So I was interested in finding the set of curves which are perpendicular to this Vector field $$\vec F(x,y) = (x-y,x+y)$$ Lets say the perpendicular curves can be expressed as level curves of the function $G(x,y)$. We know that the normal of the perpendicular curves is parallel to $\vec F(x,y)$. The general expression for finding perpendicular surfaces in 3 dimensions would be $$\nabla G\times\vec F = 0$$ Of course since we're only operating in two dimensions, the "cross product" is simply a scalar and the when applied to the given $\vec F$ the following equation results $$(x-y)\frac{\partial G}{\partial y} = (x+y)\frac{\partial G}{\partial x}$$
This is where I stop. My only known method for solving PDEs is to assume it a product of two single variable functions, rearrange the functions of each variable to their respective side, and solve two ODEs. This isn't possible here so I am not sure what to do. Can anyone solve, recommend tips, or point me to any sources?
$$ (x+y)\frac{\partial G}{\partial x}+(y-x)\frac{\partial G}{\partial y}=0$$ The Charpit-Lagrange system of characteristic equations is : $$\frac{dx}{x+y}=\frac{dy}{y-x}=\frac{dG}{0}$$ A first characteristic equation comes from $\frac{dx}{x+y}=\frac{dy}{y-x}$ . This is an homogeneous ODE easy to solve : change of variable $y(x)=xu(x)$. The solution is: $$\ln(x^2+y^2)+2\arctan(\frac{y}{x})=c_1$$ A second characteristic equation comes from $\frac{dG}{0}\neq 0$ which implies $G=c_2$.
From the relationship $\Phi(c_1,c_2)=0$ where $\Phi$ is an arbitrary function, or equivalently $c_2=F(c_1)$ where $F$ is an arbitrary function : $$G(x,y)=F\left(\ln(x^2+y^2)+2\arctan(\frac{y}{x}) \right)$$ $F$ is an arbitrary function which could be determined insofar some boundary conditions where specified.
Note : Don't confuse this notation of function $F$ with the notation $\vec F(x,y)$ used in the wording of the question.