4th Putnam 1941 Problem B3 Let y and z be any two linearly independent solutions of the differential equation y'' + p(x) y' + q(x) y = 0. Let w = y z. Find the differential equation satisfied by w.
We have w = yz, w' = y'z + yz', w'' = y''z + 2y'z' + yz''. Hence w'' + pw' + 2qw = 2y'z' (1). Now (y'z')' = y''z' + y'z'', 2py'z' = py'z' + py'z', qw' = qyz' + qy'z, so (y'z')' + 2py'z' + qw' = 0 (2). Differentiating (1) we get: w''' + p'w' + pw'' + 2q'w + 2qw' = 2(y'z')' = -4py'z' - 2qw' (using (2) ), = -2p(w'' + pw' + 2qw) - 2qw'. Rearranging, this gives: w''' + 3p w'' + (p' + 2p 2 + 4q) w' + (4pq + 2q')w = 0. 4th Putnam 1941 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002