### 44th Putnam 1983

Problem B3

y1, y2, y3 are solutions of y''' + a(x) y'' + b(x) y' + c(x) y = 0 such that y12 + y22 + y32 = 1 for all x. Find constants α, β such that y1'(x)2 + y2'(x)2 + y3'(x)2 is a solution of y' + α a(x) y + βc(x) = 0.

Solution

Answer: α = 2/3, β = -2/3.

Let z = y1'(x)2 + y2'(x)2 + y3'(x)2. Multiplying the differential equation by y, we see that y1 satisfies y1y1''' + a y1y1'' + b y1y1' + c y12 = 0. Similarly for y2 and y3. Adding, we get (y1y1''' + y2y2''' + y3y3''') + a (y1y1'' + y2y2'' + y3y3'') + b(y1y1' + y2y2' + y3y3') + c (y12 + y22 + y22) = 0 (*). We now use the relation y12 + y22 + y32 = 1 (**) to obtain alternative expressions for each of the terms in parentheses.

We already have the term following c. Differentiating (**) we get y1y1' + y2y2' + y3y3' = 0, so the term following b vanishes. Differentiating again gives (y1y1'' + y2y2'' + y3y3'') = - z, which gives the term following a. Differentiating again, we get (y1'y1'' + y2'y2'' + y3'y3'') + (y1y1''' + y2y2''' + y3y3''') = - z' . The first parenthesis here is just z'/2, so we get (y1y1''' + y2y2''' + y3y3''') = -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.