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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.

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Differential Equations exercise 21.11 question 1

Answer:Given: It is given that initial radius of balloon unit. After 3 seconds, radius of balloon units.

To find: We have to find the radius after time .

Hint: Use surface area of a balloon . Then differentiate it to find the radius.

Solution: We have radius of the balloon unit

After 3 seconds, radius of balloon units

Let be the radius and be the surface area of the balloon at any time .

where is constant.

Integrating on both sides

We are given that unit at unit

And units

By cross multiplication we get

Now substitute and in equation (i)

Differential Equations exercise 21.11 question 2

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.

Solution: Population growth rate =5% per year

Let initial population be

Then,

By cross multiplication we get,

Integrating on both sides

At we have [Putting and in equation i]

Putting in equation (i)

When we get

Hence, the required time period is years.

Differential Equations exercise 21.11 question 3

Answer: yearsGiven: Present Population

To find: When the city will have a population of

Hint: The population of city increase in time i.e. and then find the equation using integration.

Solution: The present population is and the population of a city doubled in the past years

Let be the population at any time

Then,

Integrating on both sides we get,

At we take

Then,

[Putting and in equation (i)]

Putting * in equation () we get *

When at we have

Putting in equation (ii) we get

We assume that be the time take for the population to become from

Then,

Therefore the required time is years (approximate).

Differential Equations exercise 21.11 question 4

Answer:Given: Present bacteria count

To find: The amount of hours taken for the count to reach

Hint: Use and then find the value of lambda using integration.

Solution: The bacteria count is and rate of increase in hours

Let be the bacteria count at any time ,

Then

[Where is the constant]

Integrating on both sides

Where is integral constant

At we have

Then [putting and in equation (i)]

From equation (i) we have,

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

Now, we have to find time in which the count reaches

Putting value of lambda and , we get

Hence, the time required hours

Differential Equations exercise 21.11 question 5

Answer: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.

Solution: Let P be the principal

Integrating on both sides we get,

At t=0, we have initial principal P=P

Now substituting value of C in equation (i)

Now, P

[Using anti-logarithm on both sides]

Therefore P=Rs 1822

Thus Rs 1000 will be Rs 1822 in 10 years.

Now let t

From equation (i)

[By cross multiplication]

years (approximately)

Hence, it will take 12 years to double the amount.

Differential Equations exercise 21.11 question 6

Answer: 9 times,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.

Solution: Let A be the amount of bacteria present at time t and A

Where lambda is the integrating constant

Integrating on both sides we get,

Where C is integral constant

When t=0, we have A=A

Putting the value of C in equation (i)

Since bacteria triples in 5 hours,

Putting in equation (i) we get,

Putting in equation (ii) we get,

Now let A

Hence, after 10 hours the number of bacteria will be 9 times the original.

Now, let t

Then equation (iii) is

Hence, the time needed for the bacteria to become 10 times the initial amount hours

Differential Equations exercise 21.11 question 7

Answer: 312500Given: 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 and find the value of constant lambda.

Solution: Let P be the population of the city at time t

Where lambda is the proportionality constant

Integrating on both sides we get,

Where k is integral constant

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

At

Subtracting equation (ii) and (iii)

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

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

Thus, the population of city in 2010=3,12,500

Differential Equations exercise 21.11 question 8

Answer:Given:

To find: Total cost function

Hint: Use integration and then find out the value of integral constant

Solution: Given that

Integrating on both sides we get,

We have C=100 when x=0

Putting the value of D in equation (i) we get,

Thus the total cost function

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.

Solution: Let P be the principal

Integrating on both sides we get,

[Where C is integral constant]

At t=0, we have initial principal P=P

Now substituting value of C in equation (i)

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

Subtracting 1 on both sides we get

Hence, amount increases by 8.33% in one year.

Differential Equations exercise 21.11 question 10

Answer:Given:

To find: To prove that

Hint: Integrate the linear differential equation to find integral factor.

Solution:

Dividing by L

This is a linear differential equation and comparing it with

Such that,

We know that

So, the solution is given by

Initially no current passes through the circuit so,i=0, t=0

Substituting values of C in equation (i)

Differential Equations exercise 21.11 question 11

Answer: 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.

Solution: Given that

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

Integrating on both sides

Since initial amount of radium is A

Putting value of C in equation (i) we get

Hence, the time taken 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 and integrate it to find rate of

Solution: Given that

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

[that A decrease with increase in t]

Integrating on both sides

........(i)

Since initial amount of radium is A

Putting value of C in equation (i) we get

................(ii)

Given that its half-life is 1590 years so we put and t=1590

Hence, equation (ii) becomes

Taking minus sign on both sides,

Put in equation (ii) we get

Putting t=1 in (iii) to find the amount of radium after one year, we get

Percentage amount disappeared in one year

Differential Equations exercise 21.11 question 13

Answer:Given: The slope of tangent at a point

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

Hint: First take the slope of the curve and integrate the equation.

Solution: So,

Integrating on both sides we get,

Where C is constant

Since the curve passes through the point such that satisfy equation (i)

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

[Multiplying by 2 on both sides]

Hence, found.

Differential Equations exercise 21.11 question 14

Answer:Given: The differential equation is

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

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

Solution: we have,

Integrating on both sides we get,

Since the curve passing through the point it satisfies equation (i)

Then,

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

Taking antilogarithm on both sides,

Hence, required equation is found.

Differential Equations exercise 21.11 question 15

Answer:Given: The angle is

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

Hint: First take the slope of the curve i.e. and take a linear equation

Solution:

The slope of the curve is

We have,

Differentiating with respect to we get

[From equation i]

Integrating on both sides

The curve passes through it satisfices equation (ii)

Hence, required equation is found.

Differential Equations exercise 21.11 question 16

Answer:Given: The curve is . Suppose 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. then solve the differential equation.

Solution: Equation of tangent to the curve at is

Where (X, Y) is the arbitrary point on the tangent

Putting Y=0 we get

When cut off by the tangent on the x-axis

Therefore

This is homogeneous differential equation

Putting and in (i)

We get

Integrating on both sides,

Substituting value of we get

Hence, required curve is found.

Differential Equations exercise 21.11 question 17

Answer: proved.Given: Slope at any point

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

Hint: Use the linear differential equation i.e.

Solution: Slope at any point

It is a linear differential equation comparing it with

Integrating factor

Solution of the equation is given by

[Using integration by parts]

If it passes through origin

Now equation (i) becomes

Differential Equations exercise 21.11 question 18

Answer:Given: Tangent makes an angle with x-axis

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

Hint: Find the slope of the tangent then solve using linear differential equation.

Solution: Slope of tangent and tangent makes angle

It is a linear differential equation

Comparing it with

Solution of equation is given by

Using integration by parts we get

Taking common on both sides

If it passes through

So equation (i) becomes

Differential Equations exercise 21.11 question 19

Answer:Given: Let be the point of contact of tangent with curve which intercepts on

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

Hint: Use equation of tangent at point is

Solution: Equation of tangent at is

Put Y=0

Co-ordinate of

Given( intercept on x-axis) = 2x

[separating variables]

On integrating on both sides, we get

It is passing through

Put in equation (i) we get

Hence, is required for equation of the curve.

Differential Equations exercise 21.11 question 20

Answer:Given:

To find: We have to show that the curve which satisfies and passes through

Hint: Use linear differential equation to solve.

Solution:

Dividing by

[This is a linear differential equation]

Comparing with we get

So, the solution is given by

Since curve passes through the point it satisfies the equation of the curve

Substituting value of C in equation (i) we get

Hence the required equation for the curve is found.

Differential Equations exercise 21.11 question 21

Answer:Given: Curve passes through and given slope of curve

To find: We have to show that the equation which satisfies the curve.

Hint: Solve of curve , solve this to find the equation of the curve.

Solution: Slope of curve

Integrating on both sides

As it passes through

So equation (i) becomes

Differential Equations exercise 21.11 question 22

Answer:Given: Equation of slope

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. and its solution where

Solution: Equation of slope

[It is linear differential equation]

Comparing with

Solution of the given equation is

[using integration by parts]

Taking common on both sides,

As it passes through origin

Now, substituting in equation (i)

Hence, required equation of curve is found.

Differential Equations exercise 21.11 question 23

Answer:Given: slope

Hint: x-coordinate is abscissa and y-coordinate is ordinate

Solution:

According to ques,

Integrating both sides w.r.t x

Since, the curve passes through (0,1)

Putting the value of c in eqn (i)

Differential Equations exercise 21.11 question 24

Answer: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 also we we have to find the curve

Hint: if the differential equation is homogeneous then put

Solution: we have

It is a homogeneous differential equation

[Taking x common on right hand side]

Integrating on both sides we get

Differentiating with respect to x

Let be the point where tangent passes through origin and length is equal to . So, equation of tangent at is

Length of perpendicular as tangent from origin is

Hence, is the required curve.

Differential Equations exercise 21.11 question 25

Answer:Given: Distance between foot of ordinate of the point of contact and the point of intersection of tangent and

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

Hint: Use equation of tangent at

Solution: Equation of tangent at

Put

Let co-ordinate of

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

[Distance formula ]

[Separating variables]

Integration on both sides we get

It passes through

Using antilogarithm

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

Hence is the equation of the curve.

Differential Equations exercise 21.11 question 26

Answer:Given: Equation of the normal at point on the curve

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

Hint: Use equation of normal at point on the curve is

Solution: Equation of normal on point on the curve is

It is passing through

Integrating on both sides

It passes through

Substituting in equation (i)

Multiplying by 2

Hence, the required equation is found.

Differential Equations exercise 21.11 question 28

Answer: 1567 yearsGiven: 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. 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 A

[λ is proportional constant]

Integrating on both sides

Now, when

Substituting in equation (i)

Given that in 25 years radium decomposes at 1.1%

So,

Substituting in equation (ii)

Now equation (ii) becomes

Now

Hence, time taken for radium to decay to half the original amount is 1567 years.

Differential Equations exercise 21.11 question 29

To prove: The curves which slope at any point are in rectangular hyperbola.

Hint: The homogeneous differential equation put

Solution: Shape of tangent

It is a homogeneous equation

Out

Then,

[Separating variables]

Integrating on both sides

Using antilogarithm

Hence, is the equation for a rectangular hyperbola.

Differential Equations exercise 21.11 question 30

Answer: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

To find: The curves that pass through the origin.

Hint: As then it is a linear differential equation.

Solution: Slope of tangent at

It is a linear differential equation

Comparing it with

It is a linear differential equation

Comparing it with

[using integration by parts]

Dividing by on both sides

As it passes through origin

Substituting C in equation (i)

Hence, equation of curve is found

Differential Equations exercise 21.11 question 31

Answer: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.

To find: The curves that pass through the point

Hint: Use linear differential equation to find the equation of the curve.

Solution: we have

This is a linear differential equation of the type

Where

So solution of equation is given by

Differentiating on both sides with respect to x

Substituting value of in equation (ii) we get

Dividing by

As curve passes through

Substituting value of C in equation (iii) we get

Hence, equation of required curve is found.

Differential Equations exercise 21.11 question 32

Answer: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

To find: The curves that pass through the point

Hint: First separate the given equation of slope then integrate to find the required curve.

Solution: we have

Integrate on both sides

As it passes through

Substituting value of C in equation (i) we get

Hence, the required curve is found.

Differential Equations exercise 21.11 question 33

Answer: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 .

Hint: Use to solve.

Solution: we have y(slope of tangent) = x

Integrating on both sides we get

The curve passes through

Substituting in equation (i) we get

Multiply 2 on both sides

Hence, required equation of curve is found.

Differential Equations exercise 21.11 question 34

Answer: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

Hint: Use equation of tangent i.e. and compare with

Solution: Let be the point in the curve such that the tangent at P cuts the co-ordinate at A and B. We know that the equation of tangent

Putting Y=0 then

Co-ordinate of

Here, x intercept of tangent = y

This is a linear differential equation

Now comparing with

So solution of equation is given by

As it passes through the point

Substituting value of C in equation (i) we get

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**

- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

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.

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