Let ##y(t)## be a quantity that grows from a lower bound ##d## to an upper bound ##a##. Set ##z(t) = y(t) - d## so that ##z## increases from ##0## to ##M = a - d##. Suppose we introduce a parameter ##g≠1## and impose the following governing principle: at each time ##t##, the ratio of ##dz/dt## to the remaining "distance to the top" ##M - z(t)## is proportional to ##(z(t))^g##. In particular:
$$ \frac{dz}{dt} = k \,(z(t))^g\,(M - z(t)) $$
where ##k## is a positive constant. This single ordinary differential equation already encodes the asymmetry via the exponent ##g##. To solve it, we separate variables:
$$ \int \frac{dz}{z^g(M - z)} = k\,t + C $$
Let ##w = z/(M - z)##, so ##z = wM/(1 + w)## and ##dz = M/(1 + w)^2\,dw##. We substitute into the integral:
$$ \int \frac{M/(1+w)^2\,dw}{(wM/(1+w))^g\,M/(1+w)} = \int \frac{1}{M^g}\,(1+w)^{g-1}\,w^{-g}\,dw $$
We set ##u = w/(1 + w)## so ##w = u/(1 - u)## and ##dw = 1/(1 - u)^2\,du##. Also ##1 + w = 1/(1 - u)##. This transforms the integral into:
$$ \frac{1}{M^g} \int u^{-g}\,(1 - u)^{-1}\,du $$
an expression that can be handled in terms of hypergeometric functions or related special functions. Imposing an initial condition such as ##z(0)=0## (which forces ##w(0)=0## and ##u(0)=0##) determines the constant ##C## and solves explicitly for ##u## in terms of ##t##. Reversing the substitutions, ##w(t) = u(t)/(1 - u(t))## and ##z(t)=w(t)M/(1 + w(t))##, then ##y(t)=d+z(t)##.
We find that for suitable parameter choices, the final closed-form solution takes the shape:
$$ d + \frac{M}{(1 + (\alpha\,t)^b)^g} $$
which is precisely:
$$ d + \frac{a - d}{(1 + (x/c)^b)^g} $$
after renaming constants. The exponent ##g## appears naturally from the factor ##(z(t))^g## in the differential equation and breaks the usual symmetry of the logistic curve, giving an asymmetric S-shape that approaches ##d## and ##a## at different rates.