Find the vector field associated to the PDE and define a parametrization of the curve that define a border condition $$(u+2y)u_{x}+uu_{y}=0$$ with $u(x,1)=\frac{1}{x}$.
My approach: Consider the following first-order, linear equation $$a(x,y)u_{x}+b(x,y)u_{y}=c(x,y)$$ Suppose we can find a solution $u(x,y)$. let’s start by constructing a curve $C$ parametrized by $s$ such that at each point on the curve $C$, the vector $(a(x(s), y(s)), b(x(s), y(s)), c(x(s), y(s)))$ is tangent to the curve. In particular, $$\frac{dx}{ds}=a(x(s),y(s))$$ $$\frac{dy}{ds}=b(x(s),y(s))$$ $$\frac{dz}{ds}=c(x(s),y(s))$$
Which are the integral curve for $(a(x, y), b(x, y), c(x, y))$. So, the vector field is given by $V=(a(x,y),b(x,y),c(x,y))$? In this case, we have $V=(u+2y,u,0)$?
And how define the parametrized curve that define the boundary condition?