For a research project I am carrying out I am required to solve the system:

$\frac{dp}{dt} = -lp $, $ \frac{dc}{dt} = lp - kc $ with initial conditions $p(0) = p_0 $ and $c(0) = 0 $. Here, $p,c$ denote concentrations and $l,k$ are reaction rates.

I believe that I have solved the system to reach a solution $$ c(t) = (\alpha + \frac{\beta}{l - k})e^{-kt} - \frac{lp_0}{l-k}e^{-lt} $$ where $\alpha, \beta$ are arbitrary constants.

I am quite sure (barring any typos!) that this solves the system however I can't seem to get rid of the two arbitrary constants. I don't know if I need another initial condition or am missing some relation between the two constants?

Since this is a Pharmacokinetics problem, I was able to find a relation between $ \alpha$ and $\beta$ by considering the half life. However, this resulted in something like:

$$ \alpha = \frac{((lp_0)^p - \beta^k)^{1/k}}{l-k} $$

which didn't seem to help.

Any help or feedback would be appreciated.

  • $\begingroup$ Is it known whether $k,l$ are different? $\endgroup$ Nov 24, 2014 at 19:14

1 Answer 1


To get started, observe that from $$\frac{dp}{dt}=-lp,p(0)=p_0,$$ we can show that $$p(t)=p_0e^{-lt}.$$ (I leave this to you.)

Now, we can rewrite $$\frac{dc}{dt}=lp-kc$$ as $$\frac{dc}{dt}+kc=lp=p_0le^{-lt}.$$ Multiplying this equation by the function $e^{kt},$ which is never zero, we obtain the equivalent $$\frac{dc}{dt}e^{kt}+kce^{kt}=p_0le^{(k-l)t},$$ which we can rewrite as $$\frac{dc}{dt}e^{kt}+c\frac{d\left[e^{kt}\right]}{dt}=p_0le^{(k-l)t}.$$ Applying product rule lets us rewrite this as $$\frac{d\left[ce^{kt}\right]}{dt}= p_0le^{(k-l)t},$$ whence we have for some constant $a$ that $$ce^{kt}=a+p_0l\int e^{(k-l)t}\,dt.\tag{$\star$}$$

At this point, there are two distinct possibilities we must address. First of all, we must address the case that $k=l,$ which lets us show that $c(t)=p_0lte^{-lt}.$ (Again, I leave this to you.)

Now, let's suppose that $k\ne l,$ in which case $(\star)$ becomes $$ce^{kt}=a+\frac{p_0l}{k-l}e^{(k-l)t}.$$ Since $c(0)=0,$ it follows that $a=-\frac{p_0l}{k-l},$ and so we can see that $$c(t)=\frac{p_0l}{k-l}\left(e^{-lt}-e^{-kt}\right).$$

Note: I assumed here that $k,l,$ and $p_0$ were constants. If this is not correct, or if you have trouble demonstrating any of the steps I've shown here, please let me know.

  • $\begingroup$ This is excellent. Thanks. Can I ask why you didn't consider a complementary and particular function though? As in, is that method an incorrect way to go about the problem? $\endgroup$
    – Charlie
    Nov 24, 2014 at 21:38
  • $\begingroup$ The kicker, here, is that we have a homogeneous system in disguise! Let $\vec x(t):=\begin{bmatrix}p(t)\\c(t)\end{bmatrix},$ $A:=\begin{bmatrix}-l & 0\\l & -k\end{bmatrix},$ and $\vec x_0=\begin{bmatrix}p_0\\0\end{bmatrix}.$ Then our system can be rewritten as $$\vec x'=A\vec x,\quad\vec x(0)=\vec x_0,$$ which is an initial value problem with a homogeneous ODE...that just happens to be in two dimensions. One can actually solve even non-homogeneous IVPs of this sort directly, but it can be a bit of a bear. $\endgroup$ Nov 24, 2014 at 22:55
  • $\begingroup$ For those not familiar with linear algebra, it is easier to proceed as I did above. One can certainly start by finding the general solution to $p'=-lp,$ whence the specific solution to the IVP $$p'=-lp,\quad p(0)=p_0$$ follows readily, but we ended up doing that, anyway. After that, one can find the general solution to the ODE $c'=-kc,$ then find a particular solution to $c'=lp-kc,$ whence a general solution to $c'=lp-kc$ follows, whence the specific solution to the IVP $$c'=lp-kc,\quad c(0)=0$$ follows. That takes more work, though (it seems to me). $\endgroup$ Nov 24, 2014 at 23:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .