24th Putnam 1963 Problem A3 Let D be the differential operator d/dx and E the differential operator xD(xD - 1)(xD - 2) ... (xD - n). Find an expression of the form y = ∫ 1 x g(t) dt for the solution of the differential equation Ey = f(x), with initial conditions y(1) = y'(1) = ... = y (n) (1) = 0, where f(x)is a continuous real-valued function on the reals.
The first step is to try to simplify Ey. Experimenting with small n, we soon suspect that Ey = x n+1 y (n+1) and that is easily proved by induction. So we seek a solution to x n+1 y (n+1) = f(x), subject to y(1) = y'(1) = ... = y (n) (1) = 0. The trick is to use the Taylor expansion: y(x) = y(1) + (x-1)y'(1) + (x-1) 2 /2! y''(1) + ... + (x-1) n y (n) (1) + ∫ 1 x g(t) dt, where g(t) = (x - t) n y n+1 (t)/n! = (x - t) n f(t)/(n! t n+1 ). 24th Putnam 1963 © John Scholes jscholes@kalva.demon.co.uk 5 Feb 2002