# differential equations - exponential growth and decay

The population $$P$$ of bacteria in an experiment grows according to the equation $$\frac{dP}{dt}=kP$$, where $$k$$ is a constant and $$t$$ is measured in hours. If the population of bacteria doubles every $$24$$ hours, what is the value of $$k$$?

I was given this problem and I'm not sure what to do with it. I know the formula for this kind of equation is $$ce^{kx}$$. But, how do you plug in the values given?

So you know $$P=ce^{kt}$$. The population doubles in $$24$$ hours, or $$2P=ce^{k(t+24)}$$. Now can you find $$k$$ by dividing the two equations?

As a side note, you might want to know how that formula was obtained.$$\frac{dP}{dt}=kP\implies\frac{dP}P=k~dt$$Now integrate both sides,$$\int_{P(t=0)}^{P(t=t)}\frac{dP}P=k\int_{t=0}^{t=t}dt$$giving you $$P(t)=P(0)e^{kt}$$.

• Can you explain further? I don't understand. – Burt Aug 1 at 21:43
• What part do you not understand? @burt – Shubham Johri Aug 1 at 21:45
• You were clear. I just don't understand how to do the dividing. – Burt Aug 1 at 21:48
• @burt$$\frac{2P}P=\frac{ce^{k(t+24)}}{ce^{kt}}$$ – Shubham Johri Aug 1 at 21:50
• Am I missing something here? I am completely lost. – Burt Aug 1 at 21:51

From

$$\dfrac{dP}{dt} = kP, \tag 1$$

assuming

$$P \ne 0, \tag 2$$

we deduce that

$$\dfrac{1}{P}\dfrac{dP}{dt} = k; \tag 3$$

we integrate 'twixt $$t_0$$ and $$t$$, assuming $$P$$ takes the value $$P(t_0)$$ at $$t = t_0$$:

$$\ln P(t) - \ln P(t_0) = \displaystyle \int_{t_0}^t \dfrac{1}{P(s)}\dfrac{dP(s)}{ds} \; ds = \int_{t_0}^t k \; ds = k(t - t_0), \tag 4$$

or

$$\ln \left (\dfrac{P(t)}{P(t_0)} \right ) = k(t - t_0); \tag 5$$

we apply the function $$\exp(\cdot)$$ to this to obtain

$$\dfrac{P(t)}{P(t_0)} = e^{k(t - t_0)}, \tag 6$$

whence

$$P(t) = P(t_0) e^{k(t - t_0)}. \tag 7$$

Now given that $$P(t)$$ doubles every $$24$$ hours, starting at any $$t_0$$ we take

$$t = t_0 + 24, \tag 8$$

and thus

$$2P(t_0) = P(t_0 + 24) = P(t_0) e^{24k}, \tag 9$$

$$e^{24k} = 2; \tag{10}$$
solving for $$k$$ we conclude
$$k = \dfrac{\ln 2}{24}. \tag{11}$$