Exercise 10.8 - 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 $10.8$
Question 1.
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours?
Solution:
Let ' $\mathrm{P}$ ' be the number of bacteria present at time ' $\mathrm{t}$ '
Given $\frac{d \mathbf{P}}{d t} \propto \mathbf{P}$
$
\begin{gathered}
\Rightarrow \frac{d \mathrm{P}}{d t}=k \mathrm{P} \\
\quad \int \frac{d \mathrm{P}}{\mathrm{P}}=k \int d t \\
\log \mathrm{P}=\mathrm{kt}+\mathrm{c} \\
\mathrm{P}=\mathrm{e}^{\mathrm{kt}+\mathrm{c}} \\
\mathrm{P}=\mathrm{e}^{\mathrm{kt} \mathrm{c}} \ldots(1)
\end{gathered}
$
Initially, when $\mathrm{t}=0$, let $\mathrm{P}=\mathrm{P}_0$
$
\mathrm{P}_0=\mathrm{e}^{\circ} \mathrm{c} \Rightarrow \mathrm{c}=\mathrm{P}_0
$
Given $t=5, P=3 P_0$
Substituting in (1)
$
3 \mathrm{P}_0=\mathrm{e}^{5 \mathrm{k}} \cdot \mathrm{P}_0 \Rightarrow \mathrm{e}^{5 \mathrm{k}}=3
$
Again, when $\mathrm{t}=10$
$
\begin{aligned}
& \mathrm{P}=\mathrm{e}^{10 \mathrm{k}} \cdot \mathrm{c} \\
& \mathrm{P}=\left(\mathrm{e}^{5 \mathrm{k}}\right)^2 \cdot \mathrm{P}_0=(3)^2 \mathrm{P}_0 \\
& \mathrm{P}=9 \mathrm{P}_0
\end{aligned}
$
$[\because k$ is constant]
After 10 hours, the number of bacteria is 9 times the original bacteria present.
Question 2.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from $3,00,000$ to $4,00,000$.
Solution:
Let the population of a city be ' $P$ ' at time ' $t$ '
Given $\frac{d \mathrm{P}}{d t} \propto \mathrm{P}$
$
\begin{aligned}
\Rightarrow \frac{d \mathrm{P}}{d t} & =k \mathrm{P} \\
\int \frac{d \mathrm{P}}{\mathrm{P}} & =\int k d t \\
\log \mathrm{P} & =k t+c \\
\mathrm{P} & =e^{k t} c
\end{aligned}
$
Given, when $t=0, \mathrm{P}=3,00,000$
Substituting in (1)
$
\begin{aligned}
c & =3,00,000 \\
\therefore(1) \Rightarrow \mathrm{P} & =300000 e^{k t}
\end{aligned}
$
Again, given when $t=40, \mathrm{P}=4,00,000$
$
\begin{aligned}
& \therefore 400000=300000 e^{40 k} \\
& e^{40 k}=\frac{4}{3} \\
& 40 k=\log \frac{4}{3} \\
& k=\frac{1}{40} \log \frac{4}{3}=\log \left(\frac{4}{3}\right)^{\frac{1}{40}}
\end{aligned}
$
$\therefore$ The population is $\mathrm{P}=300000 e^{\log \left(\frac{4}{3}\right)^{1 / 40}}$
$
P=300000\left(\frac{4}{3}\right)^{\frac{1}{40}}
$

Question 3.
The equation of electromotive force for an electric circuit containing resistance and self inductance is $\mathrm{E}=\mathrm{Ri}+\mathrm{L} \frac{d i}{d t}$, where $\mathrm{E}$ is the electromotive force is given to the circuit, $\mathrm{R}$ the resistance and $\mathrm{L}$, the coefficient of induction. Find the current $i$ at time $t$ when $E=0$.

Solution:
Given equation of electromotive force is
$
\mathrm{E}=\mathrm{R} i+\mathrm{L} \frac{d i}{d t}
$
Let $\mathrm{E}=\mathrm{E}_{\mathrm{o}}$
$
\begin{aligned}
\therefore \text { Equation is } \mathrm{L} \frac{d i}{d t}+\mathrm{R} i & =\mathrm{E}_{\mathrm{o}} \\
\frac{d i}{d t}+\frac{\mathrm{R} i}{\mathrm{~L}} & =\frac{\mathrm{E}_o}{\mathrm{~L}}
\end{aligned}
$
This is a Linear differential equation
$
\begin{array}{r}
\mathrm{P}(t)=\frac{\mathrm{R}}{\mathrm{L}} \text { and } \mathrm{Q}(t)=\frac{\mathrm{E}_o}{\mathrm{~L}} \\
\int \mathrm{P}(t) d t=\frac{\mathrm{R}}{\mathrm{L}} \int d t=\frac{\mathrm{R} t}{\mathrm{~L}} \\
\mathrm{I} . \mathrm{F}=e^{\int \mathrm{P}(t) d t}=e^{\mathrm{R} / / \mathrm{L}}
\end{array}
$
Now, solution is $i e^{\mathrm{R} t / \mathrm{L}}=\frac{\mathrm{E}_o}{\mathrm{~L}} \int e^{\mathrm{R} \mathrm{t} / \mathrm{L}} d t+c$
$
\begin{aligned}
& i e^{\mathrm{Rt} / \mathrm{L}}=\frac{\mathrm{E}_o}{\mathrm{~L}} \cdot \frac{e^{\mathrm{R} t / \mathrm{L}}}{\mathrm{R} / \mathrm{L}}+c \\
& i e^{\mathrm{Rt} / \mathrm{L}}=\frac{\mathrm{E}_o}{\mathrm{R}} e^{\mathrm{Rt} / \mathrm{L}+c}
\end{aligned}
$

Given $t=0$ and $i=0$
$
0=\frac{\mathrm{E}_o}{\mathrm{R}} e^0+c \Rightarrow c=-\frac{\mathrm{E}_o}{\mathrm{R}}
$
Substituting in (1)
$
\begin{aligned}
i & =\frac{\mathrm{E}_o}{\mathrm{R}} e^{\mathrm{R} t / \mathrm{L}}-\frac{\mathrm{E}_o}{\mathrm{R}} \\
i & =\frac{\mathrm{E}_o}{\mathrm{R}}\left(e^{\mathrm{R} t / \mathrm{L}}-1\right)=c\left(e^{\mathrm{R} t / \mathrm{L}}-1\right)
\end{aligned}
$
Question 4.
The engine of a motor boat moving at $10 \mathrm{~m} / \mathrm{s}$ is shut off. Given that the retardation at any
subsequent time (after shutting off the engine) equal to the velocity at that time. Find the velocity after 2 seconds of switching off the engine.
Solution:
Given, the retardation experienced by a moving motor boat after its engine is shut off is given by
$
\begin{aligned}
\frac{d v}{d t} & =v \\
\int \frac{d v}{v} & =\int d t \\
\log v & =t+c \\
v & =e^t \cdot c
\end{aligned}
$
Given, $v=10$, when $t=0$
$
\text { Now (1) } \begin{aligned}
\therefore c & =10 \\
v & =10 e^t
\end{aligned}
$
Again, given $t=2 \Rightarrow v=10 \mathrm{e}^2$

Question 5.
Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of $5 \%$ per annum compounded continuously. How much money will be in his bank account 18 months later?
Solution:
Let $\mathrm{P}$ be the principal
Rate of interest $5 \%$
$
\begin{aligned}
\Rightarrow \frac{d \mathrm{P}}{d t} & =\frac{5}{100} \mathrm{P} \\
\int \frac{d \mathrm{P}}{\mathrm{P}} & =0.05 \int d t \\
\log \mathrm{P} & =0.05 t+c \\
\mathrm{P} & =e^{0.05 \mathrm{t}} \cdot c
\end{aligned}
$
Given, when $\mathrm{t}=0, \mathrm{P}=10000$

Question 6.
Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample $10 \%$ of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of the original radioactive nuclei will remain after 1000 years?
Solution:
Let $x(t)$ denote the number of nuclei present in a sample at anytime ' $t$ '
Given $\frac{d x}{d t}=k x \quad[\because k$ is a negative constant $]$
$
\begin{aligned}
& \int \frac{d x}{x}=\int k d t \\
& \log x=k t+c
\end{aligned}
$
Let $x_0$ denote the number of nuclei present in the given sample at time $t=0$
$\therefore$ from (1) $\Rightarrow \mathrm{c}=\log \mathrm{k}_0$
$\therefore \mathrm{x}=\mathrm{x}_0 \mathrm{e}^{\mathrm{kt}}$
Given that when $t=100$ years, $x=x_{\mathrm{o}}-\frac{10}{100} x_{\mathrm{o}}=\frac{9}{10} x_{\mathrm{o}}$
$\therefore$ Required percentage after 1000 years
$
\begin{aligned}
& \begin{aligned}
=\frac{x^{\prime}}{x_0} \times 100 & =\frac{x_0 e^{1000 k}}{x_{\mathrm{o}}} \times 100 \\
& =\left(e^{100 k}\right)^{10} \times 100 \quad\left[\because e^{100 k}=\frac{9}{10}\right] \\
& =\left(\frac{9}{10}\right)^{10} \times 100=\frac{9^{10}}{10^8}
\end{aligned} \\
& \therefore \frac{9^{10}}{10^8} \% \quad \text { of radioactive element will remain after } 1000 \text { years. }
\end{aligned}
$

Question 7.
Water at temperature $100^{\circ} \mathrm{C}$ cools in 10 minutes to $80^{\circ} \mathrm{C}$ in a room temperature of $25^{\circ} \mathrm{C}$. Find
(i) The temperature of water after 20 minutes
(ii) The time when the temperature is $40^{\circ} \mathrm{C}$
$
\left[\log _e \frac{11}{15}=-0.3101 ; \log _e 5=1.6094\right]
$
Solution:

Let ' $\mathrm{T}$ ' be the temperature in time ' $t$ '
$
\begin{aligned}
\text { Given } \frac{d \mathrm{~T}}{d t} & \propto(\mathrm{T}-25) \\
\Rightarrow \frac{d \mathrm{~T}}{d t} & =-k(\mathrm{~T}-25), \quad k>0 \\
\int \frac{d \mathrm{~T}}{\mathrm{~T}-25} & =-k \int d t \\
\log (\mathrm{T}-25) & =-k t+c
\end{aligned}
$
Given when $t=0, \mathrm{~T}=100^{\circ} \mathrm{C}$
$
\begin{aligned}
\Rightarrow c & =\log 75 \\
(1) \Rightarrow \log (\mathrm{T}-25) & =-k t+\log 75 \\
\log \left(\frac{\mathrm{T}-25}{75}\right) & =-k t \\
k t & =\log \left(\frac{75}{\mathrm{~T}-25}\right)
\end{aligned}
$
Again, given $t=10, \mathrm{~T}=80$
$
\begin{aligned}
\Rightarrow 10 k & =\log \left(\frac{75}{55}\right) \\
10 k & =\log \left(\frac{15}{11}\right) \\
\therefore k & =\frac{1}{10} \log \left(\frac{15}{11}\right)=\frac{1}{10}(0.3101) \\
k & =0.03101 \\
\therefore 0.03101 t & =\log \left(\frac{75}{\mathrm{~T}-25}\right)
\end{aligned}
$

(i) When $t=20, \mathrm{~T}=$ ?
$
\begin{aligned}
0.6202 & =\log \left(\frac{75}{\mathrm{~T}-25}\right) \\
\left(\frac{75}{\mathrm{~T}-25}\right) & =e^{0.6202}=1.8593
\end{aligned}
$
$
\begin{aligned}
\mathrm{T}-25 & =40.38 \\
\mathrm{~T} & =65.38^{\circ} \mathrm{C}
\end{aligned}
$
(ii) When $\mathrm{T}=48, t=$ ?
$
\begin{aligned}
\Rightarrow 0.03101 t & =\log \left(\frac{75}{15}\right) \\
0.03101 t & =\log 5 \\
0.03101 t & =1.6094 \\
t & =\frac{1.6094}{0.03101}=52 \text { minutes. }
\end{aligned}
$

(i) When $t=20, \mathrm{~T}=$ ?
$
\begin{aligned}
0.6202 & =\log \left(\frac{75}{\mathrm{~T}-25}\right) \\
\left(\frac{75}{\mathrm{~T}-25}\right) & =e^{0.6202}=1.8593
\end{aligned}
$
$
\begin{aligned}
\mathrm{T}-25 & =40.38 \\
\mathrm{~T} & =65.38^{\circ} \mathrm{C}
\end{aligned}
$

(ii) When $\mathrm{T}=48, t=$ ?
$
\begin{aligned}
\Rightarrow 0.03101 t & =\log \left(\frac{75}{15}\right) \\
0.03101 t & =\log 5 \\
0.03101 t & =1.6094 \\
t & =\frac{1.6094}{0.03101}=52 \text { minutes. }
\end{aligned}
$

Question 8.
At $10.00$ A.M. a woman took a cup of hot instant coffee from her microwave oven placed it on a nearby Kitchen counter to cool. At this instant the temperature of the coffee was $180^{\circ} \mathrm{F}$, and 10 minutes later it was $160^{\circ} \mathrm{F}$. Assume that constant temperature of the kitchen was $70^{\circ} \mathrm{F}$.
(i) What was the temperature of the coffee at $10.15$ A.M.?
(ii) The woman likes to drink coffee when its temperature is between $130^{\circ} \mathrm{F}$ and $140^{\circ} \mathrm{F}$. between what times should she have drunk the coffee?
Solution:
Change in Temperature is proportional to the difference in temperature (Newton's Cooling Law)
$
\begin{aligned}
\therefore \frac{d \mathrm{~T}}{d t} & \propto(70-\mathrm{T}) \\
\frac{d \mathrm{~T}}{d t} & =k(70-\mathrm{T}) \\
\int \frac{d \mathrm{~T}}{70-\mathrm{T}} & =k \int d t \\
-\log (70-\mathrm{T}) & =k t+c \\
\therefore \mathrm{T} & =70+c e^{-k t} \\
\text { Given } \mathrm{T}(0) & =180 \Rightarrow 180=70+c \\
c & =180-70=110 \text { and } \mathrm{T}(10)=160 \\
\therefore 160 & =70+110 . e^{-10 k} \\
e^{-10 k} & =\frac{160-70}{110} \\
e^{-10 k} & =\frac{90}{110}=\frac{9}{11} \\
-10 k & =\log \frac{9}{11} \\
k & =\frac{1}{10} \log \frac{11}{9}
\end{aligned}
$

(i) When $t=15 \mathrm{~min}$
$
\begin{aligned}
\text { then } \mathrm{T}(15) & =70+110 e^{-15 k} \\
& =70+110 e^{-15 \frac{1}{10} \log \left(\frac{11}{9}\right)}=70+110 e^{-\frac{3}{2} \log \left(\frac{11}{9}\right)} \\
& =70+110 e^{\log \left(\frac{9}{11}\right)^{3 / 2}}=70+110\left(\frac{9}{11}\right)^{3 / 2} \\
& =70+81.39=151.39
\end{aligned}
$
(ii) Substituting $\mathrm{T}=130$ and $\mathrm{T}=140$
We get $t=22.523$

Question 9.
A pot of boiling water at $100^{\circ} \mathrm{C}$ is removed from a stove at time $t=0$ and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to $80^{\circ} \mathrm{C}$, and another 5 minutes later it has dropped to $65^{\circ} \mathrm{C}$. Determine the temperature of the kitchen.
Solution:
Let $\mathrm{A}$ be the temperature of the kitchen
$
\begin{aligned}
& \text { Given, } \frac{d \mathrm{~T}}{d t} \propto(\mathrm{A}-\mathrm{T}) \quad[\because \text { Newton's cooling law }] \\
& \frac{d \mathrm{~T}}{d t}=k(\mathrm{~A}-\mathrm{T}) \\
& \int \frac{d \mathrm{~T}}{\mathrm{~A}-\mathrm{T}}=k \int d t \\
& \log (\mathrm{A}-\mathrm{T})=\mathrm{kt}+\mathrm{c} \\
& \Rightarrow \mathrm{T}=\mathrm{A}+\mathrm{ce}^{-\mathrm{kt}} \ldots .(1)
\end{aligned}
$
Given, $\mathrm{T}(0)=100$,
$
\begin{aligned}
& \mathrm{T}(5)=80, \\
& \mathrm{~T}(10)=65
\end{aligned}
$
Substituting in (1), we get
$
\begin{aligned}
& 100=\mathrm{A}+\mathrm{c} \\
& 80=\mathrm{A}+\mathrm{ce}^{-5 \mathrm{k}} \\
& 65=\mathrm{A}+\mathrm{ce}^{-10 \mathrm{k}}
\end{aligned}
$
Solving these linear equations,
We get $\mathrm{A}=11^{\circ}$ (Kitchen Temperature)

Question 10.
A tank initially contains 50 liters of pure water. Starting at time $t=0$ a brine containing with 2 grams of dissolved salt per litre flows into the tank at the rate of 3 liters per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time $t>0$.
Solution:
Let ' $x$ ' be the amount of salt present
$
\begin{aligned}
& \frac{d x}{d t}=c(t) r(t)-\frac{x(t)}{v(t)} \cdot r_0(t) \\
& \frac{d v}{d t}=r(t)-r_0(t)
\end{aligned}
$
Given, $v(0)=50, c(t)=2, r(t)=3$ and $r_{\mathrm{o}}(t)=3$
$
\begin{aligned}
(2) \Rightarrow \frac{d v}{d t} & =3-3=0 \\
\int d v & =0+c \Rightarrow v=c \\
v(0) & =50 \Rightarrow c=50
\end{aligned}
$
$
\begin{aligned}
(1) \Rightarrow \frac{d x}{d t} & =(2)(3)-\frac{x}{50}(3) \\
\frac{d x}{d t} & =3\left(\frac{100-x}{50}\right)=\frac{-3}{50}[x-100] \\
\int \frac{d x}{x-100} & =\frac{-3}{50} \int d t \\
\log (x-100) & =\frac{-3}{50} t+c \\
x-100 & =e^{-\frac{3}{50} t} \cdot c \\
x & =100+50 e^{-\frac{3}{50} t}=50\left(2+e^{-\frac{3}{50} t}\right)
\end{aligned}
$

$\text { Since the tank initially contains only pure water, the initial amount of salt is zero. i.e., } \mathrm{x}(0)=0$