Solve $uu_x+u_y=-\frac{1}{2}u$ with characteristics I'm stuck trying to solve $uu_x+u_y=-\frac{1}{2}u$ satisfying $u(x,2x)=x^2$, this solution also needs to be in some parametrised form with $x(r,s), y(r,s),u(r,s)$.
So far I have this:
$\dfrac{dx}{ds}=u \longrightarrow x(r,s)=us+x_0$
$\dfrac{dy}{ds}=1 \longrightarrow y(r,s)=s+y_0$
$\dfrac{du}{ds}=-\dfrac{1}{2}u \longrightarrow u(r,s)=u_0e^{-s/2}$
If we have that $y(0)=0$ we can see that $y=s$, so we can replace each instance of $s$ with $y$. I know I will end up with an implicit function, but I don't know how to apply the $u(x,2x)=x^2$ condition from here. I was thinking about directly plugging in $u(x,2x)=u_0e^{-x}=x^2\Rightarrow u_0=x^2e^x$ and then after letting $x(0)=r$ (the initial curve) and some substitutions end up with $u=(2uy-x)^2e^{-x}e^{-y/2}$ but I don't think that's correct.
I feel like there's a step missing before I start doing that. I have only just been introduced to PDEs, so all of this is new, any help is welcome and I appreciate your patience with me.
As an aside, I know there is another question that is asking the exact same thing that I am, but they introduce some $\Phi$ and $F$ notations which I have never seen before and leaves me more confused than I initially was.

Let us apply the method of characteristics, with initial condition $u(x_0, 2x_0) = x_0^2$:

*

*$\frac{\text d}{\text d s} y = 1$, letting $y(0) = 2x_0$ we know $y = 2x_0 + s$.

*$\frac{\text d}{\text d s} u = -\frac12 u$, letting $u(0) = x_0^2$ we know $u = x_0^2 e^{-s/2}$

*$\frac{\text d}{\text d s} x = u$, letting $x(0) = x_0$ we know $x = x_0 - 2 x_0^2 (e^{-s/2} - 1)$
The latter is rewritten
$$
x = x_0 - 2 u\, (1 - e^{y/2-x_0}) .
$$
Resolution w.r.t. $x_0$ yields the expression
$$
x_0 = W_n\left(-2u e^{y/2-2 u - x} \right) + 2 u + x
$$
where $W_n$ is the analytic continuation of the product log function, to be substituted in $ue^{y/2} = x_0^2 e^{x_0}$ to get the final implicit expression.
