Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-2/6b/solution

The chain rule gives and . The coefficients of and are and . Their integer roots are and . Thus choose characteristic coordinates , , whose Jacobian determinant is . The coefficient of is , so the equation becomes , equivalently .
Integrating the transformed hyperbolic partial differential equation gives . Along the coordinates are , so the zero boundary condition gives for every real . Along they are , so , hence . The constants cancel and the unique resulting solution is
It directly satisfies both boundary conditions and .

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