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There are no boundary conditions. Anyway, here is what I have:

Given the PDE: $$y \dfrac{\partial u}{\partial y} - x \dfrac{\partial u}{\partial x} - 1 = 0$$ we want to solve by method characteristics. That is we want something of the form $$\dfrac{d}{ds}u(x(s),y(s)) = f(u,x(s),y(s)),$$ where $(x(s),y(s))$ is a characteristic line. First we find, $$\dfrac{d}{ds}u(x(s),y(s)) = \dfrac{du}{ds} = \dfrac{\partial u}{\partial x}\dfrac{dx}{ds} + \dfrac{\partial u}{\partial y}\dfrac{dy}{ds}$$ Thus, by setting $$\dfrac{dx}{ds} = -x \quad \quad \dfrac{dy}{ds} = y \quad \quad \dfrac{du}{ds} = 1$$ we have our desired PDE represented as a system of ODEs. Now, solving these ODEs we have: $$-\ln(x)=s \implies x=e^{-s} + x_0, \quad \text{where} \, x(0) = x_0 - 1 $$ $$\ln(y)=s \implies y=e^s + y_0, \quad \text{where} \, y(0) = y_0 - 1$$ $$u=s\implies u = s + u_0 \quad \text{where} \, u(0,0)=u_0$$

Without boundary conditions not really sure how to express the solution.

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Notice that $$\frac{x'}{x}+\frac{y'}{y}=-1+1=0$$ so $\ln x+\ln y$ is constant, that is, $xy=A$.

Similarly, $$\frac{x'}{x}+u'=0$$ so $\ln x+u=B$ constant. These constants depend on the boundary data. Eliminating them gives $\ln x+u=F(xy)$, or equivalently, $$u(x,y)= F(xy)-\ln x$$

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$$y \dfrac{\partial u}{\partial y} - x \dfrac{\partial u}{\partial x} = 1 $$ System of characteristic ODEs : $$\frac{dy}{y}=\frac{dx}{-x}=\frac{du}{1}$$ First family of characteristic curves, from $\quad\frac{dy}{y}=\frac{dx}{-x}\quad\to\quad xy=c_1$

Second family of characteristic curves, from $\quad\frac{dy}{y}=\frac{du}{1}\quad\to\quad u-\ln|y|=c_2$

General solution of the PDE expressed on the form of implicit equation : $$\Phi\left(u-\ln|y|\:,\:xy \right)=0$$ where $\Phi$ is any differentiable function of two variables.

Or equivalently, on explicit form $\quad u-\ln|y|=F\left(xy \right)$ $$u(x,y)=\ln|y|+F\left(xy \right)$$ where $F$ is any differentiable function.

Or : $$u(x,y)=-\ln|x|+G\left(xy \right)$$ where $\quad G(xy)=F(xy)+\ln|xy|\quad$ is any differentiable function.

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