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Suppose we are given the differential equation $\frac{dP(t)}{dt}=kP$ where $P(t)$ is a function of population with variable time measured in years. And say $k>0$ is the relative growth rate of the population.

The differential equation means that the rate of change of the population with respect to time is proportional to the current population. Or more plainly, the increase of population per year is equal to the value $kP$ at a given time where the population is P. Now, the problem rises. Saying increase per year is equal to $kP$ is actually not accurate. Because at time $t+0.01,$ the increase of population per year is greater than $kP.$ Thus, the actual increase per year is something greater than $kP.$ But these all sounds confusing. Am I wrong somewhere?

I apologize for any unclear sentences in this question.

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I think where you're getting mixed up is just in the semantics. $kP(t)$ is the rate of change, but notice that the rate of change is iteself dependent on the time at which one is observing the population. That's why this is a differential equation, the rate itself is changing with respect to time, and it is doing so in a smooth (differentiable) fashion. Therefore it doesn't make sense to talk about increase in population per year in terms of the quantity $kP(t)$, since this is changing with $t$. If you want to talk about population increase between times $t_1$ and $t_2$ then that quantity would be $P(t_2) - P(t_1)$.

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The solution to the differential equation is $P(t)=P(0)\exp(kt)$, where $P(0)$ is the starting population. After one year, the population has been multiplied by $e^k$. When $k \ll 1, e^k \approx 1+k$ and the fractional increase is $k$, but when $k$ is larger the exponential starts to matter more.

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