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**

Answer: 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.

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001