1. y′ + py = qyn.
Let k = -(n-1), then solve the following equation.
2. z′ + kpz = kq
Let f(x) be the kth root of g(x), and show that f is a solution to (1) iff g is a solution to (2). (We're assuming f and g are nonnegative.)
Let w be the inverse of k, whence f = gw. Substitute for y in equation (1).
(gw)′ + pgw = qgwn
wgw-1g′ + pgw = qgwn
w-1 = wn (from the definition of k and w)
wg′ + pg = q
g′ + kpg = kq
Conversely, substitute fk in equation (2) to resurrect equation (1).
The solutions to equations (1) and (2) correspond. To solve one equation is to solve the other. Since we already have a procedure to solve first order linear equations, the Bernoulli equation is under control.
If complex functions are permitted, f and g correspond n-1 to 1. Solve for g, as above, then select any of the n-1 roots of g to find f, a solution to the bernoulli equation. Like the constant c, the "correct" root may be determined by the context of the problem. For instance, if g is real, you probably want f to be the real root of g, rather than one of the n-2 complex roots.