Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 2 / 6B / Solution

Write $u(x,y)=U(\xi ,\eta )$. The chain rule gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ and ${\partial }_y=2{\partial }_{\xi }+3{\partial }_{\eta }$. Consequently the coefficients of $U_{\xi \xi }$ and $U_{\eta \eta }$ cancel, while that of $U_{\xi \eta }$ is $12-25+12=-1$. The transformed partial differential equation is $-U_{\xi \eta }=1$. Integrating once in each variable gives the most general twice differentiable solution,where $F,G$ are arbitrary twice differentiable functions. The change of variables is invertible because its Jacobian determinant is $3-2=1$.

On $y=0$, the value condition gives $F(x)+G(x)=x^2$. Differentiating it yields $F^{\prime }(x)+G^{\prime }(x)=2x$. The derivative condition gives $-5x+2F^{\prime }(x)+3G^{\prime }(x)=x$. Subtraction therefore gives $G^{\prime }(x)=2x$ and $F^{\prime }(x)=0$. The constants in $F$ and $G$ cancel in their sum, leavingIndeed $u_{xx}=0$, $u_{xy}=1$, $u_{yy}=6$, so the original operator gives $-5+6=1$, and the two initial-line conditions hold.