Characteristic method , boundry conditions

I have problem with solving following task:

Using characteristic method find solution $$u(x,y)$$ of

\begin{aligned} x^{2} u_{x}+y^{2} u_{y} &=(x+y) u \quad \text { for }(x, y) \in \mathbb{R}^{+} \times \mathbb{R}^{+} \\ u\left(x, \frac{x}{2}\right) &=1 \quad \text { for } x>0 \end{aligned}

I found out that $$y(s) = \frac{-1}{s} + C \\ x(s) = \frac{-1}{s} + C \\ z(s) = -2s + C$$

where $$z(s)$$ is function $$u(x(s), y(s))$$ How can i find boundary conditions? Thanks for help.

You solved the differential equations incorrectly, as $$-\frac1{s+C}\ne-\frac1s+C$$, and you did not use different integration constants in (as of that point) unconnected integrations.
$$ds=\frac{dx}{x^2}$$ integrates to $$s=-\frac{1}{x}+\frac1{x_0}$$ so that $$x(s)=\frac{x_0}{1-x_0s}.$$ Similarly, $$y(s)=\frac{y_0}{1-y_0s}.$$ Using these solutions in $$(x+y)ds=dz$$ gives $$-\ln|1-x_0s|-\ln|1-y_0s|=z-z_0$$ Now insert $$y_0=\frac{x_0}2$$, $$z_0=1$$ and eliminate $$s$$.
Alternatively, find that $$\frac{dx}{x^2}=\frac{dy}{y^2}\implies y^{-1}-x^{-1}=c_1$$ and $$\frac{dy-dx}{y^2-x^2}=\frac{dz}{x+y}\implies\frac{d(y-x)}{y-x}=dz \\~\\ \implies z-\ln|y-x|=c_2=\phi(c_1) \\~\\ \implies u(x,y)=z=\ln|y-x|+\phi(\frac1y-\frac1x)$$ and with the initial conditions $$1=z_0=\ln|\frac{x_0}2|+\phi(\frac1{x_0})\implies \phi(t)=1+\ln|2t|$$