13th Putnam 1953 Problem B3 k is real. Solve the differential equations y' = z(y + z) k , z' = y(y + z) k subject to y(0) = 1, z(0) = 0.
Adding, (y + z)' = (y + z) k+1 . Integrating (y + z) k = 1/(1 - kx) (1). Multiplying opposite sides of the two given equations together, we get yy'(y + z) k = zz' (y + z) k . Hence yy' = zz'. Integrating y 2 - z 2 = 1. Hence (y + z) k (y - z) k = 1, so (y - z) k = 1 - kx (2). For k non-zero, we can immediately solve (1) and (2) to get y = 1/2 ( (1 - kx) 1/k + 1/(1 - kx) 1/k ), z = 1/2 ( (1 - kx) 1/k - 1/(1 - kx) 1/k ). For k = 0, the original equations simplify to y' = z, z'= y. So y'' = y. So y = A cosh x + B sinh x, z = A cosh x + B sinh x. Applying the initial conditions, y = cosh x, z = sinh x. 13th Putnam 1953 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002