$$yu_{xx}+u_{yy}=0$$
If we look at $$\frac{b\pm\sqrt{b^2-ac}}{a}$$
We get $$\frac{\pm\sqrt{-y}}{y}$$
So we have $3$ cases, for the case of $y<0$ can we say that we have
1.$$\frac{dy}{dx}=-\frac{1}{\sqrt{y}}\Rightarrow P=\frac{2}{3}y^{1.5}+x$$
2.$$\frac{dy}{dx}=\frac{1}{\sqrt{y}}\Rightarrow Q=\frac{2}{3}y^{1.5}-x$$
If so, I get $$u(P,Q)=y(u_{PP}-2u_{PQ}+u_{QQ}+\frac{1}{2\sqrt{y}}(u_P+u_Q)+y(u_{PP}+2u_{PQ}+U_{QQ})$$
Where the answer is: $$u(P,Q)=y(u_{PP}-2u_{PQ}+u_{QQ}-\frac{1}{2\sqrt{y}}(u_P+u_Q)-y(u_{PP}+2u_{PQ}+U_{QQ})$$
Where did it went wrong?