Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 2 / 6B (Differential Equations)

Substitution of the power series, with termwise differentiation, gives the coefficient identityThus $a_0$ and $a_1$ are arbitrary, and all power series solutions of the Hermite differential equation areAn empty product is one. For a nonterminating branch the ratio of consecutive terms in its own parity subsequence is asymptotic to $|x{|}^2/j$, hence tends to zero for every fixed $x$. The ratio test proves convergence everywhere; the resulting power series can be differentiated termwise and indeed solve the differential equation. A terminating branch is already a polynomial.

The boundary condition $W(0)=0$ forces $a_0=0$; the solution is odd. Apart from the zero solution, which works for every $E$, a polynomial occurs exactly when an odd index $d$ satisfies $2d-E=0$. Before that index the odd coefficients are nonzero, and after it they vanish. Thus the odd polynomial solutions of the Hermite differential equation haveEquivalently, comparison of leading polynomial coefficients gives $E=2\mathrm{deg}W$, proving that no other nonzero polynomial is possible. Normalizing by $a_1=1$, the polynomials of degree below six areEach can be multiplied by an arbitrary nonzero constant; these are normalized odd Hermite polynomials.

Let $L=-d^2/d{x}^2+2xd/dx$. The forcing in the final inhomogeneous linear differential equation is precisely the displayed degree-five eigenfunction $P_5$, with $L{P}_5=10P_5$. HenceDirect differentiation gives $LW=x-4x^3/3+4x^5/15$, and $W(0)=0$. Among polynomial solutions this choice is unique: a polynomial in the kernel of $L$ must be constant, and its value at zero eliminates that constant.