RD Sharma Class 12 Exercise 21.11 Differential Equation Solutions Maths-Download PDF Online

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# RD Sharma Class 12 Exercise 21.11 Differential Equation Solutions Maths-Download PDF Online

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 04:19 PM IST

A heavy burden in the form of a high-end syllabus is being put upon the grade 12 students. Even though learning these subjects adds value to them in one or the other form, the easiest way of learning them is not taught everywhere. RD Sharma solution And mathematics has always been a subject that is put into the tough zone by the students. To make mathematics and the concepts like Differential Equations easier, the RD Sharma Class 12th Exercise 21.11 solution book lends a helping hand to the students.

## Differential Equations Excercise: 21.11

Differential Equations exercise 21.11 question 1

Answer: $r=\sqrt{1+\frac{1}{3} t^{2}}$
Given: It is given that initial radius of balloon$=1$ unit. After 3 seconds, radius of balloon$=2$ units.
To find: We have to find the radius after time $t$.
Hint: Use surface area of a balloon $=4\pi r^{2}$. Then differentiate it to find the radius.
Solution: We have radius of the balloon$=1$ unit
After 3 seconds, radius of balloon$=2$ units
Let $r$ be the radius and $A$ be the surface area of the balloon at any time $t$.
\begin{aligned} &=\frac{d A}{d t} \propto t \\\\ &=\frac{d A}{d t}=k t \end{aligned} where $k$ is constant.
\begin{aligned} &=\frac{d\left(4 \pi r^{2}\right)}{d t}=k t \\\\ &=8 \pi r \frac{d r}{d t}=k t \end{aligned}
Integrating on both sides
\begin{aligned} &=8 \pi \frac{r^{2}}{2}=k \frac{t^{2}}{2}+C \\\\ &=4 \pi r^{2}=k \frac{t^{2}}{2}+C \ldots(i) \end{aligned}
We are given that unit at $t=0, r=1$ unit
\begin{aligned} &=4 \pi(1)^{2}=k \times 0+C \\\\ &=C=4 \pi \end{aligned}
And $t=3, r=2$ units

\begin{aligned} &=4 \pi(2)^{2}=k \frac{(3)^{2}}{2}+C \quad \text { [From (i)] }\\ \\&=16 \pi=\frac{9}{2} k+4 \pi\\\\ &=\frac{9}{2} k=12 \pi \end{aligned}
By cross multiplication we get
\begin{aligned} &9 k=24 \pi \\\\ &k=\frac{24}{9} \pi \\\\ &k=\frac{8}{3} \pi \end{aligned}
Now substitute $C=4 \pi$ and $k=\frac{8}{3} \pi$ in equation (i)

\begin{aligned} &4 \pi r^{2}=\frac{8}{3} \pi \frac{t^{2}}{2}+4 \pi \\\\ &4 \pi r^{2}-4 \pi=\frac{8}{6} \pi t^{2} \\\\ &4 \pi\left(r^{2}-1\right)=\frac{8}{6} \pi t^{2} \\\\ &\pi\left(r^{2}-1\right)=\frac{1}{4}\cdot \frac{8}{6} \pi t^{2} \end{aligned}
\begin{aligned} &\left(r^{2}-1\right)=\frac{1}{\pi} \frac{1}{3} \pi t^{2} \\\\ &r^{2}-1=\frac{1}{3} t^{2} \\\\ &r^{2}=1+\frac{1}{3} t^{2}=\sqrt{1+\frac{1}{3} t^{2}} \end{aligned}

Differential Equations exercise 21.11 question 2

Answer: $20\log 2$ years
Given: Population growth rate =5% per year.
To find: Time taken for increase in population.
Hint: The rate of population increases with increase in time i.e. $\frac{d P}{d t}=r \% \times P$
Solution: Population growth rate =5% per year
Let initial population be $P_{0}$ and the population after time $t$ be $P$
Then,
\begin{aligned} &\frac{d P}{d t}=5 \% \times P \\\\ &=\frac{d P}{d t}=\frac{5}{100} \times P \end{aligned}
$=>\frac{d P}{d t}=\frac{P}{20}$
By cross multiplication we get,
$=>20 \frac{d P}{p}=d t$
Integrating on both sides
\begin{aligned} &=>20 \int \frac{d P}{P}=\int d t \\\\ &=>20 \log P=t+C \ldots(i) \end{aligned}
At $t=0$ we have $P=P_{0}$ [Putting $t=0$ and $P=P_{0}$ in equation i]
\begin{aligned} &=>20 \log \left(P_{0}\right)=0+C \\\\ &=>C=20 \log P_{0} \end{aligned}
Putting $C=20 \log P_{0}$ in equation (i)
\begin{aligned} &=>20 \log P=t+20 \log \left(P_{0}\right) \\\\ &=>20 \log P-20 \log \left(P_{0}\right)=t \\\\ &=>20 \log \frac{p}{p_{0}}=t \end{aligned}
When $P=2 P_{0}$ we get
\begin{aligned} &=20 \log \left(\frac{2 P_{0}}{P_{0}}\right) \\\\ &=t=20 \log 2 \end{aligned}
Hence, the required time period is $20\log 2$ years.

Differential Equations exercise 21.11 question 3

Answer: $58$ years
Given: Present Population $= 1,00,000$
To find: When the city will have a population of $5,00,000$
Hint: The population of city increase in time i.e. $\frac{d P}{d t} \propto P$ and then find the equation using integration.
Solution: The present population is $1,00,000$ and the population of a city doubled in the past $25$ years
Let $P$ be the population at any time $t$
Then, $\frac{d P}{d t} \propto P$
\begin{aligned} &\frac{d P}{d t}=k P \\\\ &\frac{d P}{d t}=k d t \end{aligned}
Integrating on both sides we get,

\begin{aligned} &=>20 \int \frac{d P}{P}=\int k d t \\\\ &=>\log P=k t+C \ldots(i) \end{aligned}
At $t=0$ we take $P=P_{0}$
Then,
$=>\log P_{0}=k \times 0+C$ [Putting $t=0$ and $P=P_{0}$ in equation (i)]
$=>C=\log P_{0}$

Putting $C=\log P_{0}$ in equation ($i$) we get

\begin{aligned} &=>\log P=k t+\log P_{0} \\\\ &=>\log P-\log P_{0}=k t \\\\ &=>\log \frac{P}{p_{0}}=k t \ldots(i i) \end{aligned}
When $P=2P_{0}$ at $t=25$ we have
\begin{aligned} &\log \left(\frac{2 P_{o}}{P}\right)=k t \\\\ &=>\log 2=25 k \end{aligned}
Putting $k=\frac{\log 2}{25}$ in equation (ii) we get
$=\log \left(\frac{p}{p_{0}}\right)=\left(\frac{\log 2}{25}\right) t$
We assume that $t_{1}$ be the time take for the population to become $5,00,000$ from $1,00,000$
Then,$\log g\left(\frac{5,00,000}{1,00,000}\right)=\left(\frac{\log 2}{25}\right) t_{1}$
\begin{aligned} &=\log 5=\left(\frac{\log 2}{25}\right) t_{1} \\\\ &=25 \log 5=\log 2 t_{1} \\\\ &=t_{1}=25\left(\frac{\log \log 5}{\log \log 2}\right) \\\\ &=t_{1}=25\left(\frac{1.609}{0.6931}\right) \end{aligned} $[\log \log 5=1.609, \log \log 2=0.6931]$
$=t_{1}=58.08 \text { Years }$
Therefore the required time is $58$ years (approximate).

Differential Equations exercise 21.11 question 4

Answer: $\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \text { hours }$
Given: Present bacteria count$=1,00,000$
To find: The amount of hours taken for the count to reach $2,00,000$
Hint: Use $\frac{d C}{d t}=\lambda C$ and then find the value of lambda using integration.
Solution: The bacteria count is $1,00,000$ and rate of increase $=10%$ in $2$ hours
Let $C$ be the bacteria count at any time $t$,
Then $\frac{d C}{d t} \propto C$
$=>\frac{d C}{d t}=\lambda C$ [Where $\lambda$ is the constant]
Integrating on both sides
\begin{aligned} &=>\int \frac{d C}{d t}=\int \lambda C \\\\ &=>\int \frac{d C}{C}=\lambda \int d t \\\\ &=>\log C=\lambda t+\log k \ldots(i) \end{aligned}
Where $\log \log k$ is integral constant
At $t=0$ we have $C=1,00,000$
Then $\log (1,00,000)=\lambda \times 0+\log k$ [putting $t=0$ and $C=1,00,000$ in equation (i)]
$=\log (1,00,000)=\log k \ldots(i i)$
\begin{aligned} &\text { At } t=2 \\ &\qquad \begin{aligned} C &=1,00,000+1,00,000\left(\frac{10}{100}\right) \\ &=1,10,000 \end{aligned} \end{aligned}
From equation (i) we have,
\begin{aligned} &=\log 1,10,000=\lambda \times 2+\log k \quad \text { [Putting } C=1,10,000 \text { and } t=2] \\\\ &=\log 1,10,000=2 \lambda+\log k \ldots(i i i) \end{aligned}
Now, subtracting equation (ii) from (iii) we have
\begin{aligned} &=\log 1,10,000-\log 1,00,000=2 \lambda+\log k-\log k \\\\ &=\log \left(\frac{1,10,000}{1,00,000}\right)=2 \lambda \\\\ &=\log \frac{11}{10}=2 \lambda \\\\ &=\lambda=\frac{1}{2} \log \left(\frac{11}{10}\right) \end{aligned}
Now, we have to find time $t$ in which the count reaches $2,00,000$
\begin{aligned} &\text { Now } C=2,00,000 \text { in equation }(i)\\\\ &=\log (2,00,000)=\lambda t+\log k \end{aligned}
Putting value of lambda and $\log k$, we get
\begin{aligned} &=\log (2,00,000)=\frac{1}{2} \log \left(\frac{11}{10}\right) t+\log 1,00,000 \\\\ &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log 2,00,000-\log 1,00,000 \end{aligned}
\begin{aligned} &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log \left(\frac{2,00,000}{1,00,000}\right) \\\\ &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log 2 \\\\ &=t=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \end{aligned}
Hence, the time required $=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}$ hours

Differential Equations exercise 21.11 question 5

Answer: $P=R s \; 1822, t=12 \text { years }$
Given: Compound interest=6% per annum,
To find: The amount after 10 years for Rs 1000 and the time in which the amount doubles.
Hint: Use compound interest i.e. $\frac{d P}{d t}=\frac{P r}{100}$
Solution: Let P be the principal
\begin{aligned} &=\frac{d P}{d t}=\frac{p r}{100} \\\\ &=\frac{d P}{P}=\frac{r}{100} d t \end{aligned}
Integrating on both sides we get,
\begin{aligned} &=>\int \frac{d P}{P}=\int \frac{r}{100} d t \\\\ &=>\log P=\frac{r}{100} \times t+C \ldots(i) \end{aligned}
At t=0, we have initial principal P=P0
\begin{aligned} &=>\log \left(P_{0}\right)=0+C \\\\ &=>C=\log P_{0} \end{aligned}
Now substituting value of C in equation (i)
\begin{aligned} &=>\log P=\frac{r}{100} t+\log P_{0} \\\\ &=>\log P-\log P_{0}=\frac{r t}{100} \\\\ &=>\log \left(\frac{p}{p_{0}}\right)=\frac{r t}{100} \end{aligned}
Now, P0=1000, t=10 years and r=6%
\begin{aligned} &=>\log \left(\frac{P}{1000}\right)=\frac{6 \times 10}{100} \\\\ &=>\log P-\log 1000=0.6 \\\\ &=>\log P=0.6+\log 1000 \end{aligned}
$=>\log P=\log e^{0.6}+\log 1000 \quad\quad\quad\left[\log \log e^{x}=x\right]$
\begin{aligned} &=>\log P=\log \left(e^{0.6} \times 1000\right) \\\\ &=>\log P=\log (1.822 \times 1000) \quad\quad\quad\left[\log \log e^{0.6}=1.822\right] \end{aligned}
\begin{aligned} &=>\log P=\log (1822)\\\\ &=>P=1822 \end{aligned}[Using anti-logarithm on both sides]
Therefore P=Rs 1822
Thus Rs 1000 will be Rs 1822 in 10 years.
Now let t1 be the time taken to double Rs 1000 then, P=2000, P0=1000, r=6%
From equation (i)
\begin{aligned} &\log \left(\frac{p}{p_{0}}\right)=\frac{r t}{100} \\\\ &=>\log \left(\frac{2000}{1000}\right)=\frac{6 t_{1}}{100} \\\\ &=>\log 2=\frac{6 t_{1}}{100} \end{aligned}
$=6 t_{1}=100 \log 2$ [By cross multiplication]

\begin{aligned} &=6 t_{1}=100 \times 0.6931\quad\quad\quad &[\log 2=0.6931]\\\\ &=t_{1}=\frac{69.31}{6}\\\\ &=t_{1}=11.55 \text { Years } \end{aligned}
$=12$ years (approximately)
Hence, it will take 12 years to double the amount.

Differential Equations exercise 21.11 question 6

Answer: 9 times, $\frac{5 \log 10}{\log 3}$
Given: The amount of bacteria triples in 5 hours.
To find: The amount of bacteria present after 10 hours and also the time for the number of bacteria to become 10 times.
Hint: Use compound interest i.e. $\frac{d P}{d t}=\frac{p r}{100}$
Solution: Let A be the amount of bacteria present at time t and A0 be the initial amount of bacteria
\text { Then, } \begin{aligned} \frac{d A}{d t} & \propto A \\ &=\frac{d A}{d t}=\lambda A \end{aligned}
Where lambda is the integrating constant
Integrating on both sides we get,
\begin{aligned} &=>\int \frac{d A}{d t}=\int \lambda A \\\\ &=>\int \frac{d A}{A}=\lambda \int d t \\\\ &=>\log A=\lambda t+C \ldots(i) \end{aligned}
Where C is integral constant
When t=0, we have A=A0
\begin{aligned} &\left.=>\log A_{0}=0+C \quad \text { [Putting } t=0 \text { and } A=A_{0} \text { in equation }(i)\right]\\\\ &=>C=\log A_{0} \end{aligned}
Putting the value of C in equation (i)
\begin{aligned} &=>\log A=\lambda t+\log A_{0} \\\\\ &=>\log A-\log A_{0}=\lambda t \\\\ &=>\log \left(\frac{A}{A_{0}}\right)=\lambda t \ldots(i i) \end{aligned}
Since bacteria triples in 5 hours,
$\text { So, } A=3 A_{0} \text { and } t=5$
Putting in equation (i) we get,
\begin{aligned} &=\log \left(\frac{4 A_{0}}{A_{0}}\right)=5 \lambda \\\\ &=\log 3=5 \lambda \\\\ &=\lambda=\frac{\log 3}{5} \end{aligned}
Putting $\lambda=\frac{\log 3}{5}$ in equation (ii) we get,
$=\log \left(\frac{A}{A_{0}}\right)=\frac{\log 3}{5} t \ldots(i i i)$
Now let A1 be the amount of bacteria present after 10 hours,
$\text { Then, } \log \left(\frac{A_{1}}{A_{0}}\right)=\frac{\log 3}{5} \times 10$
\begin{aligned} &=\log \left(\frac{A_{1}}{A_{0}}\right)=2 \log 3 \\\\ &=\log \left(\frac{A_{1}}{A_{0}}\right)=2 \times 1.0986 \quad[\log \log 3=1.0986] \end{aligned}
\begin{aligned} &=\log \left(\frac{A_{1}}{A_{0}}\right)=2.1972 \\\\ &=\frac{A_{1}}{A_{0}}=e^{2.1972} \\\\ &=A_{1}=A_{0} e^{2.1972} \\\\ &=A_{1}=9 A_{0} \quad\left[e^{2.1972}=9\right] \end{aligned}
Hence, after 10 hours the number of bacteria will be 9 times the original.
Now, let t1 be the time necessary for the bacteria to be 10 times the amount. So, A=10A0
Then equation (iii) is
\begin{aligned} &=\log \left(\frac{A}{A_{0}}\right)=\frac{\log 3}{5} \times t \\\\ &=\log \left(\frac{10 A_{0}}{A_{0}}\right)=\frac{\log 3}{5} \times t_{1} \\\\ &=\log \log 10=\frac{\log 3}{5} \times t_{1} \end{aligned}
\begin{aligned} &=5 \log 10=\log 3 \times t_{1} \\\\ &=t_{1}=\frac{5 \log 10}{\log 3} \end{aligned}
Hence, the time needed for the bacteria to become 10 times the initial amount $=\frac{5 \log 10}{\log 3}$ hours

Differential Equations exercise 21.11 question 7

Given: Population of city in 1990=2,00,000
Population on city in 2000=2,50,000
To find: Population of city in 2010
Hint: Use $\frac{d P}{d t}=\lambda P$ and find the value of constant lambda.
Solution: Let P be the population of the city at time t
\text { Then, } \begin{aligned} \frac{d P}{d t} \propto P \end{aligned}
$=\frac{d P}{d t}=\lambda P$
Where lambda is the proportionality constant
$=\frac{d P}{p}=\lambda d t$
Integrating on both sides we get,
\begin{aligned} &=>\int \frac{d P}{P}=\int \lambda d t \\\\ &=>\log P=\lambda t+\log k \ldots(i) \end{aligned}
Where k is integral constant
\begin{aligned} \text { Whent } &=1990 \text {, we have } P=2,00,000 \\\\ &=>\log 2,00,000=\lambda \times 1990+\log k \ldots(i i) \end{aligned}
[utting t=0 and A=A0 in equation (i)]
At
\begin{aligned} t=& 2000, \text { we have } P=2,50,000 \\\\ &=>\log 2,50,000=\lambda \times 2000+\log k \ldots(i i i) \end{aligned}
Subtracting equation (ii) and (iii)
\begin{aligned} &=>\log 2,50,000-\log 2,00,000=2000 \lambda-1990 \lambda \\\\ &=>\log \left(\frac{2,50,000}{2,00,000}\right)=10 \lambda \\\\ &=>\log \left(\frac{5}{4}\right)=10 \lambda \end{aligned}
$=>\lambda=\frac{1}{10} \log \left(\frac{5}{4}\right)$
Putting the value of lambda in equation (ii) we get,
\begin{aligned} &=>\log 2,00,000=\frac{1990}{10} \log \left(\frac{5}{4}\right)+\log k \\ &=>\log k=\log 2,00,000-199 \log \left(\frac{5}{4}\right) \end{aligned}
Substituting the value of lambda in equation (ii) we get
\begin{aligned} &=>\log P=\left(\frac{1}{10} \log \frac{5}{4}\right) 2010+\log 2,00,000-199 \log \frac{5}{4} \\\\ &=>\log P=201 \log \frac{5}{4}+\log 2,00,000-199 \log \frac{5}{4} \\\\ &=>\log P=\log \left(\frac{5}{3}\right)^{201}+\log 2,00,000-\log \left(\frac{5}{4}\right)^{199} \end{aligned}
\begin{aligned} &=>\log P=\log \left[\frac{\left(\frac{5}{3}\right)^{201}}{\left(\frac{5}{4}\right)^{199}}\right]+\log 2,00,000 \\\\ &=>\log P=\log \left[\left(\frac{5}{4}\right)^{201-199} \times 2,00,000\right] \\\\ &=>\log P=\log \left[\left(\frac{5}{4}\right)^{2} \times 2,00,000\right] \end{aligned}
\begin{aligned} &=>\log P=\log \left[\frac{25}{16} \times 2,00,000\right] \\\\ &=>\log P=\log 3,12,500 \\\\ &=>P=3,12,500 \end{aligned}
Thus, the population of city in 2010=3,12,500

Differential Equations exercise 21.11 question 8

Answer: $C(x)=0.075 x^{2}+2 x+100$
Given: $C^{\prime}(x)=\frac{d C}{d x}=2+0.15 x \text { and } C(0)=100$
To find: Total cost function $C(x)$
Hint: Use integration and then find out the value of integral constant
Solution: Given that
\begin{aligned} &=C^{\prime}(x)=\frac{d C}{d x}=2+0.15 x \\\\ &=d C=(2+0.15 x) d x \end{aligned}
Integrating on both sides we get,
\begin{aligned} &=\int d C=\int(2+0.15 x) d x \\\\ &=>\int d C=\int 2 \mathrm{~d} x+0.15 \int x d x \\\\ &=>C=2 x+\frac{0.15 x^{2}}{2}+D \ldots(i) \end{aligned}
We have C=100 when x=0
\begin{aligned} &=>100=2 \times 0+\frac{0.15 \times(0)^{2}}{2}+D \\ &=>D=100 \end{aligned}
Putting the value of D in equation (i) we get,
\begin{aligned} &=>C=2 x+\frac{0.15 x^{2}}{2}+100 \\\\ &=>C(x)=2 x+0.075 x^{2}+100 \end{aligned}
Thus the total cost function $C(x)=2 x+0.075 x^{2}+100$

Differential Equations exercise 21.11 question 9

Given: Compound interest = 8% per annum,
To find: The percentage increase over one year.
Hint: Use compound interest i.e. $\frac{d P}{d t}=\frac{p r}{100}$
Solution: Let P be the principal
\begin{aligned} &=\frac{d P}{d t}=\frac{p r}{100} \\\\ &=\frac{d P}{P}=\frac{r}{100} d t \end{aligned}
Integrating on both sides we get,
\begin{aligned} &=>\int \frac{d P}{P}=\int \frac{r}{100} d t \\\\ &=>\log P=\frac{r}{100} \times t+C \ldots(i) \end{aligned} [Where C is integral constant]
At t=0, we have initial principal P=P0
\begin{aligned} &=>\log \left(P_{0}\right)=0+C \\\\ &=>C=\log P_{0} \end{aligned}
Now substituting value of C in equation (i)
\begin{aligned} &=>\log P=\frac{r}{100} t+\log P_{0} \\\\ &=>\log P-\log P_{0}=\frac{r t}{100} \\\\ &=>\log \left(\frac{p}{p_{0}}\right)=\frac{r t}{100} \ldots(i i) \end{aligned}
For t=1 and r=8% in equation (ii)
\begin{aligned} &=>\log \left(\frac{p}{p_{0}}\right)=\frac{8}{100} \\\\ &=>\log \left(\frac{p}{p_{0}}\right)=0.08 \end{aligned}
\begin{aligned} &=>\frac{P}{P_{0}}=e^{0.08} \\\\ &=>\frac{P}{P_{0}}=1.0833 \quad\left[e^{0.08}=1.0833\right] \end{aligned}
Subtracting 1 on both sides we get
\begin{aligned} =&>\frac{P}{P_{0}}-1=1.0833-1 \\\\ \end{aligned}
$=\frac{P-P_{0}}{P_{0}} \times 100$
\begin{aligned} &=0.0833 \times 100 \\\\ &=8.33 \% \end{aligned}
Hence, amount increases by 8.33% in one year.

Differential Equations exercise 21.11 question 10

Answer: $i=\frac{E}{R}\left[1-e^{\left(-\frac{K}{L}\right) t}\right]$
Given: $L \frac{d i}{d t}+R i=E$
To find: To prove that $i=\frac{E}{R}\left[1-e^{\left(-\frac{R}{L}\right) t}\right]$
Hint: Integrate the linear differential equation to find integral factor.
Solution: $L \frac{d i}{d t}+R i=E$
Dividing by L
$=\frac{d t}{d t}+\frac{R}{L} i=\frac{E}{L}$
This is a linear differential equation and comparing it with
$=\frac{d y}{d x}+P y=Q$
Such that,
$=P=\frac{R}{L^{\prime}}, Q=\frac{E}{L}$
We know that
\begin{aligned} &\text { I.F. }=e^{\int p d t} \\\\ &=e^{\int \frac{R}{L} d t} \\\\ &=e^{\left(\frac{R}{L}\right) t} \end{aligned}
So, the solution is given by
\begin{aligned} &i(I f)=\int Q(I f) d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\int \frac{E}{L}\left[e^{\left(\frac{R}{L}\right) t}\right] d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{L} \times \frac{L}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \end{aligned}
\begin{aligned} &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \\\\ &=>i=\frac{\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C}{\left[e^{\left.\left(\frac{R}{L}\right) t\right]}\right.} \end{aligned}
\begin{aligned} &\text { => } i=\frac{E}{R}+\frac{C}{\left[e^{\left(\frac{R}{L}\right) t}\right]} \\\\ &=>i=\frac{E}{R}+C\left[e^{-\left(\frac{R}{L}\right) t}\right] \ldots(i) \end{aligned}
Initially no current passes through the circuit so,i=0, t=0
\begin{aligned} &=>0=\frac{E}{R}+C e^{0} \\\\ &=>0=\frac{E}{R}+C \\\\ &=>C=-\frac{E}{R} \end{aligned}
Substituting values of C in equation (i)
\begin{aligned} &=>i=\frac{E}{R}-\frac{E}{R}\left[e^{-\left(\frac{R}{L}\right) t}\right] \\\\ &=>i=\frac{E}{R}\left[1-e^{-\left(\frac{R}{L}\right) t}\right] \end{aligned}

Differential Equations exercise 21.11 question 11

Answer: $\frac{1}{k} \log 2$ where lambda is thee constant of proportionality.
Given: Decaying rate of radium at any time is proportional to its mass at that time.
To find: We have to find the time when the mass will be half of its initial mass.
Hint: The amount of radium at any time t is proportional to its mass i.e. $\frac{d A}{d t} \propto A$
Solution: Given that
$\frac{d A}{d t} \propto A$
$=>\frac{d A}{d t}=-\lambda A$ [Where ? is the constant of proportionality and minus sign indicates that A decrease with increase in t]
$=>\frac{d A}{A}=-\lambda d t$
Integrating on both sides
\begin{aligned} &=>\int \frac{d A}{A}=-\lambda \int d t \\\\ &=>\log A=-\lambda t+C \ldots(i) \end{aligned}
Since initial amount of radium is A0 then
\begin{aligned} &=>\log A_{0}=-\lambda \times 0+C \\\\ &=>\log A_{0}=C \end{aligned}
Putting value of C in equation (i) we get
\begin{aligned} &=>\log A=-\lambda t+\log A_{0} \\\\ &=>\log A-\log A_{0}=-\lambda t \\\\ &=>\log \frac{A}{A_{0}}=-\lambda t_{1} \end{aligned}
\begin{aligned} &=>\log \frac{1}{2}=-\lambda t_{1} \\\\ &=>-\log 2=-\lambda t_{1} \\\\ &=>t_{1}=\frac{1}{\lambda} \log 2 \end{aligned}

Hence, the time taken$=\frac{1}{\lambda} \log 2$ where lambda is the constant of proportionality.

Differential Equations exercise 21.11 question 12

Given: Half like of radium=1590 years.
To find: We have to find the percentage of radium decaying in one year.
Hint:T ake $\frac{d A}{d t}=-\lambda t$ and integrate it to find rate of $\lambda$
Solution: Given that
\begin{aligned} &\frac{d A}{d t} \propto A \\ &=\frac{d A}{d t}=-\lambda A \end{aligned} [Where ?the constant of proportionality and minus sign is indicates ]
$=\frac{d A}{A}=-\lambda d t$ [that A decrease with increase in t]
Integrating on both sides
\begin{aligned} &=\int \frac{d A}{A}=-\lambda \int d t\\\\ &=\log A=-\lambda t+C \end{aligned} ........(i)
Since initial amount of radium is A0 then
\begin{aligned} &=\log A_{0}=-\lambda \times 0+C \\\\ &=\log A_{0}=C \end{aligned}
Putting value of C in equation (i) we get
\begin{aligned} &=\log A=-\lambda t+\log A_{0}\\\\ &=\log A-\log A_{0}=-\lambda t\\\\ &=\log \frac{A}{A_{0}}=-\lambda t \end{aligned}................(ii)
Given that its half-life is 1590 years so we put $A=\frac{1}{2} A_{0}$ and t=1590
Hence, equation (ii) becomes
\begin{aligned} &=>\log \left(\frac{A}{A_{0}}\right)=-\lambda t \\ &=> \log \left(\frac{A_{0}}{2 A_{0}}\right)=-\lambda \times 1590 \end{aligned}
\begin{aligned} &=>\log \left(\frac{1}{2}\right)=-\lambda(1590) \\\\ &=>-\log 2=-\lambda \times 1590 \end{aligned}
Taking minus sign on both sides,
\begin{aligned} &=>\log 2=1590 \lambda \\\\ &=>\lambda=\frac{\log 2}{1590} \end{aligned}
Put $=\lambda=\frac{\log 2}{1590}$ in equation (ii) we get
\begin{aligned} &=>\log \left(\frac{A}{A_{0}}\right)=-\left(\frac{\log 2}{1590}\right) t \\\\ &=>\frac{A}{A_{0}}=e^{-\left(\frac{\log 2}{1590}\right) t} \ldots(i i i) \end{aligned}
Putting t=1 in (iii) to find the amount of radium after one year, we get
\begin{aligned} &=\frac{A}{A_{0}}=e^{-\frac{\log 2}{1590}} \\\\ &=>\frac{A}{A_{0}}=0.9996 \\\\ &=>A=A_{0} 0.9996 \end{aligned}
Percentage amount disappeared in one year
$=\frac{A_{0}-A}{A_{0}} \times 100 \\$
$=\frac{A_{0}-A_{0} 0.9996}{A_{0}} \times 100$
$=\frac{A_{0}(1-0.9996)}{A_{0}} \times 100 \\$
$=(1-0.9996) \times 100 \\$
$=0.0004 \times 100 \\$
$=0.04 \%$

Differential Equations exercise 21.11 question 13

Answer: $x^2+y^2=25$
Given: The slope of tangent at a point $P(x, y)=-\frac{x}{y}$
To find: We have to find the equation of the curve which passes through $\left ( 3,-4 \right )$
Hint: First take the slope of the curve and integrate the equation.
Solution: So,$\frac{d y}{d x}=-\frac{x}{y}$
$=y d y=-x d x$
Integrating on both sides we get,
\begin{aligned} &=\int y d y=-\int x d x \\\\ &=\frac{y^{2}}{2}=-\frac{x^{2}}{2}+C \ldots(i) \end{aligned}
Where C is constant
Since the curve passes through the point $\left ( 3,-4 \right )$ such that $\left ( 3,-4 \right )$ satisfy equation (i)
$=\frac{(-4)^{2}}{2}=-\frac{3^{2}}{2}+C$
\begin{aligned} &=>\frac{16}{2}=-\frac{9}{2}+C \\\\ &=>8=-\frac{9}{2}+C \\\\ &=>C=8+\frac{9}{2} \end{aligned}
\begin{aligned} &=>C=\frac{16+9}{2} \\\\ &=>C=\frac{25}{2} \end{aligned}
Substituting the value of C in equation (i) we get
$=>\frac{y^{2}}{2}=-\frac{x^{2}}{2}+\frac{25}{2}$
$=y^{2}=-x^{2}+25$ [Multiplying by 2 on both sides]
$=x^{2}+y^{2}=25$
Hence, found.

Differential Equations exercise 21.11 question 14

Answer: $2 x y-2 x-y-2=0$
Given: The differential equation is $y-x \frac{d y}{d x}=y^{2}+\frac{d y}{d x} .$
To find: We have to find the equation of the curve which passes through $\left ( 2,2 \right )$
Hint: We will integrate the given differential equation and then we use the points $\left ( 2,2 \right )$
Solution: we have,
$=y-x \frac{d y}{d x}=y^{2}+\frac{d y}{d x}$
Integrating on both sides we get,
\begin{aligned} &=>\int\left(y-x \frac{d y}{d x}\right)=\int\left(y^{2}+\frac{d y}{d x}\right) \\\\ &=>\int \frac{d y}{y-y^{2}}=\int \frac{d x}{1+x} \end{aligned}
\begin{aligned} &=>\int \frac{d y}{y(1-y)}=\int \frac{d x}{1+x} \\\\ &=>\int\left[\frac{1}{y}-\frac{1}{1-y}\right] d y=\int \frac{d x}{x+1} \end{aligned}
$=>\log |y|-\log |1-y|=\log |x+1|+C \ldots(i)$
Since the curve passing through the point $\left ( 2,2 \right )$ it satisfies equation (i)
Then,
\begin{aligned} &\log |2|-\log |1-2|=\log |2+1|+C \\\\ &\log |2|-\log |-1|=\log |3|+C \\\\ &C=\log |2|-\log |3| \\\\ &C=\log \left|-\frac{2}{3}\right| \end{aligned}
Putting the value of C in equation (i) we get
\begin{aligned} &\log |y|-\log |1-y|=\log |x+1|+\log \left|-\frac{2}{3}\right| \\\\ &\log \left|\frac{y}{1-y}\right|=\log \left|-\frac{2(x+1)}{3}\right| \end{aligned}
Taking antilogarithm on both sides,
\begin{aligned} &\frac{y}{1-y}=-\frac{2(x+1)}{3} \\\\ &3 y=-2(x+1)(1-y) \\\\ &3 y=-2(1+x-y-x y) \\\\ &3 y=-2-2 x+2 y+2 x y \end{aligned}
\begin{aligned} &3 y+2+2 x-2 y-2 x y=0 \\\\ &y+2+2 x-2 x y=0 \\\\ &2 x y-2 x-2-y=0 \end{aligned}
Hence, required equation is found.

Differential Equations exercise 21.11 question 15

Answer:$\tan \left(\frac{y}{x}\right)=\log \left(\frac{e}{x}\right)$
Given: The angle is $\left(\frac{y}{x}-\frac{y}{x}\right)$
To find: We have to find the equation of the curve which passes through $\left(1, \frac{\pi}{4}\right)$
Hint: First take the slope of the curve i.e.$\frac{d y}{d x}=\tan \theta$ and take a linear equation $y=v x$
Solution:
The slope of the curve is $\frac{d y}{d x}=\tan \theta$
We have,
$=\theta=\left[\frac{y}{x}-\frac{y}{x}\right]$
\begin{aligned} &\text { Then } \begin{array}{l} \frac{d y}{d x}=\text { tan inverse tan }\left\{\left[\frac{y}{x}-\frac{y}{x}\right]\right\} \\\\ \quad=\frac{d y}{d x}=\frac{y}{x}-\frac{y}{x} \end{array} \\ &\text { Let } y=v x \end{aligned}
Differentiating with respect to $x$ we get
\begin{aligned} &\frac{d y}{d x}=v+x \frac{d v}{d x} \\\\ &x \frac{d v}{d x}=\frac{d y}{d x}-v \end{aligned}
$x \frac{d v}{d x}=\frac{d y}{d x}-\frac{y}{x} \quad\quad\quad\quad\left[v=\frac{y}{x}\right]$
$x \frac{d v}{d x}=-v$ [From equation i]
Integrating on both sides
\begin{aligned} &\int v d v=\int-\frac{1}{x} d x \\\\ &\tan \frac{y}{x}=-\log |x|+C \ldots(i i) \end{aligned}
The curve passes through$\left[1, \frac{\pi}{4}\right]$ it satisfices equation (ii)
\begin{aligned} &=\log \frac{\pi}{4}=-\log |1|+C \\\\ &=C=1 \end{aligned}
\begin{aligned} &\text { Put } C=1, \tan \frac{y}{x}=-\log |x|+1 \\\\ &=\tan \frac{y}{x}=-\log |x|+1 \\\\ &=\tan \frac{y}{x}=\log \left|\frac{e}{x}\right| \end{aligned}
Hence, required equation is found.

Differential Equations exercise 21.11 question 16

Answer:$\mathrm{c} y^{4}=e^{-x / y}$
Given: The curve is $y=f(x)$. Suppose$P(x, y)$ is a part of the curve.
To find: We have to find the equation of the curve for which the intercept cut off by a tangent on x-axis.
Hint: Use equation of tangent i.e.$Y-y=\frac{d y}{d x}(X-x)$ then solve the differential equation.
Solution: Equation of tangent to the curve at $P$ is
$=Y-y=\frac{d y}{d x}(X-x)$
Where (X, Y) is the arbitrary point on the tangent
Putting Y=0 we get
\begin{aligned} &=0-y=\frac{d y}{d x}(X-x) \\\\ &=X-x=-y \frac{d x}{d y} \end{aligned}
When cut off by the tangent on the x-axis
$=x-y \frac{d x}{d y}=4 y$
Therefore $-y \frac{d x}{d y}=4 y-x$
\begin{aligned} &=\frac{d x}{d y}=\frac{x-4 y}{y} \\\\ &=\frac{d y}{d x}=\frac{y}{x-4 y} \ldots(i) \end{aligned}
This is homogeneous differential equation
Putting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$ in (i)
We get
\begin{aligned} &=v+x \frac{d v}{d x}=\frac{v x}{x-4 v x}=\frac{v x}{x(1-4 v)} \\\\ &=v+x \frac{d v}{d x}=\frac{v}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{v}{1-4 v}-v \end{aligned}
\begin{aligned} &=x \frac{d v}{d x}=\frac{v-v(1-4 v)}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{v-v+4 v^{2}}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{4 v^{2}}{1-4 v} \end{aligned}
\begin{aligned} &=\frac{1}{x} \frac{d x}{d v}=\frac{1-4 v}{4 v^{2}} \\\\ &=4 \frac{d x}{x}=\frac{1-4 v}{v^{2}} d v \end{aligned}
Integrating on both sides,
\begin{aligned} &=4 \int \frac{d x}{x}=\int \frac{1-4 v}{v^{2}} d v \\\\ &=4 \int \frac{d x}{x}=\int \frac{1}{v^{2}} d v-\int \frac{4 v}{v^{2}} d v \\\\ &=4 \int \frac{d x}{x}=\int \frac{1}{v^{2}} d v-4 \int \frac{d v}{v^{2}} \end{aligned}
\begin{aligned} &=4 \log x+\log C=-\frac{1}{v}-4 \log v \\\\ &=4 \log x+4 \log v+\log C=-\frac{1}{v} \\\\ &=4 \log (x v)+\log C=-\frac{1}{v} \end{aligned}
Substituting value of $v$ we get
\begin{aligned} &=4 \log \left(x \times \frac{y}{x}\right)+\log C=-\frac{x}{y} \\\\ &=4 \log y+\log C=-\frac{x}{y} \end{aligned}
$=\log y^{4}+\log C=-\frac{x}{y} \quad\quad\quad\quad\left[a \log x=\log x^{a}\right]$
\begin{aligned} &=\log \left(y^{4} C\right)=-\frac{x}{y} \\\\ &=y^{4} C=e^{-\frac{x}{y}} \end{aligned}
Hence, required curve is found.

Differential Equations exercise 21.11 question 17

Given: Slope at any point $=y+2 x$
To find: We have to show that the equation of curve which pass through the origin is$y+2(x+1)=2 e^{2 x}$
Hint: Use the linear differential equation i.e. $\frac{d y}{d x}+P y=Q$
Solution: Slope at any point $=y+2 x$
$\text { i.e. } \frac{d y}{d x}=y+2 x$
$=\frac{d y}{d x}-y=2 x$
It is a linear differential equation comparing it with $\frac{d y}{d c}+P y=Q$
$P=-1, Q=2 x$
Integrating factor $(I f)=e^{\int P d x}$
\begin{aligned} &=>\text { I.F. }=e^{\int(-1) d x} \\\\ &=>\text { I.F. }=e^{-x} \end{aligned}
Solution of the equation is given by
\begin{aligned} &=y \times \text { I.F. }=\int Q(\text { I.F. }) d x+C \\\\ &=>y e^{-x}=\int(2 x)\left(e^{-x}\right) d x+C \\\\ &=>y e^{-x}=2 \int(x)\left(e^{-x}\right) d x+C \end{aligned}
$=>y e^{-x}=2\left[x \int e^{-x}-\int\left(\frac{d x}{d x} \int e^{-x} d x\right) d x\right]+C$[Using integration by parts]
$=>y e^{-x}=2\left(-x e^{-x}+\int 1 e^{-x} d x\right)+C$
$=>y e^{-x}=2\left(-x e^{-x}-e^{-x}\right)+C$
\begin{aligned} &=>y e^{-x}=e^{-x}\left(-2 x-2+C e^{x}\right) \\\\ &=>y=-2 x-2+C e^{x} \\\\ &=>y+2(x+1)=C e^{x} \ldots(i) \end{aligned}
If it passes through origin
\begin{aligned} &=0+2(0+1)=C e^{0} \\\\ &=>C=2 \end{aligned}
Now equation (i) becomes
$=y+2(x+1)=2 e^{x}$

Differential Equations exercise 21.11 question 18

Answer:$y e^{-3 x}=\left(-\frac{2}{3} x-\frac{2}{9}\right) e^{-3 x}+\frac{26}{9} e^{-3}$
Given: Tangent makes an angle $(2 x+3 y)$ with x-axis
To find: We have to show that the equation of curve which pass through $(1,2)$
Hint: Find the slope of the tangent then solve using linear differential equation.
Solution: Slope of tangent $t=\tan \theta$ and tangent makes angle $(2 x+3 y)$
$=>\frac{d y}{d x}=\tan \left[\tan ^{-1}(2 x+3 y)\right]$
$=>\frac{d y}{d x}=2 x+3 y$
$=>\frac{d y}{d x}-3 y=2 x$
It is a linear differential equation
Comparing it with $\frac{d y}{d x}+P y=Q$
$=P=-3, Q=2 x$
\begin{aligned} &\text { If }=e^{\int P d x} \\\\ &=>I f=e^{-\int 3 d x} \\\\ &=>I f=e^{-3 x} \end{aligned}
Solution of equation is given by
$=y(I f)=\int Q(I f) d x+C \\$
$=>y\left(e^{-3 x}\right)=\int 2 x e^{-3 x} d x+C$
Using integration by parts we get
\begin{aligned} &=>y\left(e^{-3 x}\right)=2\left[x \int e^{-3 x}-\int\left(\frac{d x}{d x} \int e^{-3 x} d x\right) d x\right]+C \\\\ &=>y\left(e^{-3 x}\right)=2\left[-\frac{x e^{-3 x}}{3}-\frac{1}{3} \int e^{-3 x} d x\right]+C \end{aligned}
\begin{aligned} &=>y\left(e^{-3 x}\right)=-\frac{2}{3} x e^{-3 x}-\frac{2}{3} e^{-3 x} \times \frac{1}{3}+C \\\\ &=>y\left(e^{-3 x}\right)=-\frac{2}{3} x e^{-3 x}-\frac{2}{9} e^{-3 x}+C \end{aligned}
Taking $e^{-3 x}$ common on both sides
$=>y=-\frac{2}{3} x-\frac{2}{9}+C \ldots(i)$
If it passes through $\left ( 1,2 \right )$
\begin{aligned} &=>2=-\frac{2}{3} \times 1-\frac{2}{3}+C e^{3} \\\\ &=>C=\frac{26}{9} e^{-3} \end{aligned}
So equation (i) becomes
$=>y e^{-3 x}=\left(-\frac{2}{3} x-\frac{2}{9}\right) e^{-3 x}+\frac{26}{9} e^{-3}$

Differential Equations exercise 21.11 question 19

Answer:$x y=2$
Given: Let $P(x, y)$ be the point of contact of tangent with curve$y=f(x)$ which intercepts on $\mathrm{x} \text {-axis }=2 x$
To find: We have to show that the equation of curve which pass through $\left ( 1,2 \right )$
Hint: Use equation of tangent at point $P\left ( x,y \right )$ is $Y-y=\frac{d y}{d x}(X-x)$
Solution: Equation of tangent at $P\left ( x,y \right )$ is $Y-y=\frac{d y}{d x}(X-x)$
Put Y=0
\begin{aligned} &=>-y=\frac{d y}{d x}(X-x) \\\\ &=>X-x=-y \frac{d x}{d y} \\\\ &=>X=x-y \frac{d x}{d y} \end{aligned}
Co-ordinate of $B\left(x-y \frac{d x}{d y}, 0\right)$
Given( intercept on x-axis) = 2x
\begin{aligned} &=>x-y \frac{d x}{d y}=2 x \\\\ &=>-y \frac{d x}{d y}=2 x-x \\\\ &=>-y \frac{d x}{d y}=x \end{aligned}
$=>-\frac{d x}{x}=\frac{d y}{y}$ [separating variables]
On integrating on both sides, we get
\begin{aligned} &=>-\int \frac{d x}{x}=\int \frac{d y}{y} \\\\ &=>-\log x=\log y+C \ldots(i) \end{aligned}
It is passing through $\left ( 1,2 \right )$
\begin{aligned} &=>-\log 1=\log 2+C \\\\ &=>C=-\log 2 \end{aligned}
Put $C=-\log 2$ in equation (i) we get
\begin{aligned} &=>-\log x=\log y-\log 2 \\\\ &=>\log x^{-1}=\log \left(\frac{y}{2}\right) \quad\quad\quad\left[-\log x=\log x^{-1}\right] \end{aligned}
\begin{aligned} &=>\log \left(\frac{1}{x}\right)=\log \left(\frac{y}{2}\right) \\\\ &=>\frac{1}{x}=\frac{y}{2} \end{aligned}
Hence, $xy=2$ is required for equation of the curve.

Differential Equations exercise 21.11 question 20

Answer:$y=\frac{x}{x+1}(x+\log x-1)$
Given:$x(x+1) \frac{d y}{d x}-y=x(x+1)$
To find: We have to show that the curve which satisfies $x(x+1) \frac{d y}{d x}-y=x(x+1)$ and passes through $\left ( 1,0 \right )$
Hint: Use linear differential equation to solve.
Solution:$x(x+1) \frac{d y}{d x}-y=x(x+1)$
Dividing by $x(x+1)$
$=\frac{d y}{d x}-\frac{y}{x(x+1)}=1$ [This is a linear differential equation]
Comparing with $\frac{d y}{d x}+P y=Q$ we get
$P=\frac{1}{x(x+1)}, Q=1$
$\text { Now If }=e^{\int P d x}$
\begin{aligned} &=e^{-\int \frac{1}{x(x+1)} d x} \\\\ &=e^{-\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x} \\\\ &=e^{-(\log x-\log x+1)} \end{aligned}
\begin{aligned} &=e^{-\log \left(\frac{x}{x+1}\right)} \\\\ &=e^{\log \left(\frac{x}{x+1}\right)^{-1}} \quad\quad\quad\left[e^{-\log x}=\log x^{-1}\right] \end{aligned}
$=\left(\frac{x}{x+1}\right)^{-1} \quad \quad \quad\left[e^{\log x}=x\right]$
$=\frac{x+1}{x}$
So, the solution is given by
\begin{aligned} &=y(I f)=\int Q(I f) d x+C \\\\ &=y \times\left(\frac{x+1}{x}\right)=\int \frac{x+1}{x} d x+C \\\\ &=\left(\frac{x+1}{x}\right) y=\int \frac{x}{x} d x+\int \frac{1}{x} d x+C \end{aligned}
\begin{aligned} &=\left(\frac{x+1}{x}\right) y=\int 1 d x+\int \frac{1}{x} d x+C \\\\ &=\left(\frac{x+1}{x}\right) y=x+\log x+C \ldots(i) \end{aligned}
Since curve passes through the point $\left ( 1,0 \right )$ it satisfies the equation of the curve
$=\left(\frac{1+1}{1}\right) 0=1+\log 1+C$
$=0=1+0+C \quad[\log 1=0]$
$=C=-1$
Substituting value of C in equation (i) we get
\begin{aligned} &=\left(\frac{x+1}{x}\right) y=x+\log x-1 \\\\ &=y=\left(\frac{x}{x+1}\right)(x+\log x-1) \end{aligned}
Hence the required equation for the curve is found.

Differential Equations exercise 21.11 question 21

Answer: $9 y+4 x^{2}=0$
Given: Curve passes through $\left ( 3,-4 \right )$ and given slope of curve $=\frac{2y}{x}$
To find: We have to show that the equation which satisfies the curve.
Hint: Solve of curve $=\frac{2 y}{x}, \frac{d y}{d x}=\frac{2 y}{x}$, solve this to find the equation of the curve.
Solution: Slope of curve $=\frac{2 y}{x}$
\begin{aligned} &=\frac{d y}{d x}=\frac{2 y}{x} \\\\ &=\frac{d y}{y}=\frac{2}{x} d x \end{aligned}
Integrating on both sides
\begin{aligned} &=\int \frac{d y}{y}=2 \int \frac{d x}{x} \\\\ &=\log y=2 \log x+\log C \\\\ &=y=x^{2} C \ldots(i) \end{aligned}
As it passes through $\left ( 3,-4 \right )$
\begin{aligned} &=-4=(3)^{2} C \\\\ &=-4=9 C \\\\ &=C=-\frac{4}{9} \end{aligned}
So equation (i) becomes
\begin{aligned} &=y=-\frac{4}{9} x^{2} \\\\ &=9 y=-4 x^{2} \\\\ &=9 y+4 x^{2}=0 \end{aligned}

Differential Equations exercise 21.11 question 22

Answer:$3(x+3 y)=2\left(1-e^{3 x}\right)$
Given: Equation of slope $x+3 y-1$
To find: We have to show that the equation of the curve which passes through the origin.
Hint: Use linear differential equation to solve i.e. $\frac{d y}{d x}+P y=Q$ and its solution $y \times I f=\int(Q \times I f) d x$ where$If=e^{\int Pdx}$
Solution: Equation of slope $=x+3 y-1$
$=\frac{d y}{d x}=x+3 y=1$
$=\frac{d y}{d x}-3 y=x-1$ [It is linear differential equation]
Comparing with $\frac{d y}{d x}+P y=Q$
\begin{aligned} &P=-3, Q=x-1 \\\\ &=I f=e^{\int P d x} \\\\ &=e^{\int-3 d x} \end{aligned}
\\\begin{aligned} &=e^{-3 \int d x} \\\\ &=e^{-3 x} \end{aligned}
Solution of the given equation is
\begin{aligned} &=y(I f)=\int Q(I f) d x+C \\\\ &=y\left(e^{-3 x}\right)=\int(x-1)\left(x^{-3 x}\right) d x+C \\\\ &=y e^{-3 x}=(x-1) \int e^{-3 x} d x-\int\left[\frac{d}{d x}(x-1) \int e^{-3 x} d x\right] d x+C \end{aligned}[using integration by parts]
\begin{aligned} &=y e^{-3 x}=(x-1)\left(-\frac{1}{3} e^{-3 x}\right)-\int\left(-\frac{e^{-8 x}}{3}\right) d x+C \\\\ &=y e^{-3 x}=-\left(\frac{x-1}{3}\right) e^{-3 x}-\frac{1}{3} \int e^{-3 x} d x+C \end{aligned}
$=y e^{-3 x}=-\left(\frac{x-1}{3}\right) e^{-3 x}-\frac{1}{3}\left(\frac{e^{-s x}}{3}\right)+C$
Taking$e^{-3 x}$ common on both sides,
$=y=-\frac{x}{3}+\frac{1}{3}-\frac{1}{9}+\frac{c}{e^{-3 x}}$
\begin{aligned} &=y=-\frac{x}{3}+\frac{3-1}{9}+C e^{3 x} \\\\ &=y=-\frac{x}{3}+\frac{2}{9}+C e^{3 x} \ldots(i) \end{aligned}
As it passes through origin
\begin{aligned} &=0=-\frac{0}{3}+\frac{2}{9}+C e^{3(0)} \\\\ &=C=-\frac{2}{9} \end{aligned}
Now, substituting in equation (i)
\begin{aligned} &=y=-\frac{x}{3}+\frac{2}{9}-\frac{2}{9} e^{3 x} \\\\ &=y=\frac{-3 x+2-2 e^{8 x}}{9} \\\\ &=9 y=-3 x+2-2 e^{3 x} \end{aligned}
\begin{aligned} &=9 y+3 x=2\left(1-e^{3 x}\right) \\\\ &=3(3 y+x)=2\left(1-e^{3 x}\right) \end{aligned}
Hence, required equation of curve is found.

Differential Equations exercise 21.11 question 23

Answer:$y+1=2 e^{\frac{x^{2}}{2}}$
Given: slope $=x+y x$
Hint: x-coordinate is abscissa and y-coordinate is ordinate
Solution:
According to ques,
\begin{aligned} &\frac{d y}{d x}=x+y x \\\\ &\frac{d y}{d x}=x(1+y) \\\\ &\frac{d y}{1+y}=x d x \end{aligned}
Integrating both sides w.r.t x
$\int \frac{d y}{1+y}=\int x d x$
$\log |1+y|=\frac{x^{2}}{2}+c \quad \text {........(i) } \quad\left[\int \frac{d x}{x}=\log x+c\right]$
Since, the curve passes through (0,1)
\begin{aligned} &\log |1+1|=\frac{0^{2}}{2}+c \\\\ &\mathrm{c}=\log 2 \end{aligned}
Putting the value of c in eqn (i)
\begin{aligned} &\log |1+y|=\frac{x^{2}}{2}+\log 2 \\\\ &\log |1+y|-\log 2=\frac{x^{2}}{2} \quad\quad\quad\quad\left[\log m-\log n=\log _{n}^{m}\right] \end{aligned}
$\begin{gathered} \log \frac{1+y}{2}=\frac{x^{2}}{2} \\\\ \frac{1+y}{2}=e^{\frac{x^{2}}{2}} \\\\ y+1=2 e^{\frac{x^{2}}{2}} \end{gathered}$

Differential Equations exercise 21.11 question 24

Answer: $x^{2}+y^{2}=C x$
Given: A curve is such that the length of the perpendicular from the origin in the tangent at any point P of the curve is equal to the abscissa of P.
To find: The differential equation of the curve $y^{2}-2 x y \frac{d y}{d x}-x^{2}=0$ also we we have to find the curve
Hint: if the differential equation is homogeneous then put $y=v x=\frac{d y}{d x}=v+x \frac{d v}{d x}$
Solution: we have $y^{2}-2 x y \frac{d y}{d x}-x^{2}=0$
$=\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}$
It is a homogeneous differential equation
\begin{aligned} &\text { Put } y=v x \\ &\qquad=\frac{d y}{d x}=v+x \frac{d v}{d x} \end{aligned}
$\text { Now, } x \frac{d v}{d x}+v=\frac{v^{2} x^{2}-x^{2}}{2 x v x}$
$=x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v$ [Taking x common on right hand side]
\begin{aligned} &=x \frac{d v}{d x}=\frac{v^{2}-1-2 v^{2}}{2 v} \\\\ &=x \frac{d v}{d x}=\frac{-v^{2}-1}{2 v} \end{aligned}
\begin{aligned} &=x \frac{d v}{d x}=\frac{-\left(v^{2}+1\right)}{2 v} \\\\ &=\frac{2 v d v}{v^{2}+1}=-\frac{d x}{x} \end{aligned}
Integrating on both sides we get
\begin{aligned} &=\int \frac{2 v d v}{v^{2}+1}=-\int \frac{d x}{x} \\\\ &=\int \frac{2 v d v}{v^{2}+1}=-\int \frac{d x}{x} \quad\left[t=v^{2}+1, d t=2 v d v\right] \\\\ &=\int \frac{d t}{t}=-\int \frac{d x}{x} \end{aligned}
\begin{aligned} &=\log t=-\log x+\log C \\\\ &=\log \left(v^{2}+1\right)=-\log x+\log C \end{aligned}
$=\log \left(v^{2}+1\right)=\log \frac{c}{x} \quad=v^{2}+1=\frac{c}{x}$
$=\left(\frac{y}{x}\right)^{2}+1=\frac{c}{x} \quad\quad\quad\quad\left[y=v x, v=\frac{y}{x}\right]$
\begin{aligned} &=\frac{y^{2}}{x^{2}}+1=\frac{C}{x} \\\\ &=\frac{y^{2}+x^{2}}{x^{2}}=\frac{C}{x} \end{aligned}
\begin{aligned} &=y^{2}+x^{2}=\frac{C x^{2}}{x} \\\\ &=y^{2}+x^{2}=C x \end{aligned}
Differentiating with respect to x
$=2 x+2 y \frac{d y}{d x}-C=0$
$=\frac{d y}{d x}=\frac{C-2 x}{2 y}$
Let $\left ( h,k \right )$ be the point where tangent passes through origin and length is equal to $h$. So, equation of tangent at $(h, k)$ is
\begin{aligned} &=(y-k)=\left(\frac{d y}{d x}\right)_{(h, k)}(x-h) \\\\ &=y-k=\left(\frac{c-2 h}{2 k}\right)(x-h) \end{aligned}
\begin{aligned} &=2 k y-2 k^{2}=x C-2 h x-h C+2 h^{2} \\\\ &=x(C-2 h)-2 k y+2 k^{2}-h C+2 h^{2}=0 \end{aligned}
$=x(C-2 h)-2 k y+2\left(k^{2}+h^{2}\right)-h C=0$
$=x(C-2 h)-2 k y+2(C h)-h C=0 \quad\quad\quad\quad\left[h^{2}+k^{2}=C h \text { on the curve }\right]$
$=x(C-2 h)-2 k y++C h=0$
Length of perpendicular as tangent from origin is
$=L=\left[\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right]$
\begin{aligned} &=\left[\frac{(0)(C-2 h)+(0)(-2 k)+h C}{\sqrt{(C-2 h)^{2}+(-2 k)^{2}}}\right] \\\\ &=\frac{C h}{\sqrt{C^{2}+4 h^{2}+4 k^{2}-4 C h}} \end{aligned}
\begin{aligned} &=\frac{C h}{\sqrt{C^{2}+4\left(h^{2}+k^{2}-C h\right)}} \\\\ &=\frac{C h}{\sqrt{C^{2}+4(0)}} \end{aligned}
$\begin{gathered} =\frac{C h}{\sqrt{C^{2}}} \\\\ =\frac{C h}{C} \\\\ L=C \end{gathered}$
Hence, $x^{2}+y^{2}=C x$ is the required curve.

Differential Equations exercise 21.11 question 25

Answer:$y^{2}=4 x$
Given: Distance between foot of ordinate of the point of contact and the point of intersection of tangent and $x \text {-axis }=2 x$
To find: we have to find the equation of curve which passes through the point $\left ( 1,2 \right )$
Hint: Use equation of tangent at $P(x, y) \text { is }(Y-y)=\frac{d y}{d x}(X-x)$
Solution: Equation of tangent at $P(x,y)$
Put $Y=0$
\begin{aligned} &=-y=\frac{d y}{d x}(X-x) \\\\ &=-y \frac{d x}{d y}=X-x \\\\ &=X=x-y \frac{d x}{d y} \end{aligned}
Let co-ordinate of $B=\left(x-y \frac{d x}{d y}, 0\right)$
Given distance between foot of ordinate of the point of contact and the point of intersection of tangent and $\mathrm{x} \text {-axis }=2 x \text { i.e. } B C=2 x$
$=\sqrt{\left(x-y \frac{d x}{d y}-x\right)^{2}+(0)^{2}}=2 x$ [Distance formula $=\sqrt{\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}}$]
\begin{aligned} &=\sqrt{(-1)^{2}\left(y \frac{d x}{d y}\right)^{2}}=2 x \\\\ &=\sqrt{1\left(y \frac{d x}{d y}\right)^{2}}=2 x \end{aligned}
\begin{aligned} &=y \frac{d x}{d y}=2 x \\\\ &=\frac{d x}{x}=2 \frac{d y}{y} \end{aligned} [Separating variables]
Integration on both sides we get
$=\int \frac{dx}{x}=2 \int \frac{d y}{y}$
\begin{aligned} &=\log x=2 \log y+\log C \ldots(i) \end{aligned}
It passes through $\left ( 1,2 \right )$
$\begin{array}{ll} =\log 1=2 \log 2+\log C \\\\ =0=2 \log 2+\log C & {[\log 1=0]} \end{array}$
\begin{aligned} &=-2 \log 2=\log C \\\\ &=\log (2)^{-2}=\log C \quad\left[2 \log x=\log x^{2}\right] \\\\ &=\log \frac{1}{4}=\log C \end{aligned}
Using antilogarithm
$=C=\frac{1}{4}$
Substituting the value of C in equation (i) we get
\begin{aligned} &=\log x=2 \log y+\log \left(\frac{1}{4}\right) \\\\ &=\log x=\log y^{2}+\log \left(\frac{1}{4}\right) \end{aligned}
\begin{aligned} &=\log x=\log \frac{y^{2}}{4} \\\\ &=x=\frac{y^{2}}{4} \\\\ &=y^{2}=4 x \end{aligned}
Hence $y^{2}=4 x$ is the equation of the curve.

Differential Equations exercise 21.11 question 26

Answer:$x^{2}+y^{2}-6 x-7=0$
Given: Equation of the normal at point $\left ( x,y \right )$ on the curve $(Y-y)=-\frac{d x}{d y}(X-x)$
To find: we have to find the equation of curve which passes through the point $\left ( 3,0 \right )$ if the curve contains the point $\left ( 3,4 \right )$
Hint: Use equation of normal at point $\left ( x,y \right )$ on the curve is $(Y-y)=\frac{d y}{d x}(X-x)$
Solution: Equation of normal on point $\left ( x,y \right )$ on the curve is
$=Y-y=-\frac{d x}{d y}(X-x)$
It is passing through $\left ( 3,0 \right )$
\begin{aligned} &=0-y=-\frac{d x}{d y}(3-x) \\\\ &=y=\frac{d x}{d y}(3-x) \\\\ &=y d y=(3-x) d x \end{aligned}
Integrating on both sides
\begin{aligned} &=\int y d y=\int(3-x) d x \\\\ &=\int y d y=\int 3 d x-\int x d x \\\\ &=\frac{y^{2}}{2}=3 x-\frac{x^{2}}{2}+C \ldots(i) \end{aligned}
It passes through $\left ( 3,4 \right )$
\begin{aligned} &=\frac{4^{2}}{2}=3 \times 3-\frac{3^{2}}{2}+C \\\\ \end{aligned}
$=\frac{16}{2}=9-\frac{9}{2}+C$
\begin{aligned} &=\frac{16}{2}=\frac{18-9}{2}+C \\\\ &=\frac{16}{2}=\frac{9}{2}+C \end{aligned}
\begin{aligned} &=C=\frac{16-9}{2} \\\\ &=C=\frac{7}{2} \end{aligned}
Substituting $C=\frac{7}{2}$ in equation (i)
$=\frac{y^{2}}{2}=3 x-\frac{x^{2}}{2}+\frac{7}{2}$
Multiplying by 2
$=y^{2}=6 x-x^{2}+7$
Hence, the required equation is found.

Differential Equations exercise 21.11 question 28

Given: Radium decomposes at a rate proportional to the quantity of radium present.
To find: The time taken for half the amount of the original amount of radium to decompose.
Hint: The quantity of radium decomposes at the rate of time t i.e. $\frac{d A}{d t} \propto A$ then solve using integration.
Solution: Let A be the quantity of bacteria present in the culture at any time t and initial quantity of bacteria is A0 be the initial quantity of radium.
\begin{aligned} &=\frac{d A}{d t} \propto A \\\\ &=\frac{d A}{d t}=-\lambda A \end{aligned} [λ is proportional constant]
$=\frac{d A}{A}=-\lambda d t$
Integrating on both sides
$=\int \frac{d A}{A}=-\lambda \int \mathrm{dt}$
$=\log A=-\lambda t+C \ldots(i)$
Now, $A=A_{0}$ when $t=0$
\begin{aligned} &=\log A_{0}=0+C \\\\ &=C=\log A_{0} \end{aligned}
Substituting in equation (i)
\begin{aligned} &=\log A=-\lambda t+\log A_{0} \\\\ &=\log A-\log A_{0}=-\lambda t \\\\ &=\log \frac{A}{A_{0}}=-\lambda t \ldots(i i) \end{aligned}
Given that in 25 years radium decomposes at 1.1%
So,
$\begin{gathered} A=(100-1.1) \%=98.9 \% \\ A=0.989 A_{0}, t=25 \end{gathered}$
Substituting in equation (ii)
\begin{aligned} &=\log \frac{0.989 A_{0}}{A_{0}}=-25 \lambda \\\\ &=\lambda=-\frac{1}{25} \log (0.989) \end{aligned}
Now equation (ii) becomes
$=\log \frac{A}{A_{0}}=\left[-\frac{1}{25} \log (0.989)\right] t$
Now $A=\frac{1}{2} A_{0}$
\begin{aligned} &=\log \left(\frac{A}{2 A}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=\log \left(\frac{1}{2}\right)=-\frac{1}{25} \log (0.989) t \end{aligned}
\begin{aligned} &=\log \left(2^{-1}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=-1 \log (2)=-\frac{1}{25} \log (0.989) t \end{aligned}
$=t=\frac{\log 2 \times 25}{\log (0.989)}$
$=t=\frac{0.6931 \times 25}{0.01106} \quad[\log 2=0.6931 \text { And } \log (0.989)=0.01106]$
$=t=1567$
Hence, time taken for radium to decay to half the original amount is 1567 years.

Differential Equations exercise 21.11 question 29

Given: Slope of tangent $=\frac{x^{2}+y^{2}}{2 x y}$.
To prove: The curves which slope at any point $(x, y) \text { is } \frac{x^{2}+y^{2}}{2 x y}$ are in rectangular hyperbola.
Hint: The homogeneous differential equation put $y=v x \frac{d y}{d x}=v+x \frac{d v}{d x}$
Solution: Shape of tangent $=\frac{x^{2}+y^{2}}{2 x y}$
$=\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}$
It is a homogeneous equation
Out $y=vx$
Then,
\begin{aligned} &=v+x \frac{d v}{d x}=\frac{x^{2}+v^{2} x^{2}}{2 v x^{2}} \\\\ &=v+x \frac{d v}{d x}=\frac{x^{2}\left(1+v^{2}\right)}{2 v x^{2}} \\\\ &=v+x \frac{d v}{d x}=\frac{1+v^{2}}{2 v} \end{aligned}
\begin{aligned} &=x \frac{d v}{d x}=\frac{1+v^{2}}{2 v}-v \\\\ &=x \frac{d v}{d x}=\frac{1+v^{2}-2 v^{2}}{2 v} \\\\ &=x \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \end{aligned}
$=\frac{2 v}{1-v^{2}}=\frac{d x}{x}$
$=-\left(\frac{2 v}{v^{2}-1}\right)=\frac{d x}{x}$
$=\frac{2 v}{v^{2}-1}=-\frac{d x}{x}$ [Separating variables]
Integrating on both sides
\begin{aligned} &=\int \frac{2 v}{1-v^{2}} d v=-\int \frac{1}{x} d x \\\\ &=\int \frac{1}{t} d t=-\int \frac{1}{x} d x \quad\left[1-v^{2}=t, 2 v d v=d t\right] \end{aligned}
\begin{aligned} &=\log t=-\log x+\log C \\\\ &=\log \left(1-v^{2}\right)=-\log x+\log C \\\\ &=-\log x=\log \left(1-v^{2}\right)-\log C \\\\ &=-\log x=\log \frac{\left(1-v^{2}\right)}{c} \end{aligned}
\begin{aligned} &=\log x^{-1}=\log \frac{\left(1-v^{2}\right)}{c} \\\\ &=\log \frac{1}{x}=\log \frac{\left(1-v^{2}\right)}{c} \end{aligned}
Using antilogarithm
\begin{aligned} &=\frac{1}{x}=\frac{1-v^{2}}{c} \\\\ &=1-v^{2}=\frac{c}{x} \\\\ &=1-\left(\frac{y}{x}\right)^{2}=\frac{c}{x} \\\\ &=1-\frac{y^{2}}{x^{2}}=\frac{c}{x} \end{aligned}
\begin{aligned} &=\frac{x^{2}-y^{2}}{x^{2}}=\frac{c}{x} \\\\ &=x^{2}-y^{2}=\frac{c x^{2}}{x} \\\\ &=x^{2}-y^{2}=C x \end{aligned}
Hence, $x^{2}-y^{2}=C x$ is the equation for a rectangular hyperbola.

Differential Equations exercise 21.11 question 30

Answer:$x+y=e^{x}-1$
Given: The slope of tangent at each point of a curve is equal to the sum of the co-ordinate of the point i.e. slope of tangent at $(x, y)=x+y$
To find: The curves that pass through the origin.
Hint: As $\frac{d y}{d x}+P y=Q$ then it is a linear differential equation.
Solution: Slope of tangent at $(x, y)=x+y$
\begin{aligned} &=\frac{d y}{d x}=x+y \\\\ &=\frac{d y}{d x}-y=x \end{aligned}
It is a linear differential equation
Comparing it with $\frac{d y}{d x}+P y=Q$
$=P=-1, Q=x$
$=I f=e^{\int \mathrm{P} d x}$
\begin{aligned} &=I f=e^{-\int 1 d x} \\\\ &=I f=e^{-x} \end{aligned}
It is a linear differential equation
Comparing it with $\frac{d y}{d x}+P y=Q$
\begin{aligned} &=y(I f)=\int \mathrm{Q}(I f) d x+C \\\\ &=y\left(e^{-x}\right)=\int x e^{-x} d x+C \end{aligned}
$=y e^{-x}=x \int e^{-x} d x-\left(\frac{d x}{d x} \int e^{-x} d x\right) d x+C$ [using integration by parts]
\begin{aligned} &=y e^{-x}=-x e^{-x}-\int e^{-x} d x+C \\\\ &=y e^{-x}=-x e^{-x}+e^{-x}+C \end{aligned}
Dividing by $e^{-x}$ on both sides
$=y=-x-1+C e^{x} \ldots(i)$
As it passes through origin $(x, y)=(0,0)$
\begin{aligned} &=0=0-1+C e^{0} \\\\ &=C=1 \end{aligned}
Substituting C in equation (i)
\begin{aligned} &=y=-x-1+e^{x} \\\\ &=x+y=e^{x}-1 \end{aligned}
Hence, equation of curve is found

Differential Equations exercise 21.11 question 31

Answer:$y+1=2 e^{\frac{x^{2}}{2}}$
Given: The slope of tangent at each point of a curve is equal to the sum of abscissa and product of abscissa and ordinate of the point i.e. $\frac{d y}{d x}=x+y$
To find: The curves that pass through the point $\left ( 0,1 \right )$
Hint: Use linear differential equation to find the equation of the curve.
Solution: we have $\frac{d y}{d x}=x+y$
$=\frac{d y}{d x}-x y=x \ldots(i)$
This is a linear differential equation of the type
$=\frac{d y}{d x}-P y=Q$
Where $P=-x, Q=x$
\begin{aligned} &=I f=e^{\int p d x} \\\\ &=I f=e^{-\int x d x} \\\\ &=I f=e^{-\frac{x^{2}}{2}} \end{aligned}
So solution of equation is given by
$=y e^{-\frac{x^{2}}{2}}=\int \mathrm{x} e^{-\frac{x^{2}}{2}} d x+C \ldots(i i)$
\begin{aligned} &\text { Let } I=\int \mathrm{x} e^{-\frac{x^{2}}{2}} d x \\\\ &\text { Let }-\frac{x^{2}}{2}=t \end{aligned}
Differentiating on both sides with respect to x
\begin{aligned} &=-\frac{2 x}{2} d x=d t \\\\ &=-x d x=d t \\\\ &=x d x=-d t \end{aligned}
\begin{aligned} &=I=\int e^{-t} d t \\\\ &=I=-e^{-t} \\\\ &=I=-e^{-\frac{x^{2}}{2}} \end{aligned}
Substituting value of $I$ in equation (ii) we get
$=y\left(e^{-\frac{x^{2}}{2}}\right)=-e^{-\frac{x^{2}}{2}}+C$
Dividing by $e^{-\frac{x^{2}}{2}}$
$=y=-1+C e^{\frac{x^{2}}{2}} \ldots(i i i)$
As curve passes through $\left ( 0,1 \right )$
\begin{aligned} &=1=-1+C e^{0} \\\\ &=C=2 \end{aligned}
Substituting value of C in equation (iii) we get
\begin{aligned} &=y=-1+2 e^{\frac{x^{2}}{2}} \\\\ &=y+1=2 e^{\frac{x^{2}}{2}} \end{aligned}
Hence, equation of required curve is found.

Differential Equations exercise 21.11 question 32

Answer:$3 y=x^{3}+4$
Given: The slope of tangent at each point of a curve is equal to the square of the abscissa of the point i.e. slope of tangent at $(x, y)=x^{2}, \frac{d y}{d x}=x+y$
To find: The curves that pass through the point $\left ( -1,1 \right )$
Hint: First separate the given equation of slope then integrate to find the required curve.
Solution: we have $\frac{d y}{d x}=x^{2}$
$=d y=x^{2} d x$
Integrate on both sides
\begin{aligned} &=>\int \mathrm{dy}=\int x^{2} d x \\\\ &=y=\frac{x^{3}}{3}+C \ldots(i) \end{aligned}
As it passes through$\left ( -1,1 \right )$
\begin{aligned} &=1=\frac{(-1)^{3}}{3}+C \\\\ &=1=-\frac{1}{3}+C \end{aligned}
\begin{aligned} &=C=1+\frac{1}{3} \\\\ &=C=\frac{4}{3} \end{aligned}
Substituting value of C in equation (i) we get
\begin{aligned} &=y=\frac{x^{3}}{3}+\frac{4}{3} \\\\ &=3 y=x^{2}+4 \end{aligned}
Hence, the required curve is found.

Differential Equations exercise 21.11 question 33

Answer:$x^{2}-y^{2}=-a^{2}$
Given: The product of the slope and the ordinate is equal to the abscissa i.e. y(slope of tangent)
= x
To find: The equation of the curves that pass through the point $\left ( 0,a \right )$.
Hint: Use $y \times \text { slope of tangent }=x$ to solve.
Solution: we have y(slope of tangent) = x
\begin{aligned} &=y \frac{d y}{d x}=x \\\\ &=y d y=x d x \end{aligned}
Integrating on both sides we get
\begin{aligned} &=\int y d y=\int x d x \\\\ &=\frac{y^{2}}{2}=\frac{x^{2}}{2}+C \ldots(i) \end{aligned}
The curve passes through $\left ( 0,a \right )$
\begin{aligned} &=\frac{a^{2}}{2}=\frac{0^{2}}{2}+C \\\\ &=\frac{a^{2}}{2}=C \\\\ &=C=\frac{a^{2}}{2} \end{aligned}
Substituting in equation (i) we get
$=\frac{y^{2}}{2}=\frac{x^{2}}{2}+\frac{a^{2}}{2}$
Multiply 2 on both sides
\begin{aligned} &=y^{2}=x^{2}+a^{2} \\\\ &=x^{2}-y^{2}=a^{2} \end{aligned}
Hence, required equation of curve is found.

Differential Equations exercise 21.11 question 34

Answer:$x+y \log y=y$
Given: The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact.
To find: The particular curve through the point $\left ( 1,1 \right )$
Hint: Use equation of tangent i.e. $Y-y=\frac{d y}{d x}(X-x)$ and compare with $\frac{d x}{d y}+P y=Q$
Solution: Let $P(x,y)$ be the point in the curve $y=y(x)$ such that the tangent at P cuts the co-ordinate at A and B. We know that the equation of tangent $Y-t=\frac{d y}{d x}(X-x)$
Putting Y=0 then
\begin{aligned} &=-y=\frac{d y}{d x}(X-x) \\\\ &=-y \frac{d x}{d y}+x=X \end{aligned}
Co-ordinate of $B=\left(-y \frac{d x}{d y}+x, 0\right)$
Here, x intercept of tangent = y
$=-y \frac{d x}{d y}+x=y$
$=\frac{d x}{d y}-\frac{x}{y}-1$
This is a linear differential equation
Now comparing with $\frac{d x}{d y}+P y=Q$
$P=\frac{1}{y}, Q=-1$
\begin{aligned} &=I f=e^{\int P d y} \\\\ &=I f=e^{\int \frac{1}{y} d y} \end{aligned}
\begin{aligned} &=I f=e^{\log y} \\\\ &=I f=\frac{1}{y} \end{aligned}
So solution of equation is given by
\begin{aligned} &=x(I f)=\int \mathrm{Q}(I f) d y+C \\\\ &=x\left(\frac{1}{y}\right)=\int(-1)\left(\frac{1}{y}\right) d y+C \\\\ &=\frac{x}{y}=-\log y+C \ldots(i) \end{aligned}
As it passes through the point $\left ( 1,1 \right )$
\begin{aligned} &=\frac{1}{1}=-\log 1+C \\\\ &=1=C \quad[\log 1=0] \end{aligned}
Substituting value of C in equation (i) we get
\begin{aligned} &=\frac{x}{y}=-\log y+1 \\\\ &=x=y(-\log y+1) \\\\ &=x=y-y \log y \\\\ &=x+y \log y=y \end{aligned}
Hence, required equation of curve is found.

The only set of solution books that most schools have availed is the RD Sharma books. Class 12, mathematics, chapter 21, Differential Equations, contains around eleven exercises. The last exercise, ex 21.11, has about 34 questions, including its subparts. The concepts covered in this exercise include basic ideas on differential equations, solving a first-order equation, forming a differential equation, first degree equations, etc. The RD Sharma Class 12 Chapter 21 Exercise 21.11 reference material is enough to solve this section's doubts. Therefore, the students need not worry regarding the challenges they would face while doing their homework.

As it is the last exercise of the chapter, the students must be aware of the previous exercises' sums to understand this exercise effectively. The students can make use of the other RD Sharma books to revise those exercises and the RD Sharma Class 12th Exercise 21.11 book to refer to the sums in this exercise. Everyone who has scored well in their public exams has admitted that the benefits of the RD Sharma books are ineffable. As many questions were picked from the practice section of this book, students who use it get trained for their public exams directly.

The solutions are given in the Class 12 RD Sharma Chapter 21 Exercise 21.11 Solution book updates according to the latest edition of NCERT books. Therefore, students can work out the Differential equations sums by themselves and later recheck the answer with the help of the RD Sharma Class 12th Exercise 21.11 guide. Else, rechecking the sums would be a bit hectic, and they tend to lose marks because of this.

The best way to own the RD Sharma books is by downloading them from the Career 360 website. Be it the RD Sharma Class 12 Solutions Differential Equation Ex 21.11 or the other books; everything is available for free of cost at this site. Many students have reaped the goodness of this book, and now it's your turn to own a copy of the RD Sharma Class 12 Solutions Chapter 21 Ex 21.11 and prepare for your exams.

RD Sharma Chapter wise Solutions

1. Which is the most recommended mathematics guide to refer to the Differential Equations concept?

The RD Sharma Class 12th Exercise 21.11 is the most recommended reference guide for the students to refer to the Differential Equations concept.

2. How can I own a copy of the RD Sharma books for free?

3. Is the last exercise of the Differential Equations Chapter complex for the students to learn?

Exercise 21.11 is the last exercise of the Differential Equations chapter. Students can use the RD Sharma Class 12th Exercise 21.11 reference book to work out the sums effortlessly in this exercise.

4. Are the solutions given in the RD Sharma books well verified?

The answer key present in the RD Sharma solution guides is rechecked and verified to assure its accuracy. Therefore, the students need not worry about such issues

5. Do the RD Sharma books provide solutions for Level 2 questions?

The RD Sharma books contain the solved sums for every question in the Level 1, Level 2, MCQ, FBQ, RE, and even CSBQ section asked in the textbook.

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##### Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
##### Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available
##### Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction.

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions.

3 Jobs Available
##### Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available
##### Bank Probationary Officer (PO)

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

3 Jobs Available
##### Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
##### Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
##### Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.

3 Jobs Available
##### Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available
##### Bank Branch Manager

Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

3 Jobs Available
##### Treasurer

Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
##### Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
##### Architect

Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

2 Jobs Available
##### Landscape Architect

Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared.

2 Jobs Available
##### Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
##### Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
##### Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available
##### Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
##### Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
##### Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

3 Jobs Available
##### Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
##### Chemical Pathologist

Are you searching for a chemical pathologist job description? A chemical pathologist is a skilled professional in healthcare who utilises biochemical laboratory tests to diagnose disease by analysing the levels of various components or constituents in the patient’s body fluid.

2 Jobs Available
##### Biochemical Engineer

A Biochemical Engineer is a professional involved in the study of proteins, viruses, cells and other biological substances. He or she utilises his or her scientific knowledge to develop products, medicines or ways to improve quality and refine processes. A Biochemical Engineer studies chemical functions occurring in a living organism’s body. He or she utilises the observed knowledge to alter the composition of products and develop new processes. A Biochemical Engineer may develop biofuels or environmentally friendly methods to dispose of waste generated by industries.

2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
##### Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
##### Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
##### Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Engineer

A career as Production Engineer is crucial in the manufacturing industry. He or she ensures the functionality of production equipment and machinery to improve productivity and minimize production costs in order to drive revenues and increase profitability.

2 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available
##### Product Designer

Individuals who opt for a career as product designers are responsible for designing the components and overall product concerning its shape, size, and material used in manufacturing. They are responsible for the aesthetic appearance of the product. A product designer uses his or her creative skills to give a product its final outlook and ensures the functionality of the design.

Students can opt for various product design degrees such as B.Des and M.Des to become product designers. Industrial product designer prepares 3D models of designs for approval and discusses them with clients and other colleagues. Individuals who opt for a career as a product designer estimate the total cost involved in designing.

2 Jobs Available
##### R&D Personnel

A career as R&D Personnel requires researching, planning, and implementing new programs and protocols into their organization and overseeing new products’ development. He or she uses his or her creative abilities to improve the existing products as per the requirements of the target market.

2 Jobs Available
##### Commercial Manager

A Commercial Manager negotiates, advises and secures information about pricing for commercial contracts. He or she is responsible for developing financial plans in order to maximise the business's profitability.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Computer System Analyst

Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

2 Jobs Available
##### Test Manager

A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures.

2 Jobs Available
##### Azure Developer

A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service.

2 Jobs Available
##### Deep Learning Engineer

A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

2 Jobs Available