Problem 116 - A Complete Resource Book in Mathematics for JEE Main

Question

Let the population of rabbits surviving at a time \(t\) be governed by the differential equation \(\frac{d p(t)}{d t}=\frac{1}{2} p(t)-200 .\) If initially \(p(0)=100\), then \(p(t)\) equals [2014] (A) \(400-300 e^{[/ 2}\) (B) \(300-200 e^{-t / 2}\) (C) \(600-500 e^{t / 2}\) (D) \(400-300 e^{-t / 2}\)

Step-by-Step Solution

Answer

The answer is (D) \(400 - 300 e^{-t/2}\).

Step 1: Identify the Type of Differential Equation

The differential equation \( \frac{dp(t)}{dt} = \frac{1}{2} p(t) - 200 \) is a first-order linear differential equation.

Step 2: Arrange in Standard Form

We can rewrite the equation as \( \frac{dp(t)}{dt} - \frac{1}{2} p(t) = -200 \), which is in the form \( \frac{dp}{dt} + P(t)p = Q(t) \) with \(P(t)= -\frac{1}{2}\) and \(Q(t) = -200\).

Step 3: Find the Integrating Factor

The integrating factor is given by \( e^{\int P(t)\, dt} = e^{\int -\frac{1}{2}\, dt} = e^{-\frac{t}{2}} \).

Step 4: Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor: \(e^{-\frac{t}{2}} \left(\frac{dp(t)}{dt} - \frac{1}{2}p(t)\right) = -200e^{-\frac{t}{2}} \).

Step 5: Integrate Both Sides

The left-hand side becomes the derivative of \( p(t)e^{-\frac{t}{2}} \), so integrate: \( \int d(p(t) e^{-\frac{t}{2}}) = \int -200e^{-\frac{t}{2}}\, dt \). This can be simplified to: \( p(t)e^{-\frac{t}{2}} = 400e^{-\frac{t}{2}} + C \).

Step 6: Solve for \\(p(t)\\)

Multiply both sides by \(e^{\frac{t}{2}}\) to solve for \(p(t)\): \( p(t) = 400 + Ce^{\frac{t}{2}} \).

Step 7: Apply Initial Condition to Find C

Use the initial condition \( p(0) = 100 \). Plug this into the equation: \( 100 = 400 + C \cdot 1 \). Solving gives \( C = -300 \).

Step 8: Substitute C Back

Substitute \( C = -300 \) back into the expression for \( p(t) \): \( p(t) = 400 - 300e^{-\frac{t}{2}} \).

Step 9: Compare with Given Options

The finalized expression \( p(t) = 400 - 300 e^{-t / 2} \) matches option (D).

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is an equation that expresses the rate of change of a function in terms of the function itself and time. It is characterized by the presence of the first derivative of a function. The general form of a first-order linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( P(t) \) and \( Q(t) \) are functions of time \( t \).
In this concept, we focus on the problem involving rabbit population growth. The differential equation \( \frac{dp(t)}{dt} = \frac{1}{2} p(t) - 200 \) fits the form above, where \( P(t) = -\frac{1}{2} \) and \( Q(t) = -200 \). The goal is to find the specific solution to this equation given the initial population condition.

Integrating Factor Method

The Integrating Factor Method is a powerful technique used to solve first-order linear differential equations. It involves multiplying both sides of the differential equation by a specially chosen function called the integrating factor.
The integrating factor is determined by the expression \( e^{\int P(t) \ dt} \). In the rabbit population growth example, since \( P(t) = -\frac{1}{2} \), the integrating factor is \( e^{-t/2} \).
By multiplying through the entire differential equation by this integrating factor, the left-hand side of the equation becomes an easily integrable expression. This allows us to solve for the unknown function. Thus, the integrating factor transforms our equation into a form that simplifies direct integration.

Initial Condition Application

Applying the initial condition in solving differential equations helps to find a particular solution. It provides additional information that determines a constant in the general solution.
In our problem, we apply the initial condition \( p(0) = 100 \), which states that the rabbit population at time \( t=0 \) is 100. This piece of information is crucial because it allows us to find the constant \( C \) in the solution derived from the Integrating Factor Method.
Substituting \( t = 0 \) and \( p(0) = 100 \) into the equation \( p(t) = 400 + Ce^{-t/2} \), we solve for \( C \), which helps in accurately defining the expression for \( p(t) \) that predicts the population over time.

Population Growth Modeling

Population growth modeling is an essential application of differential equations. It helps predict how populations will change over time based on known variables.
In the context of our example, the differential equation represents how a population of rabbits changes with respect to time. The term \( \frac{1}{2} p(t) \) reflects the natural growth rate, while the constant \(-200\) might imply a steady state or limiting factor that decreases the growth, like resource shortage or predation.
The end solution provides a mathematical model \( p(t) = 400 - 300e^{-t/2} \), which describes not only the growth trend but also stabilization at a long-term population. Understanding such models helps ecologists and planners in making informed decisions about wildlife management and conservation efforts.