Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 9 / b / v / Solution

Put $W=1+x+y$. The identity $W=(s+1{)}^2$ gives $s=-1\pm \sqrt{W}$, and hence $z=-1\pm 2\sqrt{W}$. On the initial half-line $W=1$ and the prescribed value is $z=1$, which selects the positive sign. The branch containing the data has $s>-1$, soFor $W>0$, its derivatives are $z_x=z_y=W^{-1/2}$. Substitution checks the partial differential equation:On the data $y=-x$, the value is one. The other algebraic branch gives $z=-3$ there and does not meet these data. The positive-branch formula defines a solution on the larger half-plane $W>0$, but the part uniquely fixed by the prescribed half-line also requires a nonnegative characteristic footpoint $t$, as part (vi) explains.