Let S be the set of all curves satisfying y' + a(x) y = b(x), where a(x) and b(x) are never zero. Show that if C ∈ S, then the tangent at the point x = k on C passes through a point Pk which is independent of C.
Solution
Easy.
Let Ch be the curve with y(k) = h. Then the gradient of Ch at (k, h) is b(k) - a(k) h, so the tangent at (k, h) is (y - h) = (b(k) - a(k) h)(x - k). This may be written as (a(k) x - a(k) k - 1) h = (b(k) (x - k) - y). Evidently, this always passes through the point (k + 1/a(k), b(k)/a(k) ) whatever the value of h.
© John Scholes
jscholes@kalva.demon.co.uk
26 Nov 1999