The function $u(x,y)$ satisfies the partial differential eqaution $x^{3}\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=0$
Show that the characteristic that passes through the point $(x,y)$ is given by $y(x)=\frac{1}{2}(x^{-2}-x_{0}^{-2})$ for $x_{0}\ne 0$
This is what I have so far
$\frac{dx}{ds}=x^{3}$ $\frac{dy}{ds}=-1$
Therefore $\frac{dy}{dx} = -\frac{1}{x^{3}}$
$\Rightarrow \int 1 dy = \int -\frac{1}{x^{3}} dx \Rightarrow y=\frac{1}{2x^{2}}+c$
I know I now need to find the constant in terms of $x_0$ but I don't know how
Maybe there is some misprint in your problem, you should revise the given point $(x,y)$ which the characteristic you fixed pass through as $(x_0,0)$. Actually, the characteristic lines of a linear first order PDE are a family of integral curve,just like what you have done. See bellow,we take $(x_0,0)$ into your the characteristics $$0=\frac{1}{2x_0^2}+C,$$ solve the integral constant $$C=-\frac{1}{2x_0^2}.$$