Chapter 4: Problem 4

Consider a bacterial population whose growth rate is $d \mathbf{N} / d t=K(t) N$. Show that $$ N(t)=N_{0} \exp \left(\int_{0}^{t} K d s\right). $$

Short Answer

Expert verified

The solution is $N(t) = N_0 \exp \left( \int_{0}^{t} K(s) \, ds \right)$.

Step by step solution

Understand the Differential Equation

We start with the differential equation describing the growth of a bacterial population: $ \frac{d \mathbf{N}}{d t} = K(t) N $ where $\mathbf{N}$ is the population and $K(t)$ is the growth rate at time $t$.

Separate the Variables

Separate the variables $N$ and $t$ to simplify the equation. $ \frac{d \mathbf{N}}{N} = K(t) \ d t $

Integrate Both Sides

Integrate both sides to find the solution for $N$. $ \int \frac{1}{N} \, dN = \int K(t) \, dt $

Solve the Integrals

The integral of $\frac{1}{N}$ with respect to $N$ is $\ln(N)$, and the integral of $K(t)$ with respect to $t$ is simply the integral of the function $K(t)$. \[ \ln(N) = \int_{0}^{t} K(s) \, ds + C \] where $C$ is the constant of integration.

Determine the Constant

To find the constant $C$, use the initial condition. At $t = 0$, assume $N = N_0$: \[ \ln(N_0) = C \] So $C = \ln(N_0)$.

Simplify the Expression

Substituting $C = \ln(N_0)$ into the equation: \[ \ln(N) = \int_{0}^{t} K(s) \, ds + \ln(N_0) \] Exponentiate both sides to solve for $N$: \[ N = N_0 \exp \left( \int_{0}^{t} K(s) \, ds \right) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

The given problem deals with a differential equation, which is a mathematical equation involving an unknown function and its derivatives. Here, the unknown function is the bacterial population size, represented by $ N(t) $.
The equation to start with is: $\frac{dN}{dt} = K(t)N$.
This equation suggests the rate of change of the population size over time depends on the current population size and a time-dependent growth rate, $K(t)$.
Differential equations are widely used in various fields such as physics, engineering, and biology to model behaviors and phenomena.

Separation of Variables

To solve the differential equation, we use a technique called 'separation of variables'. This method involves rearranging the equation so that all terms involving the dependent variable (\Thunkwards tickled{N}👟

Consider a bacterial population whose growth rate is \(d \mathbf{N} / d t=K(t) N\). Show that $$ N(t)=N_{0} \exp \left(\int_{0}^{t} K d s\right). $$