44th Putnam 1983 Problem B3 y 1 , y 2 , y 3 are solutions of y''' + a(x) y'' + b(x) y' + c(x) y = 0 such that y 1 2 + y 2 2 + y 3 2 = 1 for all x. Find constants α, β such that y 1 '(x) 2 + y 2 '(x) 2 + y 3 '(x) 2 is a solution of y' + α a(x) y + βc(x) = 0. Solution
: α = 2/3, β = -2/3. Let z = y 1 '(x) 2 + y 2 '(x) 2 + y 3 '(x) 2 . Multiplying the differential equation by y, we see that y 1 satisfies y 1 y 1 ''' + a y 1 y 1 '' + b y 1 y 1 ' + c y 1 2 = 0. Similarly for y 2 and y 3 . Adding, we get (y 1 y 1 ''' + y 2 y 2 ''' + y 3 y 3 ''') + a (y 1 y 1 '' + y 2 y 2 '' + y 3 y 3 '') + b(y 1 y 1 ' + y 2 y 2 ' + y 3 y 3 ') + c (y 1 2 + y 2 2 + y 2 2 ) = 0 (*). We now use the relation y 1 2 + y 2 2 + y 3 2 = 1 (**) to obtain alternative expressions for each of the terms in parentheses. We already have the term following c. Differentiating (**) we get y 1 y 1 ' + y 2 y 2 ' + y 3 y 3 ' = 0, so the term following b vanishes. Differentiating again gives (y 1 y 1 '' + y 2 y 2 '' + y 3 y 3 '') = - z, which gives the term following a. Differentiating again, we get (y 1 'y 1 '' + y 2 'y 2 '' + y 3 'y 3 '') + (y 1 y 1 ''' + y 2 y 2 ''' + y 3 y 3 ''') = - z' . The first parenthesis here is just z'/2, so we get (y 1 y 1 ''' + y 2 y 2 ''' + y 3 y 3 ''') = -3/2 z', which deals with the first term in (*). Putting all this together we have -3/2 z' - az + c = 0, or z' + 2/3 az - 2/3 c = 0. 44th Putnam 1983 © John Scholes jscholes@kalva.demon.co.uk 16 Jan 2001