr2023-3} and \eqref{eq:28Apr2023-4} and the condition that $u(x,0) = \sin x$, we have \begin{align*} (u-1)e^y = \varphi \left( xe^y \right) & \implies (\sin x -1) e^0 = \varphi \left( x \right) \\ & \implies \varphi (x) = \sin x - 1. \end{align*} Therefore, the solution of \eqref{eq:28Apr2023-1} will be \[ (u-1)e^y = \sin \left( xe^y \right) -1 \] Hence, \[ (u(0,1) - 1)e^1 = \sin (0) - 1 \implies u(0,1) = 1-\frac{1}{e}. \]