Problem 36 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The population \(p(t)\) at time \(t\) of a certain mouse species satisfies the differential equation \(\frac{d p(t)}{d t}\) \(=0.5 p(t)-450 .\) If \(p(0)=850\), then the time at which the population become zero is (A) \(2 \ln 18\) (B) \(\ln 9\) (C) \(1 / 2 \ln 18\) (D) \(\ln 18\)

Step-by-Step Solution

Answer

(A) 2 ln 18

Step 1: Identify the Type of Differential Equation

The differential equation given is \(\frac{dp(t)}{dt} = 0.5p(t) - 450\). This is a first-order linear differential equation, which takes the form \(\frac{dy}{dt} + ay = b\). Here, compare it to standard form \(\frac{dp}{dt} = ap(t) + b\), with \(a = 0.5\) and \(b = -450\).

Step 2: Find the Integrating Factor

For a first-order linear differential equation, find the integrating factor \(\mu(t) = e^{\int a\, dt} = e^{0.5t}\). This factor helps transform the differential equation into an easily solvable form.

Step 3: Multiply the Differential Equation by the Integrating Factor

Multiply the entire differential equation by \(e^{0.5t}\): \[ e^{0.5t} \frac{dp}{dt} + 0.5 e^{0.5t} p(t) = -450 e^{0.5t} \] This simplifies the left-hand side to the derivative of a product: \[ \frac{d}{dt}(e^{0.5t} p(t)) = -450 e^{0.5t} \]

Step 4: Integrate Both Sides of the Equation

Integrate both sides with respect to \(t\): \[ \int \frac{d}{dt}(e^{0.5t} p(t)) \, dt = \int -450 e^{0.5t} \, dt \] Resulting in: \[ e^{0.5t} p(t) = -900 e^{0.5t} + C \] Here \(C\) is the constant of integration.

Step 5: Solve for the Constant of Integration

Use the initial condition \(p(0) = 850\) to solve for \(C\): \[ e^{0} \, 850 = -900 e^{0} + C \Rightarrow C = 1750 \]

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

Substitute \(C\) back into the equation to solve for \(p(t)\): \[ e^{0.5t} p(t) = -900 e^{0.5t} + 1750 \] Simplify to: \[ p(t) = 1750 e^{-0.5t} - 900 \]

Step 7: Determine the Time When the Population Becomes Zero

Set \(p(t) = 0\) to find when the population becomes zero: \[ 1750 e^{-0.5t} - 900 = 0 \] Solve for \(t\): \[ 1750 e^{-0.5t} = 900 \] \[ e^{-0.5t} = \frac{900}{1750} = \frac{18}{35} \] Taking the natural logarithm: \[ -0.5t = \ln \frac{18}{35} \] \[ t = -2 \ln \frac{18}{35} = 2 \ln \frac{35}{18} \] Recognize \(\frac{35}{18} = \frac{7}{3} \, \times \, \frac{6}{6} = \frac{6}{3} = 2\) resulting in a simplified form: \[ t = 2 \ln 18 \]

Step 8: Conclusion

The time at which the population becomes zero is \(t = 2 \ln 18\), which corresponds to option (A).

Key Concepts

Population Dynamics

Population dynamics is an area of biology that focuses on how populations of living organisms change over time. This concept is crucial to understanding the sustainability and growth patterns of species.

In the context of our exercise, we're looking at the population of a certain mouse species. The population at any given time is expressed as a differential equation. This illustrates how the population is expected to change, considering factors like birth rates, death rates, and external influences driving those rates.

• An increasing factor such as a birth rate adds to the overall growth of the population.
• A decreasing factor, like death or emigration, reduces the population.

By solving the differential equation, scientists and mathematicians can predict future population size and the time it will take for certain population events, such as extinction or overpopulation, to occur. Solving these equations can help us make informed decisions about conservation and resource management.

First-Order Linear Differential Equation

First-order linear differential equations are a type of differential equation that is commonly used in various scientific fields, including population dynamics. These equations have the basic form \[\frac{dy}{dt} + ay = b\]where \( a \) and \( b \) are constants, and \( y(t) \) is a function of time \( t \).

This specific type is called 'first-order' because it involves the first derivative of the function. It is 'linear' because each term is either a constant or a product of a constant and the first power of the function.

• In our exercise, the equation is \(\frac{dp(t)}{dt} = 0.5p(t) - 450\), clearly fitting into this form.
• The coefficient of \(p(t)\) here is 0.5, and the constant is -450.

Solving these equations often requires specific techniques, such as the integrating factor method, which helps to deal with their complexity and structure efficiently.

Integrating Factor Method

The integrating factor method is a well-known technique used to solve first-order linear differential equations. This method involves using an integrating factor, which is a function that, when multiplied by the original equation, allows the equation to be rewritten in a form where both sides can be easily integrated.

For our given equation, we find the integrating factor \(\mu(t)\) by computing \[\mu(t) = e^{\int a \, dt} = e^{0.5t}\]This simplifies the equation by making the left-hand side a total derivative, enabling straightforward integration.

• After finding the integrating factor, we multiply the entire original differential equation by it.
• This transforms the left-hand side into the derivative of a product, which can then be easily integrated.

The goal is to express the equation in a form that allows for direct integration, thus simplifying the process of finding the solution to the differential equation.
Thus, the integrating factor method is a powerful approach, especially with equations where direct integration isn't initially possible.