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I am trying to answer the question

Species A decays at rate α into species B and species B decays at rate β into species C.

Show that if $ a \neq b $, then...

$ A(t) = A_0e^{-\alpha t}, \quad B(t) = B_0e^{-\beta t} + \alpha A_0(\frac{e^{-\alpha t}-e^{-\beta t}}{\beta - \alpha}) $

The first part I found extremely easy, however the second part I can't see where I'm going wrong, perhaps because I am extremely tired! :P

Here's my working so far:

$ \frac{dB}{dt} = \alpha A - \beta B \rightarrow \int\frac{1}{\alpha A - \beta B} dB = \int dt \rightarrow -\frac{1}{\beta} ln(\alpha A - \beta B) = t+c \rightarrow \alpha A - \beta B = e^{-\beta (t+c)}$

Setting t=0 we get:

$ \alpha A_0 - \beta B_0 = e^{-\beta c} \rightarrow c = \frac{ln(\alpha A_0 - \beta B_0)}{-\beta} $

Subbing in c, I then get the equation to be:

$B\beta = \alpha A - e^{-\beta t}e^{ln(\alpha A_0 - \beta B_0))}$

Which then along with the equation for A simplifies to

$B = \frac{\alpha A_0}{\beta}(e^{-\alpha t}-e^{-\beta t}) + B_0e^{-\beta t}$

So the only thing I cannot get right is the $\beta - \alpha$ on the denominator... I must be making a silly error somewhere, but I can't see where! Can anyone point me in the right direction? I don't want a full work-through, just pointers please :D

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There is an error, unfortunately not of the mechanical kind. When you integrated with respect to $B$ (first line), the function you were integrating had an $A$ in it. This is a function of $t$, so the integral is not right. You are treating $\alpha A$ as if it were constant. – André Nicolas Apr 12 '12 at 19:08

Hint: We have the differential equation $$\frac{dB}{dt}+\beta B=\alpha A.$$ We know $A$, so we know the right-hand side. We have a linear differential equation. Solve the homogeneous DE, and find a particular solution of the inhomogeneous equation.

But it may be that the problem is intended to be trivial. You are asked to show (verify?) that the given answer is right. So it might be enough to show that it works (very easy), and then appeal to uniqueness.

Details: The homogeneous differential equation $\frac{dB}{dt}+\beta B=0$ has general solution $$B=Ce^{-\beta t}.$$ We find a particular solution of the equation $$\frac{dB}{dt}+\beta B=\alpha A_0e^{-\alpha t}.$$ We are told that $\alpha \ne \beta$. So we try for a solution of the shape $ke^{-\alpha t}$. Substituting in the DE, we get $$-k \alpha e^{-\alpha t} +k\beta e^{-\alpha t}=\alpha A_0e^{-\alpha t}.$$ This gives $k=\frac{\alpha A_0}{\beta-\alpha}$, so the general solution is $$B=Ce^{-\beta t}+ \frac{\alpha A_0}{\beta-\alpha}e^{-\alpha t}.$$ Finally, we take care of the initial condition $B(0)=B_0$. Substitute $t=0$ in the general solution. We get $B_0=C+\frac{\alpha A_0}{\beta -\alpha}$, so $$C=B_0-\frac{\alpha A_0}{\beta -\alpha}.$$ Thus $$B=\left(B_0-\frac{\alpha A_0}{\beta -\alpha}\right)e^{-\beta t} +\frac{\alpha A_0}{\beta-\alpha}e^{-\alpha t}.$$ Rearrange parentheses. We get $$B=B_0e^{-\beta t}+\frac{\alpha A_0}{\beta -\alpha}\left(e^{-\alpha t}- e^{-\beta t}\right).$$

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Will I need to use an integrating factor to get the particular solution? – simonthumper Apr 12 '12 at 19:15
If $\alpha=\beta$ need separate treatment, as you can see from the answer given, which implicitly assumes $\alpha\ne \beta$. If $\alpha\ne \beta$, try for a solution of shape $ke^{-\alpha t}$. Should work, tell me if you run into problems. – André Nicolas Apr 12 '12 at 19:21
This is confusing me soooo much! I've only ever done particular solutions and general solutions for second order DE's, for first order I've only ever done using integrating factors, but that doesn't work in this case because you don't know what B is, so you can't use it as an integrating factor! :/ Graghhh I'm soooo frustrated with this question :/ – simonthumper Apr 12 '12 at 19:35
Oooop, just realised I'm being stupid... One second! – simonthumper Apr 12 '12 at 19:39
Ok I've followed your method, however even the way your saying to do it doesn't account for the $\alpha A_0 \frac{e^{−\alpha t}−e^{−\beta t}}{\beta − \alpha}$ because you don't get the $ e^{-\beta t} $ in the numerator... :/ – simonthumper Apr 13 '12 at 11:40
up vote 0 down vote accepted

The easiest way to do this is to times both sides by an integration factor and integrate. If we rearrange the equation slightly we get this:

$ \frac{dB}{dt} + \beta B = \alpha A_0 e^{-\alpha t} $ (The right hand side comes from the first part of the question)

This is then a standard equation in the form $ \frac{dy}{dx} + g(x)y = p(x) $ And can be solved by multiplying both sides by the integrating factor $ u(x) = e^{\int -g(x) dx} $ and then integrating both sides.

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