Q.The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1,00,000, when will the city have population 4,00,000?
Let ‘p’ be the population at time ‘t’ years.
∴ $\dfrac{dp}{dt} \propto p$
∴ Differential equation can be written as $\dfrac{dp}{dt} = kp$
where k is constant of proportionality.
∴ $\dfrac{dp}{p} = k\,dt$
On integrating we get
$\square = kt + c$ ...(i)

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 $t = 4$:

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