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

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.

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RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

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 π 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 (4 π r^{2})}{d t} = k t \\ = 8 π r \frac{d r}{d t} = k t \end{aligned}

Integrating on both sides

\begin{aligned} = 8 π \frac{r^{2}}{2} = k \frac{t^{2}}{2} + C \\ = 4 π r^{2} = k \frac{t^{2}}{2} + C \dots (i) \end{aligned}

We are given that unit at

t = 0, r = 1

unit

\begin{aligned} = 4 π (1)^{2} = k \times 0 + C \\ = C = 4 π \end{aligned}

And

t = 3, r = 2

units

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

$\begin{aligned} 4 π r^{2} = \frac{8}{3} π \frac{t^{2}}{2} + 4 π \\ 4 π r^{2} - 4 π = \frac{8}{6} π t^{2} \\ 4 π (r^{2} - 1) = \frac{8}{6} π t^{2} \\ π (r^{2} - 1) = \frac{1}{4} \cdot \frac{8}{6} π t^{2} \end{aligned}$
$\begin{aligned} (r^{2} - 1) = \frac{1}{π} \frac{1}{3} π t^{2} \\ r^{2} - 1 = \frac{1}{3} t^{2} \\ r^{2} = 1 + \frac{1}{3} t^{2} = \sqrt{1 + \frac{1}{3} t^{2}} \end{aligned}$

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

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 \dots (i) \end{aligned}

t = 0

we have

P = P_{0}

[Putting

t = 0

and

P = P_{0}

in equation i]

\begin{aligned} => 20 \log (P_{0}) = 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 (P_{0}) \\ => 20 \log P - 20 \log (P_{0}) = t \\ => 20 \log \frac{p}{p_{0}} = t \end{aligned}

When

P = 2 P_{0}

we get

\begin{aligned} = 20 \log (\frac{2 P_{0}}{P_{0}}) \\ = 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,

$\begin{aligned} => 20 \int \frac{d P}{P} = \int k d t \\ => \log P = k t + C \dots (i) \end{aligned}$
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

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

Differential Equations exercise 21.11 question 4

Answer:

\frac{2 \log 2}{\log (\frac{11}{10})} 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} = λ C

and then find the value of lambda using integration.
Solution: The bacteria count is

1, 00, 000

and rate of increase

= 10

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} = λ C

[Where

λ

is the constant]
Integrating on both sides

\begin{aligned} => \int \frac{d C}{d t} = \int λ C \\ => \int \frac{d C}{C} = λ \int d t \\ => \log C = λ t + \log k \dots (i) \end{aligned}

Where

\log \log k

is integral constant
At

t = 0

we have

C = 1, 00, 000

Then

\log (1, 00, 000) = λ \times 0 + \log k

[putting

t = 0

and

C = 1, 00, 000

in equation (i)]

= \log (1, 00, 000) = \log k \dots (i i)

\begin{aligned} At t = 2 \\ \begin{aligned} C & = 1, 00, 000 + 1, 00, 000 (\frac{10}{100}) \\ = 1, 10, 000 \end{aligned} \end{aligned}

From equation (i) we have,

\begin{aligned} = \log 1, 10, 000 = λ \times 2 + \log k [Putting C = 1, 10, 000 and t = 2] \\ = \log 1, 10, 000 = 2 λ + \log k \dots (i i i) \end{aligned}

Now, subtracting equation (ii) from (iii) we have

\begin{aligned} = \log 1, 10, 000 - \log 1, 00, 000 = 2 λ + \log k - \log k \\ = \log (\frac{1, 10, 000}{1, 00, 000}) = 2 λ \\ = \log \frac{11}{10} = 2 λ \\ = λ = \frac{1}{2} \log (\frac{11}{10}) \end{aligned}

Now, we have to find time

t

in which the count reaches

2, 00, 000

\begin{aligned} Now C = 2, 00, 000 in equation (i) \\ = \log (2, 00, 000) = λ t + \log k \end{aligned}

Putting value of lambda and

\log k

, we get

\begin{aligned} = \log (2, 00, 000) = \frac{1}{2} \log (\frac{11}{10}) t + \log 1, 00, 000 \\ = \frac{1}{2} \log (\frac{11}{10}) t = \log 2, 00, 000 - \log 1, 00, 000 \end{aligned}

\begin{aligned} = \frac{1}{2} \log (\frac{11}{10}) t = \log (\frac{2, 00, 000}{1, 00, 000}) \\ = \frac{1}{2} \log (\frac{11}{10}) t = \log 2 \\ = t = \frac{2 \log 2}{\log (\frac{11}{10})} \end{aligned}

Hence, the time required

= \frac{2 \log 2}{\log (\frac{11}{10})}

hours

Differential Equations exercise 21.11 question 5

Answer:

P = R s 1822, t = 12 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 \dots (i) \end{aligned}

At t=0, we have initial principal P=P₀
_{$\begin{aligned} => \log (P_{0}) = 0 + C \\ => C = \log P_{0} \end{aligned}$}
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 (\frac{p}{p_{0}}) = \frac{r t}{100} \end{aligned}

Now, P₀=1000, t=10 years and r=6%

\begin{aligned} => \log (\frac{P}{1000}) = \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 [\log \log e^{x} = x]

\begin{aligned} => \log P = \log (e^{0.6} \times 1000) \\ => \log P = \log (1.822 \times 1000) [\log \log e^{0.6} = 1.822] \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 t₁be the time taken to double Rs 1000 then, P=2000, P0=1000, r=6%
From equation (i)

\begin{aligned} \log (\frac{p}{p_{0}}) = \frac{r t}{100} \\ => \log (\frac{2000}{1000}) = \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 & [\log 2 = 0.6931] \\ = t_{1} = \frac{69.31}{6} \\ = t_{1} = 11.55 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

Then, \begin{aligned} \frac{d A}{d t} & \propto A \\ = \frac{d A}{d t} = λ A \end{aligned}

Where lambda is the integrating constant
Integrating on both sides we get,

\begin{aligned} => \int \frac{d A}{d t} = \int λ A \\ => \int \frac{d A}{A} = λ \int d t \\ => \log A = λ t + C \dots (i) \end{aligned}

Where C is integral constant
When t=0, we have A=A₀
_{$\begin{aligned} => \log A_{0} = 0 + C [Putting t = 0 and A = A_{0} in equation (i)] \\ => C = \log A_{0} \end{aligned}$}
Putting the value of C in equation (i)

\begin{aligned} => \log A = λ t + \log A_{0} \\ => \log A - \log A_{0} = λ t \\ => \log (\frac{A}{A_{0}}) = λ t \dots (i i) \end{aligned}

Since bacteria triples in 5 hours,

So, A = 3 A_{0} and t = 5

Putting in equation (i) we get,

\begin{aligned} = \log (\frac{4 A_{0}}{A_{0}}) = 5 λ \\ = \log 3 = 5 λ \\ = λ = \frac{\log 3}{5} \end{aligned}

Putting

λ = \frac{\log 3}{5}

in equation (ii) we get,

= \log (\frac{A}{A_{0}}) = \frac{\log 3}{5} t \dots (i i i)

Now let A₁ be the amount of bacteria present after 10 hours,

Then, \log (\frac{A_{1}}{A_{0}}) = \frac{\log 3}{5} \times 10

\begin{aligned} = \log (\frac{A_{1}}{A_{0}}) = 2 \log 3 \\ = \log (\frac{A_{1}}{A_{0}}) = 2 \times 1.0986 [\log \log 3 = 1.0986] \end{aligned}

\begin{aligned} = \log (\frac{A_{1}}{A_{0}}) = 2.1972 \\ = \frac{A_{1}}{A_{0}} = e^{2.1972} \\ = A_{1} = A_{0} e^{2.1972} \\ = A_{1} = 9 A_{0} [e^{2.1972} = 9] \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 (\frac{A}{A_{0}}) = \frac{\log 3}{5} \times t \\ = \log (\frac{10 A_{0}}{A_{0}}) = \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

Answer: 312500
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} = λ P

and find the value of constant lambda.
Solution: Let P be the population of the city at time t

Then, \begin{array}{r} \frac{d P}{d t} \propto P \end{array}

= \frac{d P}{d t} = λ P

Where lambda is the proportionality constant

= \frac{d P}{p} = λ d t

Integrating on both sides we get,

\begin{aligned} => \int \frac{d P}{P} = \int λ d t \\ => \log P = λ t + \log k \dots (i) \end{aligned}

Where k is integral constant

\begin{aligned} Whent & = 1990, we have P = 2, 00, 000 \\ => \log 2, 00, 000 = λ \times 1990 + \log k \dots (i i) \end{aligned}

[utting t=0 and A=A0 in equation (i)]
At

\begin{aligned} t = & 2000, we have P = 2, 50, 000 \\ => \log 2, 50, 000 = λ \times 2000 + \log k \dots (i i i) \end{aligned}

Subtracting equation (ii) and (iii)

\begin{aligned} => \log 2, 50, 000 - \log 2, 00, 000 = 2000 λ - 1990 λ \\ => \log (\frac{2, 50, 000}{2, 00, 000}) = 10 λ \\ => \log (\frac{5}{4}) = 10 λ \end{aligned}

=> λ = \frac{1}{10} \log (\frac{5}{4})

Putting the value of lambda in equation (ii) we get,

\begin{aligned} => \log 2, 00, 000 = \frac{1990}{10} \log (\frac{5}{4}) + \log k \\ => \log k = \log 2, 00, 000 - 199 \log (\frac{5}{4}) \end{aligned}

Substituting the value of lambda in equation (ii) we get

\begin{aligned} => \log P = (\frac{1}{10} \log \frac{5}{4}) 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 {(\frac{5}{3})}^{201} + \log 2, 00, 000 - \log {(\frac{5}{4})}^{199} \end{aligned}

\begin{aligned} => \log P = \log [\frac{{(\frac{5}{3})}^{201}}{{(\frac{5}{4})}^{199}}] + \log 2, 00, 000 \\ => \log P = \log [{(\frac{5}{4})}^{201 - 199} \times 2, 00, 000] \\ => \log P = \log [{(\frac{5}{4})}^{2} \times 2, 00, 000] \end{aligned}

\begin{aligned} => \log P = \log [\frac{25}{16} \times 2, 00, 000] \\ => \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^{'} (x) = \frac{d C}{d x} = 2 + 0.15 x 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^{'} (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 d x + 0.15 \int x d x \\ => C = 2 x + \frac{0.15 x^{2}}{2} + D \dots (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

Answer: 8.33%
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 \dots (i) \end{aligned}

[Where C is integral constant]
At t=0, we have initial principal P=P₀
_{$\begin{aligned} => \log (P_{0}) = 0 + C \\ => C = \log P_{0} \end{aligned}$}
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 (\frac{p}{p_{0}}) = \frac{r t}{100} \dots (i i) \end{aligned}

For t=1 and r=8% in equation (ii)

\begin{aligned} => \log (\frac{p}{p_{0}}) = \frac{8}{100} \\ => \log (\frac{p}{p_{0}}) = 0.08 \end{aligned}

\begin{aligned} => \frac{P}{P_{0}} = e^{0.08} \\ => \frac{P}{P_{0}} = 1.0833 [e^{0.08} = 1.0833] \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} [1 - e^{(- \frac{K}{L}) t}]

Given:

L \frac{d i}{d t} + R i = E

To find: To prove that

i = \frac{E}{R} [1 - e^{(- \frac{R}{L}) t}]

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^{'}}, Q = \frac{E}{L}

We know that

\begin{aligned} I.F. = e^{\int p d t} \\ = e^{\int \frac{R}{L} d t} \\ = e^{(\frac{R}{L}) t} \end{aligned}

So, the solution is given by

\begin{aligned} i (I f) = \int Q (I f) d t + C \\ => i [e^{(\frac{R}{L}) t}] = \int \frac{E}{L} [e^{(\frac{R}{L}) t}] d t + C \\ => i [e^{(\frac{R}{L}) t}] = \frac{E}{L} \times \frac{L}{R} [e^{(\frac{R}{L}) t}] + C \end{aligned}

\begin{aligned} => i [e^{(\frac{R}{L}) t}] = \frac{E}{R} [e^{(\frac{R}{L}) t}] + C \\ => i = \frac{\frac{E}{R} [e^{(\frac{R}{L}) t}] + C}{[e^{(\frac{R}{L}) t]}} \end{aligned}

\begin{aligned} => i = \frac{E}{R} + \frac{C}{[e^{(\frac{R}{L}) t}]} \\ => i = \frac{E}{R} + C [e^{- (\frac{R}{L}) t}] \dots (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} [e^{- (\frac{R}{L}) t}] \\ => i = \frac{E}{R} [1 - e^{- (\frac{R}{L}) t}] \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} = - λ A

[Where ? is the constant of proportionality and minus sign indicates that A decrease with increase in t]

=> \frac{d A}{A} = - λ d t

Integrating on both sides

\begin{aligned} => \int \frac{d A}{A} = - λ \int d t \\ => \log A = - λ t + C \dots (i) \end{aligned}

Since initial amount of radium is A₀ then

\begin{aligned} => \log A_{0} = - λ \times 0 + C \\ => \log A_{0} = C \end{aligned}

Putting value of C in equation (i) we get

\begin{aligned} => \log A = - λ t + \log A_{0} \\ => \log A - \log A_{0} = - λ t \\ => \log \frac{A}{A_{0}} = - λ t_{1} \end{aligned}

\begin{aligned} => \log \frac{1}{2} = - λ t_{1} \\ => - \log 2 = - λ t_{1} \\ => t_{1} = \frac{1}{λ} \log 2 \end{aligned}

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

Differential Equations exercise 21.11 question 12

Answer: 0.04%
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} = - λ t

and integrate it to find rate of

λ

Solution: Given that

\begin{aligned} \frac{d A}{d t} \propto A \\ = \frac{d A}{d t} = - λ A \end{aligned}

[Where ?the constant of proportionality and minus sign is indicates ]

= \frac{d A}{A} = - λ d t

[that A decrease with increase in t]
Integrating on both sides

\begin{aligned} = \int \frac{d A}{A} = - λ \int d t \\ = \log A = - λ t + C \end{aligned}

........(i)
Since initial amount of radium is A₀ then

\begin{aligned} = \log A_{0} = - λ \times 0 + C \\ = \log A_{0} = C \end{aligned}

Putting value of C in equation (i) we get

\begin{aligned} = \log A = - λ t + \log A_{0} \\ = \log A - \log A_{0} = - λ t \\ = \log \frac{A}{A_{0}} = - λ 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 (\frac{A}{A_{0}}) = - λ t \\ => \log (\frac{A_{0}}{2 A_{0}}) = - λ \times 1590 \end{aligned}

\begin{aligned} => \log (\frac{1}{2}) = - λ (1590) \\ => - \log 2 = - λ \times 1590 \end{aligned}

Taking minus sign on both sides,

\begin{aligned} => \log 2 = 1590 λ \\ => λ = \frac{\log 2}{1590} \end{aligned}

Put

= λ = \frac{\log 2}{1590}

in equation (ii) we get

\begin{aligned} => \log (\frac{A}{A_{0}}) = - (\frac{\log 2}{1590}) t \\ => \frac{A}{A_{0}} = e^{- (\frac{\log 2}{1590}) t} \dots (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

(3, - 4)

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 \dots (i) \end{aligned}

Where C is constant
Since the curve passes through the point

(3, - 4)

such that

(3, - 4)

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

(2, 2)

Hint: We will integrate the given differential equation and then we use the points

(2, 2)

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 (y - x \frac{d y}{d x}) = \int (y^{2} + \frac{d y}{d x}) \\ => \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 [\frac{1}{y} - \frac{1}{1 - y}] d y = \int \frac{d x}{x + 1} \end{aligned}

=> \log | y | - \log | 1 - y | = \log | x + 1 | + C \dots (i)

Since the curve passing through the point

(2, 2)

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 | - \frac{2}{3} | \end{aligned}

Putting the value of C in equation (i) we get

\begin{aligned} \log | y | - \log | 1 - y | = \log | x + 1 | + \log | - \frac{2}{3} | \\ \log | \frac{y}{1 - y} | = \log | - \frac{2 (x + 1)}{3} | \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 (\frac{y}{x}) = \log (\frac{e}{x})

Given: The angle is

(\frac{y}{x} - \frac{y}{x})

To find: We have to find the equation of the curve which passes through

(1, \frac{π}{4})

Hint: First take the slope of the curve i.e.

\frac{d y}{d x} = \tan θ

and take a linear equation

y = v x

Solution:
The slope of the curve is

\frac{d y}{d x} = \tan θ

We have,

= θ = [\frac{y}{x} - \frac{y}{x}]

\begin{aligned} Then \begin{matrix} \frac{d y}{d x} = tan inverse tan {[\frac{y}{x} - \frac{y}{x}]} \\ = \frac{d y}{d x} = \frac{y}{x} - \frac{y}{x} \end{matrix} \\ 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} [v = \frac{y}{x}]

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 \dots (i i) \end{aligned}

The curve passes through

[1, \frac{π}{4}]

it satisfices equation (ii)

\begin{aligned} = \log \frac{π}{4} = - \log | 1 | + C \\ = C = 1 \end{aligned}

\begin{aligned} Put C = 1, \tan \frac{y}{x} = - \log | x | + 1 \\ = \tan \frac{y}{x} = - \log | x | + 1 \\ = \tan \frac{y}{x} = \log | \frac{e}{x} | \end{aligned}

Hence, required equation is found.

Differential Equations exercise 21.11 question 16

Answer:

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

= 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} \dots (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 (x \times \frac{y}{x}) + \log C = - \frac{x}{y} \\ = 4 \log y + \log C = - \frac{x}{y} \end{aligned}

= \log y^{4} + \log C = - \frac{x}{y} [a \log x = \log x^{a}]

\begin{aligned} = \log (y^{4} C) = - \frac{x}{y} \\ = y^{4} C = e^{- \frac{x}{y}} \end{aligned}

Hence, required curve is found.

Differential Equations exercise 21.11 question 17

Answer: proved.
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

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} => I.F. = e^{\int (- 1) d x} \\ => I.F. = e^{- x} \end{aligned}

Solution of the equation is given by

\begin{aligned} = y \times I.F. = \int Q (I.F.) d x + C \\ => y e^{- x} = \int (2 x) (e^{- x}) d x + C \\ => y e^{- x} = 2 \int (x) (e^{- x}) d x + C \end{aligned}

=> y e^{- x} = 2 [x \int e^{- x} - \int (\frac{d x}{d x} \int e^{- x} d x) d x] + C

[Using integration by parts]

=> y e^{- x} = 2 (- x e^{- x} + \int 1 e^{- x} d x) + C

=> y e^{- x} = 2 (- x e^{- x} - e^{- x}) + C

\begin{aligned} => y e^{- x} = e^{- x} (- 2 x - 2 + C e^{x}) \\ => y = - 2 x - 2 + C e^{x} \\ => y + 2 (x + 1) = C e^{x} \dots (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} = (- \frac{2}{3} x - \frac{2}{9}) 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 θ

and tangent makes angle

(2 x + 3 y)

=> \frac{d y}{d x} = \tan [\tan^{- 1} (2 x + 3 y)]

=> \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} 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 (e^{- 3 x}) = \int 2 x e^{- 3 x} d x + C

Using integration by parts we get

\begin{aligned} => y (e^{- 3 x}) = 2 [x \int e^{- 3 x} - \int (\frac{d x}{d x} \int e^{- 3 x} d x) d x] + C \\ => y (e^{- 3 x}) = 2 [- \frac{x e^{- 3 x}}{3} - \frac{1}{3} \int e^{- 3 x} d x] + C \end{aligned}

\begin{aligned} => y (e^{- 3 x}) = - \frac{2}{3} x e^{- 3 x} - \frac{2}{3} e^{- 3 x} \times \frac{1}{3} + C \\ => y (e^{- 3 x}) = - \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 \dots (i)

If it passes through

(1, 2)

\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} = (- \frac{2}{3} x - \frac{2}{9}) 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

x -axis = 2 x

To find: We have to show that the equation of curve which pass through

(1, 2)

Hint: Use equation of tangent at point

P (x, y)

Y - y = \frac{d y}{d x} (X - x)

Solution: Equation of tangent at

P (x, y)

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 (x - y \frac{d x}{d y}, 0)

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 \dots (i) \end{aligned}

It is passing through

(1, 2)

\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 (\frac{y}{2}) [- \log x = \log x^{- 1}] \end{aligned}

\begin{aligned} => \log (\frac{1}{x}) = \log (\frac{y}{2}) \\ => \frac{1}{x} = \frac{y}{2} \end{aligned}

Hence,

x y = 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

(1, 0)

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

Now If = e^{\int P d x}

\begin{aligned} = e^{- \int \frac{1}{x (x + 1)} d x} \\ = e^{- \int (\frac{1}{x} - \frac{1}{x + 1}) d x} \\ = e^{- (\log x - \log x + 1)} \end{aligned}

\begin{aligned} = e^{- \log (\frac{x}{x + 1})} \\ = e^{\log {(\frac{x}{x + 1})}^{- 1}} [e^{- \log x} = \log x^{- 1}] \end{aligned}

= {(\frac{x}{x + 1})}^{- 1} [e^{\log x} = x]

= \frac{x + 1}{x}

So, the solution is given by

\begin{aligned} = y (I f) = \int Q (I f) d x + C \\ = y \times (\frac{x + 1}{x}) = \int \frac{x + 1}{x} d x + C \\ = (\frac{x + 1}{x}) y = \int \frac{x}{x} d x + \int \frac{1}{x} d x + C \end{aligned}

\begin{aligned} = (\frac{x + 1}{x}) y = \int 1 d x + \int \frac{1}{x} d x + C \\ = (\frac{x + 1}{x}) y = x + \log x + C \dots (i) \end{aligned}

Since curve passes through the point

(1, 0)

it satisfies the equation of the curve

= (\frac{1 + 1}{1}) 0 = 1 + \log 1 + C

= 0 = 1 + 0 + C [\log 1 = 0]

= C = - 1

Substituting value of C in equation (i) we get

\begin{aligned} = (\frac{x + 1}{x}) y = x + \log x - 1 \\ = y = (\frac{x}{x + 1}) (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

(3, - 4)

and given slope of curve

= \frac{2 y}{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 \dots (i) \end{aligned}

As it passes through

(3, - 4)

\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 (1 - e^{3 x})

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

I f = e^{\int P d x}

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 (e^{- 3 x}) = \int (x - 1) (x^{- 3 x}) d x + C \\ = y e^{- 3 x} = (x - 1) \int e^{- 3 x} d x - \int [\frac{d}{d x} (x - 1) \int e^{- 3 x} d x] d x + C \end{aligned}

[using integration by parts]

\begin{aligned} = y e^{- 3 x} = (x - 1) (- \frac{1}{3} e^{- 3 x}) - \int (- \frac{e^{- 8 x}}{3}) d x + C \\ = y e^{- 3 x} = - (\frac{x - 1}{3}) e^{- 3 x} - \frac{1}{3} \int e^{- 3 x} d x + C \end{aligned}

= y e^{- 3 x} = - (\frac{x - 1}{3}) e^{- 3 x} - \frac{1}{3} (\frac{e^{- s x}}{3}) + 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} \dots (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 (1 - e^{3 x}) \\ = 3 (3 y + x) = 2 (1 - e^{3 x}) \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 ........(i) [\int \frac{d x}{x} = \log x + c]

Since, the curve passes through (0,1)

\begin{aligned} \log | 1 + 1 | = \frac{0^{2}}{2} + c \\ 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} [\log m - \log n = \log_{n}^{m}] \end{aligned}

\begin{matrix} \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{matrix}

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} Put y = v x \\ = \frac{d y}{d x} = v + x \frac{d v}{d x} \end{aligned}

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{- (v^{2} + 1)}{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} [t = v^{2} + 1, d t = 2 v d v] \\ = \int \frac{d t}{t} = - \int \frac{d x}{x} \end{aligned}

\begin{aligned} = \log t = - \log x + \log C \\ = \log (v^{2} + 1) = - \log x + \log C \end{aligned}

= \log (v^{2} + 1) = \log \frac{c}{x} = v^{2} + 1 = \frac{c}{x}

= {(\frac{y}{x})}^{2} + 1 = \frac{c}{x} [y = v x, v = \frac{y}{x}]

\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

(h, k)

be the point where tangent passes through origin and length is equal to

h

. So, equation of tangent at

(h, k)

\begin{aligned} = (y - k) = {(\frac{d y}{d x})}_{(h, k)} (x - h) \\ = y - k = (\frac{c - 2 h}{2 k}) (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 (k^{2} + h^{2}) - h C = 0

= x (C - 2 h) - 2 k y + 2 (C h) - h C = 0 [h^{2} + k^{2} = C h on the curve]

= x (C - 2 h) - 2 k y + + C h = 0

Length of perpendicular as tangent from origin is

= L = [\frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}}]

\begin{aligned} = [\frac{(0) (C - 2 h) + (0) (- 2 k) + h C}{\sqrt{(C - 2 h)^{2} + (- 2 k)^{2}}}] \\ = \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 (h^{2} + k^{2} - C h)}} \\ = \frac{C h}{\sqrt{C^{2} + 4 (0)}} \end{aligned}

\begin{matrix} = \frac{C h}{\sqrt{C^{2}}} \\ = \frac{C h}{C} \\ L = C \end{matrix}

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 -axis = 2 x

To find: we have to find the equation of curve which passes through the point

(1, 2)

Hint: Use equation of tangent at

P (x, y) 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 = (x - y \frac{d x}{d y}, 0)

Given distance between foot of ordinate of the point of contact and the point of intersection of tangent and

x -axis = 2 x i.e. B C = 2 x

= \sqrt{{(x - y \frac{d x}{d y} - x)}^{2} + (0)^{2}} = 2 x

[Distance formula

= \sqrt{{(x - x_{1})}^{2} + {(y - y_{1})}^{2}}

]

\begin{aligned} = \sqrt{(- 1)^{2} {(y \frac{d x}{d y})}^{2}} = 2 x \\ = \sqrt{1 {(y \frac{d x}{d y})}^{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{d x}{x} = 2 \int \frac{d y}{y}

\begin{aligned} = \log x = 2 \log y + \log C \dots (i) \end{aligned}

It passes through

(1, 2)

\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 [2 \log x = \log x^{2}] \\ = \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 (\frac{1}{4}) \\ = \log x = \log y^{2} + \log (\frac{1}{4}) \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

(x, y)

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

(3, 0)

if the curve contains the point

(3, 4)

Hint: Use equation of normal at point

(x, y)

on the curve is

(Y - y) = \frac{d y}{d x} (X - x)

Solution: Equation of normal on point

(x, y)

on the curve is

= Y - y = - \frac{d x}{d y} (X - x)

It is passing through

(3, 0)

\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 \dots (i) \end{aligned}

It passes through

(3, 4)

\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

Answer: 1567 years
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} = - λ A \end{aligned}

[λ is proportional constant]

= \frac{d A}{A} = - λ d t

Integrating on both sides

= \int \frac{d A}{A} = - λ \int dt

= \log A = - λ t + C \dots (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 = - λ t + \log A_{0} \\ = \log A - \log A_{0} = - λ t \\ = \log \frac{A}{A_{0}} = - λ t \dots (i i) \end{aligned}

Given that in 25 years radium decomposes at 1.1%
So,

\begin{matrix} A = (100 - 1.1) % = 98.9 % \\ A = 0.989 A_{0}, t = 25 \end{matrix}

Substituting in equation (ii)

\begin{aligned} = \log \frac{0.989 A_{0}}{A_{0}} = - 25 λ \\ = λ = - \frac{1}{25} \log (0.989) \end{aligned}

Now equation (ii) becomes

= \log \frac{A}{A_{0}} = [- \frac{1}{25} \log (0.989)] t

Now

A = \frac{1}{2} A_{0}

\begin{aligned} = \log (\frac{A}{2 A}) = - \frac{1}{25} \log (0.989) t \\ = \log (\frac{1}{2}) = - \frac{1}{25} \log (0.989) t \end{aligned}

\begin{aligned} = \log (2^{- 1}) = - \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} [\log 2 = 0.6931 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

Answer:

Given: Slope of tangent

= \frac{x^{2} + y^{2}}{2 x y}

.
To prove: The curves which slope at any point

(x, y) 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 = v x

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} (1 + v^{2})}{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}

= - (\frac{2 v}{v^{2} - 1}) = \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 [1 - v^{2} = t, 2 v d v = d t] \end{aligned}

\begin{aligned} = \log t = - \log x + \log C \\ = \log (1 - v^{2}) = - \log x + \log C \\ = - \log x = \log (1 - v^{2}) - \log C \\ = - \log x = \log \frac{(1 - v^{2})}{c} \end{aligned}

\begin{aligned} = \log x^{- 1} = \log \frac{(1 - v^{2})}{c} \\ = \log \frac{1}{x} = \log \frac{(1 - v^{2})}{c} \end{aligned}

Using antilogarithm

\begin{aligned} = \frac{1}{x} = \frac{1 - v^{2}}{c} \\ = 1 - v^{2} = \frac{c}{x} \\ = 1 - {(\frac{y}{x})}^{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 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 Q (I f) d x + C \\ = y (e^{- x}) = \int x e^{- x} d x + C \end{aligned}

= y e^{- x} = x \int e^{- x} d x - (\frac{d x}{d x} \int e^{- x} d x) 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} \dots (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

(0, 1)

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 \dots (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 x e^{- \frac{x^{2}}{2}} d x + C \dots (i i)

\begin{aligned} Let I = \int x e^{- \frac{x^{2}}{2}} d x \\ 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 (e^{- \frac{x^{2}}{2}}) = - e^{- \frac{x^{2}}{2}} + C

Dividing by

e^{- \frac{x^{2}}{2}}

= y = - 1 + C e^{\frac{x^{2}}{2}} \dots (i i i)

As curve passes through

(0, 1)

\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

(- 1, 1)

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 dy = \int x^{2} d x \\ = y = \frac{x^{3}}{3} + C \dots (i) \end{aligned}

As it passes through

(- 1, 1)

\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

(0, a)

.
Hint: Use

y \times 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 \dots (i) \end{aligned}

The curve passes through

(0, a)

\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

(1, 1)

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 = (- y \frac{d x}{d y} + x, 0)

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 Q (I f) d y + C \\ = x (\frac{1}{y}) = \int (- 1) (\frac{1}{y}) d y + C \\ = \frac{x}{y} = - \log y + C \dots (i) \end{aligned}

As it passes through the point

(1, 1)

\begin{aligned} = \frac{1}{1} = - \log 1 + C \\ = 1 = C [\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

Frequently Asked Questions (FAQs)

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?

To get a copy of the RD Sharma books for free of cost, you can download it from the Career 360 website.

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.