Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 20 / a / ii / Solution

The first boundary condition requires $u(x,0)=a{x}^2+d=0$ 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