: a = 6, b = 14, c = 1. The differential equation looks fairly horrible; there is no obvious systematic way of solving it. So we try various types of solutions. Trying simple polynomials leads to x 2 + x. Trying exponentials leads to e 3x . Obviously, a particular solution to F(y) = 36x + 6 is y = 2, so the general solution is y = 2 + Ae 3x + Bx + Bx 2 . f(0) = 1 implies A = -1. ( f(-1) - 2)( f(1) - 6) = (- e -3 )( -4 - e 3 + 2B) = 1, so B = 2. f(-2) = 2 - e -6 + 4, f(2) = 2 - e 6 + 12, so we need a = 6, b = 14, c = 1 (and then the relation reduces to ( - e -6 )( - e 6 ) = 1, as required). 40th Putnam 1979 © John Scholes jscholes@kalva.demon.co.uk 4 Dec 1999
40th Putnam 1979 Problem B4 Find a non-trivial solution of the differential equation F(y) ≡ (3x 2 + x - 1)y'' - (9x 2 + 9x - 2)y' + (18x + 3)y = 0. y = f(x) is the solution of F(y) = 6(6x + 1) such that f(0) = 1, and ( f(-1) - 2)( f(1) - 6) = 1. Find a relation of the form ( f(-2) - a)( f(2) - b) = c. Solution