### 16th Putnam 1956

**Problem B1**

The differential equation a(x, y) dx + b(x, y) dy = 0 is homogeneous and exact (meaning that a(x, y) and b(x, y) are homogeneous polynomials of the same degree and that ∂a/∂y = ∂b/∂x). Show that the solution y = y(x) satisfies x a(x, y) + y b(x, y) = c, for some constant c.

**Solution**

d(x a + y b) = (a dx + b dy) + x( ∂a/∂x dx + ∂a/∂y dy) + y( ∂b/∂x dx + ∂b/∂y dy).

Using exactness, this becomes: (a dx + b dy) + x( ∂a/∂x dx + ∂b/∂x dy) + y( ∂a/∂y dx + ∂b/∂y dy). Homogeneity implies that for some integer k we have x ∂a/∂x + y ∂a/∂y = k a, and x ∂b/∂x + y ∂b/∂y = k b. So we get finally d(x a + y b) = (a dx + b dy) + k (a dx + b dy). So if y is a solution, then d(x a + y b) = 0 and hence x a + y b = c for some constant c.

16th Putnam 1956

© John Scholes

jscholes@kalva.demon.co.uk

4 Dec 1999