Let I be an interval and f(x) a continuous real-valued function on I. Let y_{1} and y_{2} be linearly independent solutions of y'' = f(x) y, which take positive values on I. Show that from some positive constant k, k √(y_{1} y_{2}) is a solution of y'' + 1/y^{3} = f(x) y.

**Solution**

Answer: k = √(2/d) where d is the Wronksian.

The key is to show that y_{1}y_{2}' - y_{1}'y_{2} is a constant. This expression is known as the Wronskian and it is well-known that it is non-zero iff y_{1} and y_{2} are linearly independent.

The derivative is y_{1}y_{2}'' - y_{1}y_{2}'' = f y_{1}y_{2} - f y_{1}y_{2} = 0, which shows that the expression is constant. Let its value be d.

Let us write z = k √(y_{1}y_{2}), where k is a constant to be determined later. It is also convenient to put c = k^{2}/2, so that z^{2} = 2c y_{1}y_{2}. Differentiating we get: z z' = c(y_{1}y_{2}' + y_{1}'y_{2}) (*). Differentiating again: z z'' + (z')^{2} = c y_{1}y_{2}'' + c y_{1}''y_{2} + 2c y_{1}'y_{2}' = 2c f y_{1}y_{2} + 2c y_{1}'y_{2}' = f z^{2} + 2c y_{1}'y_{2}'.

Multiplying by z^{2}: z^{3}z'' + (z z')^{2} = f z^{4} + 2z^{2} c y_{1}'y_{2}' = f z^{4} + 4 c^{2} y_{1}y_{2}y_{1}'y_{2}'. Using (*), z^{3} z'' + c^{2}(y_{1}y_{2}' + y_{1}'y_{2})^{2} = f z^{4} + 4c^{2} y_{1}y_{2}y_{1}'y_{2}'. Hence z^{3}z'' + c^{2}(y_{1}y_{2}' - y_{1}'y_{2})^{2} = f z^{4}, or z^{3} z'' + c^{2}d^{2} = f z^{4}. So if we set c = 1/d, then z'' + 1/z^{3} = f z.

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2002