Substituting the particular value of f(t) given in equation (10) and integrating, it can readily be deduced that
Q/Q_{T} = (ae^{-λ_{2}t} - be^{-λ_{1}t})/(a - b) (11), where a = (1 - e^{-λ_{2}T})/λ_{2}, b = (1 - e^{-λ_{1}T})/λ_{1}.
In a similar way, the number of particles R of the matter C present at any time can be deduced by substitution of the value of f(t) in equation (5). These equations are, however, too complicated in form for simple application to experiment, and will not be considered here. 200. Case 4. The matter A is supplied at a constant rate from a primary source. Required to find the number of particles of A, B, C at any subsequent time t, when initially A, B, C are absent.
The solution can be simply obtained in the following way. Suppose that the conditions of Case 2 are fulfilled. The products A, B, C are in radio-active equilibrium and let P_{0}, Q_{0}, R_{0} be the number of particles of each present. Suppose the source is removed. The values of P, Q, R at any subsequent time are given by equations (7), (8) and (9) respectively. Now suppose the source, which has been removed, still continues to supply A at the same constant rate and let P_{1}, Q_{1}, R_{1} be the number of particles of A, B, C again present with the source at any subsequent time. Now we have seen, that the rate of change of any individual product, considered by itself, is independent of conditions and is the same whether the matter is mixed with the parent substance or removed from it. Since the values of P_{0}, Q_{0}, R_{0} represent a steady state where the rate of supply of each kind of matter is equal to its rate of change, the sum of the number of particles A, B, C present at any time with the source, and in the matter from which it was removed, must at all times be equal to P_{0}, Q_{0}, R_{0}, . . ., that is
P_{1} + P = P_{0},
Q_{1} + Q = Q_{0},
R_{1} + R = R_{0}.
