$$(1+x^2)\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=0$$
integrating to give: $\arctan x=\ln y+A$
I managed to find the general solution as $u(x,y)=F(A)=F(\ln y -\arctan x)$
I have the boundary conditions
$u(0, y) = \ln y$, for the domain $x ≥ 0, y > 0;$
$ u(x, 0) = \ln x$, for the domain $x > 0, y ≥ 0$.
I substitute the boundary conditions into the general solution, but the question asks for each of the boundary conditions, find the particular if it exist which I dont understand what to do because of the domains.
$u(x,y)=f(\tan^{-1}(x)-\ln y)$. The second condition $u(x,0)=\ln x$ clearly makes the problem ill posed, since $\ln y$ is not defined at $y=0$. So the solution does not exist for the second condition. The first condition $u(0,y)=\ln y$ gives $f(-\ln y)=\ln y$, ie $f(z)=-z$. This then gives $u(x,y)=\ln y-\tan^{-1}(x)$.