"Secondary exponential decay" - differential equation

Because of an unhealthy interest in mathematical problems I’ve been trying to figure out an equation for the amount of a radioactive product of radioactive decay.

The equation for the amount left after primary decay is of course:

x(t) = x0 * e^-Lt

where x0 is the initial amount and L is the decay constant. But if the result of this decay is a radioactive isotope y with decay constant l, what is the equation for the amount of y at time t?

I think it should be a solution to the differential equation:

y’ = -ly + x0 * L * e^-Lt

but I’m not sure I can solve that, and I’m not sure I haven’t made any wrong assumptions, and I should have spent the last two hours doing something else besides playing around with this. So any knowledgable input would be very welcome, or I might waste most of tomorrow’s workday as well. :smiley:

Why aren’t you doing it numerically, with an iterative solver? What’s wrong with you?

Your equation looks right, and equations of this form can be solved by the method of undetermined coefficients, in which (essentially) you solve the equation first as if only the -ly part were there, and second, you solve the equation if only the x0 (etc.) part were there, and then combine the results (carefully). Example 2 in this article http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients seems to have exactly the information you need.

Pish tosh. If you want it done right, do it analytically.

Andy L’s article is pretty much what you need, but if you just want a plug-and-chug formula, the general solution of the equation

y’ + l y = q(t)


y(t) = e[sup]-l t[/sup] ( integral ( q(t) e[sup]l t[/sup] ) + C)

Since q(t) is itself an exponential in your case, the result will be a sum of two exponentials. C is a constant of integration determined by the condition that y(0) = 0 (or whatever you want it to be.) In the general case (with l != L), you end up with a solution that’s the difference between two decaying exponentials; the amount of isotope y peaks at some particular time, and then decays away thereafter. How quickly the amount of isotope y decreases depends on which is larger, l or L; in general, after a sufficiently long time, the amount of y will be decrease exponentially, with a time constant equal to the larger of l or L.

Bumping the thread to add - if the l=L, the solution becomes y=X0texp(-L*t).

How bout when l!=L?

I haven’t got sufficient need, curiousity and/or time to work it out. :smiley:

y=L*X0/(L-l) * (e^-l t - e^-Lt)