### 36th Putnam 1975

Problem A5

Let I be an interval and f(x) a continuous real-valued function on I. Let y1 and y2 be linearly independent solutions of y'' = f(x) y, which take positive values on I. Show that from some positive constant k, k √(y1 y2) is a solution of y'' + 1/y3 = f(x) y.

Solution

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

The key is to show that y1y2' - y1'y2 is a constant. This expression is known as the Wronskian and it is well-known that it is non-zero iff y1 and y2 are linearly independent.

The derivative is y1y2'' - y1y2'' = f y1y2 - f y1y2 = 0, which shows that the expression is constant. Let its value be d.

Let us write z = k √(y1y2), where k is a constant to be determined later. It is also convenient to put c = k2/2, so that z2 = 2c y1y2. Differentiating we get: z z' = c(y1y2' + y1'y2) (*). Differentiating again: z z'' + (z')2 = c y1y2'' + c y1''y2 + 2c y1'y2' = 2c f y1y2 + 2c y1'y2' = f z2 + 2c y1'y2'.

Multiplying by z2: z3z'' + (z z')2 = f z4 + 2z2 c y1'y2' = f z4 + 4 c2 y1y2y1'y2'. Using (*), z3 z'' + c2(y1y2' + y1'y2)2 = f z4 + 4c2 y1y2y1'y2'. Hence z3z'' + c2(y1y2' - y1'y2)2 = f z4, or z3 z'' + c2d2 = f z4. So if we set c = 1/d, then z'' + 1/z3 = f z.