Exercise 10.2 - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $10.2$
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.
Solution:
Radium decays at a rate proportional to the amount $Q$ present.
$
\begin{aligned}
& \frac{d \mathrm{Q}}{d t} \propto \mathrm{Q} \\
& \frac{d \mathrm{Q}}{d t}=k \mathrm{Q}[\because k \text { decay constant }]
\end{aligned}
$
(ii) The population $\mathrm{P}$ of a city increases at a rate proportional to the product of population and to the difference between $5,00,000$ and the population.
Solution:
Rate of change of $\mathrm{P}$ with respect to ' $\mathrm{t}$ ' is $\frac{d \mathrm{P}}{d t}$
Product of population and the difference between 50,000 and the population is $\mathrm{P}(50,000-\mathrm{P})$
Given $\frac{d \mathrm{P}}{d t} \propto \mathrm{P}(50000-\mathrm{P})$
$\therefore \quad \frac{d \mathrm{P}}{d t}=k \mathrm{P}(50000-\mathrm{P}) \quad[\because k$ is a constant $]$
(iii) For a certain substance, the rate of change of vapor pressure $P$ with respect to temperature $T$ is proportional to the vapor pressure and inversely proportional to the square of the temperature.
Solution:
Rate of change of $\mathrm{P}$ with respect to ${ }^{\prime} t^{\prime}$ is $\frac{d \mathrm{P}}{d t}$
Given, $\frac{d P}{d t} \propto \frac{P}{T^2}$
$\therefore \quad \frac{d \mathrm{P}}{d t}=k \frac{\mathrm{P}}{\mathrm{T}^2} \quad[\because k$ is a constant $]$

(iv) A saving amount pays $8 \%$ interest per year, compounded continuously. In addition, the income from another investment is credited to the amount continuously at the rate of ₹ 400 per year.
Solution:
Let ' $\mathrm{x}$ ' be the amount invested.
Interest $=8 \%$ of $x$
$
=\frac{8}{100} x=\frac{2}{25} x
$
$\therefore$ The required equation is $\frac{d x}{d t}=\frac{2 x}{25}+400$.
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.
Solution:
Volume of sphere $\begin{aligned} \mathrm{V} & =\frac{4}{3} \pi r^3 \\ \frac{d \mathrm{~V}}{d t} & =4 \pi r^2 \cdot \frac{d r}{d t}\end{aligned}$
Surface area $=4 \pi r^2$
Given, $\frac{d \mathrm{~V}}{d t} \propto(\mathrm{S} . \mathrm{A})$
$
\begin{aligned}
\frac{d \mathrm{~V}}{d t} & =-k(\mathrm{~S} . \mathrm{A}) \quad[k-\mathrm{constant}] \\
4 \pi r^2 \cdot \frac{d r}{d t} & =-k\left(4 \pi r^2\right) \\
\frac{d r}{d t} & =-k
\end{aligned}
$