# "Secondary exponential decay" - differential equation **URL:** \ **Category:** Factual Questions\ **Created:** [November 24, 2009, 4:11pm UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782 "2009-11-24T16:11:17Z")\ **Posts on this page:** 7\ **Page:** 1
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [November 24, 2009, 4:11pm UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782/1 "2009-11-24T16:11:17Z")
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. 😃 ---
**Author:** ![Napier](https://avatars.discourse-cdn.com/v4/letter/n/ce73a5/32.png) [@Napier](https://boards.straightdope.com/u/Napier)\ **Post date:** [November 24, 2009, 5:19pm UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782/2 "2009-11-24T17:19:06Z")
Why aren’t you doing it numerically, with an iterative solver? What’s wrong with you? ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [November 24, 2009, 5:28pm UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782/3 "2009-11-24T17:28:42Z")
> [@naita](#): > > 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. 😃 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 [Method of undetermined coefficients - Wikipedia](http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients) seems to have exactly the information you need. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [November 24, 2009, 7:18pm UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782/4 "2009-11-24T19:18:02Z")
> [@Napier](#): > > Why aren’t you doing it numerically, with an iterative solver? What’s wrong with you? 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) is 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. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [December 2, 2009, 2:38am UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782/5 "2009-12-02T02:38:34Z")
> [@MikeS](#): > > 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) > > is > > 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=X0_t_exp(-L\*t). ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [December 2, 2009, 8:27am UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782/6 "2009-12-02T08:27:29Z")
> [@Andy\_L](#): > > Bumping the thread to add - if the l=L, the solution becomes y=X0_t_exp(-L\*t). How bout when l!=L? I haven’t got sufficient need, curiousity and/or time to work it out. 😃 ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [December 2, 2009, 8:50pm UTC](https://boards.straightdope.com/t/secondary-exponential-decay-differential-equation/518782/7 "2009-12-02T20:50:54Z")
> [@naita](#): > > How bout when l!=L? > > I haven’t got sufficient need, curiousity and/or time to work it out. 😃 y=L\*X0/(L-l) \* (e^-l t - e^-Lt)