<s>[INST] Characteristic curves : Why do I get different results? With a given $h(x)$ we want to solve $$xu_y-yu_x=u \\ u(x,0)=h(x)$$ 
I have solved it using two ways. 
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First way: 
For $x\neq 0$ we get $u_y-\frac{y}{x}u_x=\frac{u}{x}$. 
We have that $$\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}$$ 
Therefore, we have $\frac{dy}{ds}=1$, $\frac{dx}{ds}=-\frac{y}{x}$ and $\frac{du}{ds}=\frac{u}{x}$. 
From $\displaystyle{\frac{dy}{ds}=1}$ we get $\displaystyle{y=s}$. 
From $\displaystyle{\frac{dx}{ds}=-\frac{y}{x}}$ with $x(0)=x_0$ we get $\displaystyle{\frac{dx}{dy}=-\frac{y}{x} \Rightarrow x dx=-y dy \Rightarrow x^2=-y^2+x_0^2 \Rightarrow x=\pm \sqrt{-y^2+x_0^2}}$. 
From $\displaystyle{\frac{du}{ds}=\frac{u}{x}}$ we get \begin{align*}&\frac{du}{dy}=\frac{u}{\pm \sqrt{-y^2+x_0^2}} \Rightarrow \frac{1}{u}du=\frac{1}{\pm \sqrt{x_0^2-y^2}}dy\Rightarrow \frac{1}{u}du=\frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy \\ &  \Rightarrow \int \frac{1}{u}du=\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy\end{align*} 
We set $\displaystyle{\frac{y}{x_0}=\cos k \Rightarrow \frac{1}{x_0}dy=-\sin k dk \Rightarrow dy=-x_0\sin k \ dk}$. 
Therefore we get: 
\begin{align*}\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy&=\int \frac{-x_0\sin k}{\pm |x_0|\sqrt{1-\cos^2k}}dk=\int \frac{-x_0\sin k}{\pm |x_0|\sqrt{\sin^2 k}}dk=\int \frac{-x_0\sin k}{\pm |x_0| |\sin k|}dk \\ & =\mp \int \frac{x_0\sin k}{|x_0 \sin k|}dk=\mp \int  \pm 1 dk=k+c\end{align*}
So, we get \begin{align*}&\int \frac{1}{u}du=\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy \Rightarrow \ln |u|=k+c \Rightarrow e^{\ln |u|}=e^{k+c} \Rightarrow |u|=e^ce^k \Rightarrow u=\pm e^ce^k \Rightarrow u=Ce^k \\ &  \Rightarrow u=Ce^{\arccos \left (\frac{y}{x_0}\right )}\end{align*} 
For $y=0$ we have $\displaystyle{u=h(x_0) \Rightarrow Ce^{\arccos \left (\frac{y}{x_0}\right )}=h(x_0) \Rightarrow  C=\frac{2h(x_0)}{\pi}}$. 
Therefore, the solution is $$u=\frac{2h(x_0)}{\pi}e^{\arccos \left (\frac{y}{x_0}\right )} \Rightarrow u=\frac{2h(\pm \sqrt{x^2+y^2
})}{\pi}e^{\arccos \left (\frac{y}{\pm \sqrt{x^2+y^2
}}\right )}$$ 
$$$$ 
Second way: 
From $\displaystyle{xu_y-yu_x=u}$ and $\displaystyle{\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}}$ we get : $\displaystyle{\frac{dx}{ds}=-y, \ \frac{dy}{ds}=x, \ \text{ and } \ \frac{du}{ds}=u}$. 
From $\displaystyle{\frac{dx}{ds}=-y}$ we get that $\displaystyle{ds=-\frac{1}{y}dx}$ and from $\displaystyle{\frac{dy}{ds}=x}$ we get $\displaystyle{ds=\frac{1}{x}dy}$. 
Therefore we get (using that $x(0)=x_0$) $\displaystyle{-\frac{1}{y}dx=\frac{1}{x}dy \Rightarrow -xdx=ydy \Rightarrow -\frac{x^2}{2}=\frac{y^2}{2}-\frac{x_0^2}{2} \Rightarrow x^2=-y^2+x_0^2 \Rightarrow x=\pm \sqrt{x_0^2-y^2}}$. 
From $\displaystyle{\frac{du}{ds}=u}$ using that $ds=\frac{1}{x}dy$ we get that \begin{align*}\frac{xdu}{dy}&=u \Rightarrow \frac{1}{u}du=\frac{1}{x}dy \Rightarrow \frac{1}{u}du=\frac{1}{\left (\pm \sqrt{x_0^2-y^2}\right )}dy \Rightarrow \ln |u|=\pm \arctan \left (\frac{y}{x_0^2-y^2}\right )+C \\ & \Rightarrow u=\pm e^{\pm C}e^{\arctan \left (\frac{y}{x_0^2-y^2}\right )}\end{align*} 
Using the condition for $y=0$ then $\displaystyle{u=h(x_0)}$ we get that $\displaystyle{h(x_0)=\pm e^{\pm C}}$. 
Therefore the solution is $$u=h(x_0)e^{\arctan \left (\frac{y}{x_0^2-y^2}\right )} \Rightarrow u=h(\pm \sqrt{x^2+y^2})e^{\arctan \left (\frac{y}{x^2}\right )}$$ 
Why do I get different results? Have I done something wrong? 
 [/INST] (The two solutions are the same. The notation tends to obscure the identity of the solutions. You have $h(\pm \sqrt{x^2+y^2})$ and $\dfrac{2h(\pm \sqrt{x^2+y^2})}{\pi}$. Nevertheless, they are not the same function. We need to make the following stipulation: $\dfrac{2h(\pm \sqrt{x^2+y^2})}{\pi}=g(\pm \sqrt{x^2+y^2})$, that it's legitimate as both are functions of the same argument, $\pm \sqrt{x^2+y^2}$) True, but irrelevant now.
Further, $\arctan \left (\dfrac{y}{x^2}\right )$ is not correct, it has to be $\arctan \left (\dfrac{y}{x}\right )$ as the integral of $1/\sqrt{x_0^2-y^2}$ is $\arctan\left(y/\sqrt{x_0^2-y^2}\right)$
And finally,
Corrected (added the $\pm\pi/2$ phase)
$\arctan \left (\pm\dfrac{y}{x}\right )=\arccos \left (\dfrac{y}{\pm \sqrt{x^2+y^2}}\right )\pm\pi/2$ being careful with the signs in order to the identity to hold.
Postcriptum: I don't detelete the answer as it's neat now and I prefer it here than in the never empty trash bin.
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