Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/8/a/solution

The origin is an ordinary point of the Hermite differential equation, so put . Substitution and matching powers gives
Thus the linear recurrence relation is
Even and odd coefficients evolve independently from arbitrary . The general power series solution is
with an empty product equal to one. Its first terms are
Both series converge at every finite , since the ratio of successive terms in each parity subsequence tends to zero. For a nonzero polynomial of degree , the highest coefficient in the differential equation requires . Conversely, at such a value the appropriate parity series terminates, while its opposite-parity initial coefficient can be set to zero. Hence
Here is a physicists' Hermite polynomial. Two explicit choices are
Direct differentiation verifies both. These are polynomial solutions up to an arbitrary overall factor. Without a polynomial requirement or additional boundary conditions, the equation also has nonpolynomial solutions for other real ; discreteness here comes from termination.

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