Understanding how to find the General Solution of a PDE using the method of characteristics

Question

I've been trying and failing to understand how to find the "general solution" for PDEs, as in an answer with an arbitrary function $F(x,y,u)$, for a PDE with no boundary conditions given. I understand the concept, just not really how to get this arbitrary function.

Example 4: For the PDE $$yu\frac{\partial u}{\partial x} - xu\frac{\partial u}{\partial y} = x-y,$$ the characteristic equations $$\frac{dx}{d\tau} = yu,\quad \frac{dy}{d\tau} = - xu,\quad \frac{du}{d\tau} = x-y,$$ may be rearranged to give [Exercise] $$\frac{d}{d\tau}(x^2+y^2) = \frac{d}{d\tau}(u^2+2x+2y) = 0.$$ It follows that the general solution is $$u^2 = -2x-2y + F(x^2+y^2)$$ where $F$ is an arbitrary function.

In this example, I'm struggling to understand where the $F(x^2 + y^2)$ has come from. Why is it $(x^2 + y^2)$ in the argument? I've tried changing of variables but that seems too complicated for something like this.

A second example is this:

Example 5: The PDE $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1,$$ has the general solution $$u = \frac{x+y}{2} + F(x-y)$$ where $F$ is an arbitrary function.

Answer 1

$$yu\frac{\partial u}{\partial x} - xu\frac{\partial u}{\partial y} = x-y,$$ the characteristic equations $$\frac{dx}{d\tau} = yu,\quad \frac{dy}{d\tau} = - xu,\quad \frac{du}{d\tau} = x-y,$$ Also, you can rearrange the set of equations as : $$\frac{dx}{yu} = \frac{dy}{-xu} = \frac{du}{x-y} $$ The equation of a first family of characteristic curves comes from : $$\frac{dx}{yu} = \frac{dy}{-xu} \quad\to\quad xdx+ydy=0 \quad\to\quad x^2+y^2=c_1$$ The equation of a second family of characteristic curves comes from : $\frac{dx}{yu} = \frac{dy}{-xu} = \frac{dx+dy}{yu-xu}$ $$\frac{dx+dy}{(y-x)u}=\frac{du}{x-y} \quad\to\quad udu+dy+dx=0 \quad\to\quad u^2+2(y+x)=c_2$$ This is valid any $c_1$ and $c_2$ on the characteristic curves. Elsewhere, $c_1$ and $c_2$ are related on the form of $\Phi(c_1,c_2)=0$ any differentiable function $\Phi$ of two variables. Or on the form of an equivalent relationship $c_1=f(c_2)$ or $c_2=F(c_1)$. So, a form of the general solution is : $$u^2+2(y+x)=F(x^2+y^2)$$ any differentiable function $F$. $$u(x,y)=\pm\sqrt{-2(x+y)+F(x^2+y^2)}$$ With the same method, solving the other PDE is even simpler.