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.
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Sign up to join this communityWho 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.
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.
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 $P$.