Q.Case Study - 2 A bacteria sample of certain number of bacteria is observed to grow exponentially in a given amount of time. Using exponential growth model, the rate of growth of this sample of bacteria is calculated. The differential equation representing the growth of bacteria is given as $\dfrac{dP}{dt} = kP$, where P is the population of bacteria at any time $t$. Based on the above information, answer the following questions:

The core idea here is solving a first-order separable differential equation to model exponential growth, then using given conditions to find specific constants.

(i) General solution of the differential equation

The differential equation $\frac{dP}{dt} = kP$ describes a rate of change proportional to the current quantity, which is characteristic of exponential growth. To find the general solution, we separate variables and integrate.

1. Concept: Separation of Variables.
2. Reasoning:
   • Rearrange the equation to group $P$ terms with $dP$ and $t$ terms with $dt$: $\frac{dP}{P} = k dt$.
   • Integrate both sides: $\int \frac{dP}{P} = \int k dt$.
   • This yields $\ln|P| = kt + C_1$, where $C_1$ is the integration constant.
   • Exponentiate both sides to solve for $P$: $P = e^{kt+C_1} = e^{C_1}e^{kt}$. Let $A = e^{C_1}$ (since $e^{C_1}$ is an arbitrary positive constant, we can write $A$ as any non-zero constant, absorbing the absolute value).

(ii) Value of $k$

We use the given initial conditions to determine the constants in our general solution.

1. Concept: Using initial conditions to determine constants.
2. Reasoning: …

The problem describes a bacteria sample growing exponentially, modeled by the differential equation $\frac{dP}{dt} = kP$. We need to find the general solution to this equation and then use given population data to determine the specific growth constant $k$.

The core idea behind exponential growth is that the rate at which a quantity grows is directly proportional to the quantity itself. This means the more there is, the faster it grows. This phenomenon is observed in many natural processes, like population growth (bacteria, humans under ideal conditions), compound interest, and radioactive decay (though decay is negative growth).

Mathematically, this relationship is expressed as a differential equation: $\frac{dP}{dt} = kP$.

Here, $\frac{dP}{dt}$ represents the instantaneous rate of change of the population $P$ with respect to time $t$. The constant $k$ is the proportionality constant, often called the growth rate constant. If $k > 0$, it's growth; if $k < 0$, it's decay.

To "solve" this differential equation means to find a function $P(t)$ that satisfies this relationship. We expect an exponential function because only an exponential function has a derivative that is proportional to itself (e.g., $\frac{d}{dt}(e^{kt}) = k e^{kt}$).

Part (i): Obtain the general solution of the given differential equation and express it as an exponential function of $t$.

1. Identify the type of differential equation: The given equation is $\frac{dP}{dt} = kP$. This is a first-order, linear, homogeneous differential equation. More importantly for solving, it is a separable differential equation, meaning we can rearrange it to have all terms involving $P$ on one side and all terms involving $t$ on the other.

2. Separate the variables: To integrate, we need to isolate $P$ terms with $dP$ and $t$ terms with $dt$.

   Divide both sides by $P$ (assuming $P \neq 0$) and multiply by $dt$:

> [!WARNING]
> When separating variables by dividing by $P$, we implicitly assume $P \neq 0$. If $P=0$, then $\frac{dP}{dt}=0$, which means $P(t)=0$ is a valid (trivial) solution. Our general solution will encompass this case if we allow the constant of integration to be zero.

3. Integrate both sides: Now, integrate both sides of the separated equation.

The integral of $\frac{1}{P}$ with respect to $P$ is $\ln|P|$, and the integral of $k$ with respect to $t$ is $kt$. We must add a constant of integration, say $C_1$, on one side (conventionally, the side with the independent variable).

4. Solve for $P$ by exponentiating: To isolate $P$, we raise $e$ to the power of both sides of the equation.

Using the properties $e^{\ln x} = x$ and $e^{a+b} = e^a e^b$:

Since $P$ represents a population, it must be non-negative. We can remove the absolute value by introducing a new constant $A = \pm e^{C_1}$. Since $e^{C_1}$ is always positive, $A$ can be any non-zero real number. For population growth, $P$ is typically positive, so $A$ will be positive.
Let $A = e^{C_1}$ (assuming $P > 0$).
$$ P(t) = A e^{kt} $$ …