the velocity of a skydiver falling to the ground is governed by:
m dv/dt = mg - kv^2
solve for v(t) with initial conditions v(0) = 0
i rearranged equation into standard form:
v' + (k/m)v^2 = g
so i got an integrating factor, mu(x) = e^integ(k/m), but does the fact that the v is squared prohibit me from using an integrating factor?