R is the reals. f, g : R2 → R have continuous partial derivatives of all orders. What conditions must they satisfy for the differential equation f(x, y) dx + g(x, y) dy = 0 to have an integrating factor h(xy)?
Solution
If h(xy) is an integrating factor, then f(x, y) h(xy) dx + g(x, y) h(xy) dy is an exact form. In other words, ∂/∂y (f(x, y) h(xy) ) = ∂/∂x (g(x, y) h(xy). So h ∂f/∂y + x f h' = h ∂g/∂x + y g h'. Hence h'/h = (∂f/∂y - ∂g/∂x) / (yg - xf).
The lhs is a function of xy, so the rhs must be also. Equally if the rhs is a function of xy, then we can immediately integrate to get h(xy). Thus the required necessary and sufficient condition for the existence of an integrating factor is that (∂f/∂y - ∂g/∂x) / (yg - xf) should be a function of xy.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002