Solution
= Solution
The first <boundary condition> requires $u(x,0)=ax^2+d=0$ for all $x$, so $a=d=0$. Since $u_y=bx+2cy$, the second <boundary condition> gives $b=0$. The <partial differential equation> then reads $2a-2c=1$, fixing $c=-1/2$. Thus
$$
\boxed{a=0,\quad b=0,\quad c=-\frac12,\quad d=0,\qquad u(x,y)=-\frac{y^2}{2}.}
$$