Q.The population of a town increases at a rate proportional to its current population. If the population doubles every $25$ years and the present population is $1{,}00{,}000$, find the population after $50$ years.

The population satisfies $\dfrac{dp}{dt}=kp$, a variables-separable equation. Integrating gives $\log p = kt + c$; the conditions fix $c=\log(1,00,000)$ and $k=\dfrac{1}{25}\log 2$. Setting $p=4,00,000$ gives $\log 4 = \dfrac{t}{25}\log 2$, and since $\log 4 = 2\log 2$, we get $t=50$ years.

The rate of growth of population is proportional to the number present, so with $p$ the population at time $t$ (in years):

$\frac{dp}{dt}\propto p \ \Rightarrow\ \frac{dp}{dt}=kp$

where $k$ is the constant of proportionality. Separating the variables:

$\frac{dp}{p}=k\,dt$

Integrating both sides:

$\log p = kt + c \quad \text{...(i)}$

Finding $c$: When $t=0$, $p=1,00,000$. Substituting in (i):

$\log(1,00,000) = k(0) + c \ \Rightarrow\ c = \log(1,00,000)$

Putting this back in (i) and shifting $c$:

$\log p - \log(1,00,000) = kt \ \Rightarrow\ \log\!\left(\frac{p}{1,00,000}\right)=kt \quad \text{...(ii)}$

Finding $k$: The population doubled in $25$ years, so when $t=25$, $p=2,00,000$. From (ii):

$\log\!\left(\frac{2,00,000}{1,00,000}\right)=25k \ \Rightarrow\ \log 2 = 25k \ \Rightarrow\ k=\frac{1}{25}\log 2$

Hence: …

With $\frac{dN}{dt} = kN$, integration gives $\log N = kt + \log N_0$. Doubling in $4$ h gives $k = \frac{1}{4}\log 2$, and at $t = 12$ h, $N = N_0 e^{3\log 2} = 8N_0$, i.e. the bacteria increase $8$ times.

Let $N$ be the number of bacteria at time $t$. Since the rate of increase is proportional to the number present,

$\dfrac{dN}{dt} \propto N \quad\Rightarrow\quad \dfrac{dN}{dt} = kN$

Separating variables and integrating:

$\dfrac{dN}{N} = k\,dt \quad\Rightarrow\quad \int \dfrac{1}{N}\,dN = k\int dt \quad\Rightarrow\quad \log N = kt + C \quad\ldots(1)$

At $t = 0$, $N = N_0$: $\log N_0 = 0 + C \Rightarrow C = \log N_0$.

At $t = 4$, $N = 2N_0$ (doubling):

$\log(2N_0) = 4k + \log N_0 \Rightarrow \log\!\left(\dfrac{2N_0}{N_0}\right) = 4k \Rightarrow \log 2 = 4k$

$\therefore k = \dfrac{1}{4}\log 2 \quad\ldots(2)$

…

Model the growth as $\dfrac{dN}{dt} = kN$, giving $N = N_0 e^{kt}$. Doubling in $4$ hours fixes $k = \dfrac{\log 2}{4}$, and at $t = 12$ we get $N = 8N_0$.

Let $N$ be the number of bacteria at time $t$ hours, with $N_0$ present initially. The rate of increase is proportional to the number present:

$\dfrac{dN}{dt} = kN$, where $k > 0$ is the growth constant.

Separating the variables and integrating,

$\displaystyle \int \dfrac{dN}{N} = \int k\, dt \;\Rightarrow\; \log N = kt + c$.

At $t = 0$, $N = N_0$, so $c = \log N_0$. Hence

$\log N = kt + \log N_0 \;\Rightarrow\; \log\!\left(\dfrac{N}{N_0}\right) = kt \;\Rightarrow\; N = N_0 e^{kt}$.

Use the doubling condition. The number doubles in $4$ hours, i.e. $N = 2N_0$ when $t = 4$:

…