Method of characteristics: $xu_t -tu_x = u$ I am just learning the method of characteristics. Suppose I want to solve
$$xu_t - tu_x = u \quad u(x,0) = h(x)$$
I write
$$\dot{t}(s) = x\\
\dot{x}(s) = -t \\
\dot{z}(s) = z$$
If $t(0) = 0$ and $x(0) = x_0$, we have (I think)
$$t(s) = x_0\sin(s)\\
x(s) = x_0\cos(s)\\
z(s) = h(x_0)e^s$$
Where do I go from here? First I'd like to get rid of the $s$, then I'd like to get rid of the $x_0$, but I'm not seeing how to do that.

$xu_t-tu_x=u$
$\dfrac{u_t}{t}-\dfrac{u_x}{x}=\dfrac{u}{xt}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=\dfrac{1}{t}$ , letting $t(0)=0$ , we have $t^2=2s$
$\dfrac{dx}{ds}=-\dfrac{1}{x}$ , letting $x(0)=x_0$ , we have $x^2=-2s+x_0^2=-t^2+x_0^2$
$\dfrac{du}{ds}=\dfrac{u}{xt}=\dfrac{u}{\sqrt{2s(x_0^2-2s)}}$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)e^{\tan^{-1}\frac{\sqrt{2s}}{\sqrt{x_0^2-2s}}}=f(\sqrt{x^2+t^2})e^{\tan^{-1}\frac{t}{x}}$
$u(x,0)=h(x)$ :
$f(x)=h(x)$
$\therefore u(x,t)=h(\sqrt{x^2+t^2})e^{\tan^{-1}\frac{t}{x}}$
