Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 2 / 8C / a / Solution

The chain rule gives ${\partial }_x={\partial }_u+{\partial }_v$ and ${\partial }_t={\partial }_u-{\partial }_v$. Hence ${\partial }_x^2-{\partial }_t^2=4{\partial }_u{\partial }_v$, and the forced wave equation becomes $y_{uv}=1$. Integrating twice gives $y=uv+F(u)+G(v)$. At $t=0$ the initial conditions areThus $F-G$ is constant, which cancels from the final expression. Combining the remaining sum yieldsDifferentiation checks the forcing and both initial conditions.