Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.2

Question 1. Express each of the following physical statements in the form of differential equation.
(i) Radium decays at a rate proportional to the amount Q present.
(ii) The population P of a city increases at a rate propotional to the product of population and to the difference between 5,00,000 and the population.
(iii) For a certain substance, the rate of change of vapor pressure P with respect proportional to the vapor pressure and inversely proportional to the square of the temperature.
(iv) A saving amount pays 8% interest per year compound continuously. In addition, the income from another investment is credited to the amount continuously at the rate of Rs 400 per year.
Answer:
(i) Let \( Q \) be the amount of Radium present at any time \( t \). The rate at which \( Q \) decreases is given by \( \frac{dQ}{dt} \). Since it is decreasing, a negative sign is used.
This rate of decrease is directly proportional to the amount \( Q \) itself, meaning the more radium there is, the faster it decays.
So, we can write the relationship as \( \frac{dQ}{dt} = -kQ \), where \( k \) is a positive constant.
Rearranging this, the differential equation is \( \frac{dQ}{dt} + kQ = 0 \).
(ii) Let \( P \) be the population of the city at time \( t \). The rate of change of population is \( \frac{dP}{dt} \).
This rate is proportional to two factors: the current population \( P \) and the difference between the maximum population (5,00,000) and the current population \( (5,00,000 - P) \). This accounts for a limited growth.
So, the differential equation representing this is \( \frac{dP}{dt} = kP(5,00,000 - P) \), where \( k \) is a constant of proportionality.
(iii) Let \( P \) be the vapor pressure and \( T \) be the temperature. The rate of change of vapor pressure with respect to time \( t \) is \( \frac{dP}{dt} \).
This rate is directly proportional to the vapor pressure \( P \). It is also inversely proportional to the square of the temperature \( T \), meaning higher temperatures reduce this rate significantly.
Combining these proportionalities, we get \( \frac{dP}{dt} = \frac{kP}{T^2} \), where \( k \) is a constant.
If we let \( \frac{k}{T^2} = \alpha \), then the equation becomes \( \frac{dP}{dt} = \alpha P \).
\( \implies \frac{dP}{dt} - \alpha P = 0 \)
This is the required differential equation.
(iv) Let \( x \) be the total saving amount at time \( t \). The account earns 8% interest compounded continuously, which adds \( \frac{8}{100}x \) to the amount per year.
In addition, Rs 400 is continuously credited to the amount each year from another investment, which is a constant addition.
The total rate of change of the amount \( x \) with respect to time \( t \) is the sum of these two contributions.
So, the differential equation is \( \frac{dx}{dt} = \frac{8}{100}x + 400 \).
Simplifying the fraction, we get \( \frac{dx}{dt} = \frac{2x}{25} + 400 \).
Rearranging, the required differential equation is \( \frac{dx}{dt} - \frac{2}{25}x - 400 = 0 \).
In simple words: We translate each physical statement into a mathematical equation involving rates of change. "Proportional" means multiplying by a constant, "inversely proportional" means dividing, and "decay" or "decrease" implies a negative sign. Each part describes how a quantity changes over time.

🎯 Exam Tip: Pay close attention to keywords like "proportional to," "inversely proportional to," "rate of change," and "decreases" or "increases" to correctly form the terms and signs in the differential equation.

Question 2. Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
Answer: Let \( r \) be the radius of the spherical rain drop at time \( t \).
The volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \).
The surface area of a sphere is given by \( S = 4\pi r^2 \).
The problem states that the rate of evaporation, which is the rate of change of volume \( \frac{dV}{dt} \), is proportional to its surface area \( S \).
Since the raindrop is evaporating, its volume is decreasing, so we introduce a negative constant \( k \).
Thus, we have \( \frac{dV}{dt} = -kS \).
We can also express \( \frac{dV}{dt} \) using the chain rule with respect to \( r \):
\( \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = 4\pi r^2 \frac{dr}{dt} \). This shows how the volume change is linked to the radius change.
Now, substitute the expressions for \( S \) and \( \frac{dV}{dt} \) into the proportional equation:
\( 4\pi r^2 \frac{dr}{dt} = -k (4\pi r^2) \)
Assuming \( r \ne 0 \) (since it's a raindrop), we can divide both sides by \( 4\pi r^2 \):
\( \frac{dr}{dt} = -k \)
This is the required differential equation, showing that the radius decreases at a constant rate.
In simple words: When a round raindrop evaporates, its size shrinks. The speed at which its radius gets smaller is constant, meaning it shrinks at the same rate regardless of how big or small it is.

🎯 Exam Tip: For problems involving geometric shapes, always start by listing the relevant formulas for volume and surface area, then apply the chain rule to relate the rates of change.