Take for instance dP/dt = kP

We get after separating: dP/P = kdt, but why shouldn't it be

dP/kP = dt instead, mathematically it doesn't make sense to say that k must belong absolutely to the right hand side of the equation.


1 Answer 1


Who said $k$ has to be on the right hand side? It is a constant, so it can be on either side. The important point is that anything in terms of $P$ is on the side with $dP$ and everything in terms of $t$ is on the side with $dt$. Since a constant is independent of $P$ and $t$, it may go on either side.

Let's solve $\dfrac{dP}{dt} = kP$ both ways to illustrate why it doesn't matter.

  1. First we separate the equation as $$\frac{dP}{P} = k \, dt.$$ Integrating both sides, we get $$\ln |P| = kt + C.$$ Exponentiating both sides, we get the solution $$P = e^{kt + C} = Ae^{kt},$$ where $A$ is an arbitrary constant.

  2. Now separate the equation as $$\frac{dP}{kP} = dt.$$ Integrating both sides, we get $$\frac{1}{k} \ln |P| = t + C.$$ Multiplying both side by $k$ results in $$\ln |P| = kt + C',$$ where $C' = kC$. Since $C$ is an arbitrary constant, this is an irrelevant detail. Now exponentiating both sides gives $$P = e^{kt + C'} = Ae^{kt},$$ where $A$ is an arbitrary constant.

So both methods give the same general solution. The reason for this is that we can factor out a constant from an integral, so the side that the constant is on doesn't affect the integration.

The reason every book will leave the $k$ on the right-hand side for this problem is because we want a solution for $P$; we'll just have to move $k$ back to the right-hand side after integrating if we start with it on the left-hand side with $P$.

  • $\begingroup$ but for Newton's formula to work it must be on the left hand side. $\endgroup$ May 14, 2013 at 23:38
  • $\begingroup$ not only Newton's formula, but many other formulas we use in engineering. $\endgroup$ May 14, 2013 at 23:38
  • $\begingroup$ "k" doesn't "belong" on either side. The question is simply where is it most useful to put it? If Newton's method works with "k" on the LHS, then put it there. If some other method works with "k" on the RHS, put it there. $\endgroup$ May 14, 2013 at 23:40
  • $\begingroup$ Can you be more specific? Which "Newton's formula" are you referring to? $\endgroup$
    – Glen O
    May 14, 2013 at 23:42
  • 1
    $\begingroup$ @GladstoneAsder it's not that it "must be"; the equation will come out exactly the same no matter what side $k$ is on. A lot of equations have a standard form that they're always written in, even though others are still correct. I've never seen $\pm\sqrt{\left(\frac b{2a}\right)^2-\frac ca}-\frac{b}{2a}$ written in any textbook, but it's still the quadratic formula. $\endgroup$ May 14, 2013 at 23:53

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