1. ## Decay Chains

There is a decay chain with three substances, and I need to solve for each (F, G, and H)

dF/dt = -BF
dG/dt = BF - CG
dH/dt = CG

F(0) = 100
G(0) = 30
H(0) = 15

The half-life of F is 13 hours
The half-life of G is 22 hours

B and C are constants

I solved the first part as:

B = 0.053319
F(t) = 100e^(-0.053319t)

However, I am not sure what to do next, should I just plug the new B and F into the dG/dt equation? Is C = to ln(2)/22 or does this change since the DE is different for G?

Any help would be appreciated.

2. Yeah, it seems that you solved correctly for B, F, and C. I got the same results. But I also stuck on the same step solving for G, and separation of variables doesn't help much either.... I believe that you work out next equation with G the same way you did solving for F. You must have dG/G on the one side to solve for G and I couldn't figure out how. Will try again...

3. If my C is correct, I think I can work out dG/dt with an integrating factor and then I should get a simple dH/dt equal to a function of t.

4. Originally Posted by machi4velli
There is a decay chain with three substances, and I need to solve for each (F, G, and H)

dF/dt = -BF
dG/dt = BF - CG
dH/dt = CG

F(0) = 100
G(0) = 30
H(0) = 15

The half-life of F is 13 hours
The half-life of G is 22 hours

B and C are constants

I solved the first part as:

B = 0.053319
F(t) = 100e^(-0.053319t)

However, I am not sure what to do next, should I just plug the new B and F into the dG/dt equation? Is C = to ln(2)/22 or does this change since the DE is different for G?

Any help would be appreciated.
It is long but straightforward.

You solved for F(t) correctly.

Now you have:
G' = BF - CG

G' + CG = BF = 100Be^{-Bt}

Typically there would be little problem in solving this. The difficulty arises when we consider the half-life of G. How do we deal with this?

If we weren't adding any G then we'd simply have the equation
G' + CG = 0
just like the first part of the problem. But we are constantly adding an amount of G from the decay of F, represented by the 100Be^{-Bt} term. This means that at 22 hours we DON'T have merely 15 (units? grams?) of G left, we should have more. So we should have the condition:
30*e^{-Dt} < G(t), where D = (1/22)*ln(2)

This is the only restriction I am going to put on the solution to the G equation.

G' + CG = 100Be^{-Bt}

The solution to the homogeneous equation is
Gh(t) = (beta)e^{-Ct}

The particular solution will be of the form:
Gp' + CGp = 100Be^{-Bt}
and the homogeneous solution is (beta)e^{-Ct}, so we try a form like:

Gp(t) = (gamma)te^{-Bt}
Gp'(t) = (gamma)e^{-Bt} - (gamma)Bte^{-Bt}

Thus:
(gamma)e^{-Bt} - (gamma)Bte^{-Bt} + (gamma)Cte^{-Bt} = 100Be^{-Bt}

The only way to have this true for arbitrary t is for like terms to cancel. So:
(gamma) = 100B
-B + C = 0 ==> C = B.

Thus
G(t) = Gh(t) + Gp(t) = (beta)e^{-Bt} + 100Bte^{-Bt}
(recall that C = B!)

Now G(0) = 30, so
G(0) = (beta) = 30, thus

G(t) = 30e^{-Bt} + 100Bte^{-Bt} > 30e^{-Bt}
as we needed.

Now to solve the H equation:
H' = CG

H' = 30Be^{-Bt} + 100B^2te^{-Bt} <-- C = B

Just integrate this from 0 to t:

H(t) = 30B(-1/B)(e^{-Bt} - 1) + 100B^2[-*(t/B)*e^{-Bt} - (t/B^2)*e^{-Bt} + 1/B^2] + (delta)
(Typically we don't need a constant of integration for a definite integral, but in this we are solving a differential equation, not simply integrating.)

H(t) = -30(e^{-Bt} - 1) + 100B[-te^{-Bt} - (t/B)*e^{-Bt} + 1/B] + (delta)

So at t = 0:
H(0) = 15 = -30(1 - 1) + 100B[-0 - 0 + 1/B] + (delta)

15 = 100 + (delta)

(delta) = -85

Thus
F(t) = 100e^(-Bt)
G(t) = 30e^{-Bt} + 100Bte^{-Bt} > 30e^{-Bt}
H(t) = -30(e^{-Bt} - 1) + 100B[-te^{-Bt} - (t/B)*e^{-Bt} + 1/B] - 85
where
B = (1/13)*ln(2)

-Dan