51st Putnam 1990 Problem B1 R is the real line. Find all possible functions f: R → R with continuous derivative such that f(α) 2 = 1990 + ∫ 0 α ( f(x) 2 + f '(x) 2 ) dx for all α.
Putting y = f(x) and differentiating the relationship gives 2y y' = y 2 + y' 2 or (y - y') 2 = 0. So y = y' or y = - y'. Integrating, y = Ae x or y = - Ae x . But y(0) = ±√1990, so f(x) = ±√1990 e x . The function is continuous, so we cannot "mix" the two solutions: either f(x) = √1990 e x for all x, or f(x) = - √1990 e x for all x. 51st Putnam 1990 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2001