 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v/5/b/i/solution

\partial_{x} = \partial_{ξ} + \partial_{η}

and

\partial_{t} = α \partial_{ξ} + β \partial_{η}

. Substitution into the differential equation yields

(α^{2} + 2 B α + C) y_{ξξ} + 2 {α β + B (α + β) + C} y_{ξ η} + (β^{2} + 2 Bβ + C) y_{ηη} = 0.

α \neq = β

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