# NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

The NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations have been provided here with the best possible explanations for every question available in the chapter. In this chapter, students learn about order and degree of differential equations, method of solving a differential equation, their properties and much more. Solving the problems in the different exercises present in the chapter can help the students create a strong grasp over the concept of Differential Equations.

The chapter covers the concepts of Differential Equations in an easy to understand technique, which favours the students in their exam preparation. Students can download these NCERT Solutions for free to practise them whenever they wish to. The answers for each question provided at CoolGyan’S can help the students in understanding how the questions have to be actually answered.

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations

### Access Answers to NCERT Class 12 Maths Chapter 9 – Differential Equations

Exercise 9.1 Page: 382

Determine order and degree (if defined) of differential equations given in Exercises 1 to 10

Solution:

The given differential equation is,

⇒ y”” + sin (y’’’) = 0

The highest order derivative present in the differential equation is y’’’’, so its order is three. Hence, the given differential equation is not a polynomial equation in its derivatives and so, its degree is not defined.

2. y’ + 5y = 0

Solution:

The given differential equation is, y’ + 5y = 0

The highest order derivative present in the differential equation is y’, so its order is one.

Therefore, the given differential equation is a polynomial equation in its derivatives.

So, its degree is one.

So, its degree is one.

So, its degree is not defined.

Therefore, its degree is one.

6. (y’’’)2 + (y’’)3 + (y’)4 + y5 = 0

Solution:

The given differential equation is, (y’’’)2 + (y’’)3 + (y’)4 + y5 = 0

The highest order derivative present in the differential equation is y’’’.

The order is three. Therefore, the given differential equation is a polynomial

equation in y’’’, y’’ and y’.

Then the power raised to y’’’ is 2.

Therefore, its degree is two.

7. y’’’ + 2y’’ + y’ = 0

Solution:

The given differential equation is, y’’’ + 2y’’ + y’ = 0

The highest order derivative present in the differential equation is y’’’.

The order is three. Therefore, the given differential equation is a polynomial

equation in y’’’, y’’ and y’.

Then the power raised to y’’’ is 1.

Therefore, its degree is one.

8. y’ + y = ex

Solution:

The given differential equation is, y’ + y = ex

= y’ + y – ex = 0

The highest order derivative present in the differential equation is y’.

The order is one. Therefore, the given differential equation is a polynomial equation in y’.

Then the power raised to y’ is 1.

Therefore, its degree is one.

9. y’’’ + (y’)2 + 2y = 0

Solution:

The given differential equation is, y’’’ + (y’)2 + 2y = 0

The highest order derivative present in the differential equation is y’’.

The order is two. Therefore, the given differential equation is a polynomial equation in y’’ and y’.

Then the power raised to y’’ is 1.

Therefore, its degree is one.

10. y’’’ + 2y’ + sin y = 0

Solution:-

The given differential equation is, y’’’ + 2y’ + sin y = 0

The highest order derivative present in the differential equation is y’’.

The order is two. Therefore, the given differential equation is a polynomial equation in y’’ and y’.

Then the power raised to y’’ is 1.

Therefore, its degree is one.

11. The degree of the differential equation.

(A) 3 (B) 2 (C) 1 (D) not defined

Solution:-

(D) not defined

The given differential equation is,

The highest order derivative present in the differential equation is .

The order is three. Therefore, the given differential equation is not a polynomial.

Therefore, its degree is not defined.

12. The order of the differential equation

(A) 2 (B) 1 (C) 0 (D) not defined

Solution:-

(A) 2

The given differential equation is,

The highest order derivative present in the differential equation is .

Therefore, its order is two.

Exercise 9.2 Page: 385

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

1. y = ex + 1 : y″ – y′ = 0

Solution:-

From the question it is given that y = ex + 1

Differentiating both sides with respect to x, we get,

⇒ y” = ex

Then,

Substituting the values of y’ and y” in the given differential equations, we get,

y” – y’ = ex – ex = RHS.

Therefore, the given function is a solution of the given differential equation.

2. y = x2 + 2x + C : y′ – 2x – 2 = 0

Solution:-

From the question it is given that y = x2 + 2x + C

Differentiating both sides with respect to x, we get,

y’ = 2x + 2

Then,

Substituting the values of y’ in the given differential equations, we get,

= y’ – 2x -2

= 2x + 2 – 2x – 2

= 0

= RHS

Therefore, the given function is a solution of the given differential equation.

3. y = cos x + C : y′ + sin x = 0

Solution:-

From the question it is given that y = cos x + C

Differentiating both sides with respect to x, we get,

y’ = -sinx

Then,

Substituting the values of y’ in the given differential equations, we get,

= y’ + sinx

= – sinx + sinx

= 0

= RHS

Therefore, the given function is a solution of the given differential equation.

4. y = √(1 + x2): y’ = ((xy)/(1 + x2))

Solution:-

5. y = Ax : xy′ = y (x ≠ 0)

Solution:-

From the question it is given that y = Ax

Differentiating both sides with respect to x, we get,

y’ = A

Then,

Substituting the values of y’ in the given differential equations, we get,

= xy’

= x × A

= Ax

= Y … [from the question]

= RHS

Therefore, the given function is a solution of the given differential equation

6. y = x sinx : xy’ = y + x (√(x2 – y2)) (x ≠ 0 and x>y or x< – y)

Solution:-

Solution:-

8. y – cos y = x : (y sin y + cos y + x) y′ = y

Solution:-

9. x + y = tan-1y : y2 y′ + y2 + 1 = 0

Solution:-

Therefore, the given function is the solution of the corresponding differential equation.

Solution:-

11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Solution:-

(D) 4

The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation.

12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Solution:-

(D) 0

The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.

Exercise 9.3 Page: 391

In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

Solution:-

2. y2 = a (b2 – x2)

Solution:-

3. y = ae3x + be-2x

Solution:-

4. y = e2x (a + bx)

Solution:-

From the question it is given that y = e2x (a + b x)  … [we call it as equation (i)]

Differentiating both sides with respect to x, we get,

y’ = 2e2x(a + b x) + e2x × b … [equation (ii)]

Then, multiply equation (i) by 2 and afterwards subtract it to equation (ii),

We have,

y’ – 2y = e2x(2a + 2bx + b) – e2x (2a + 2bx)

y’ – 2y = 2ae2x + 2e2xbx + e2xb – 2ae2x – 2bxe2x

y’ – 2y = be2x … [equation (iii)]

Now, differentiating equation (iii) both sides,

We have,

⇒ y’’ – 2y = 2be2x … [equation (iv)]

Then,

5. y = ex (a cos x + b sin x)

Solution:

From the question it is given that y = ex(a cos x + b sin x)

… [we call it as equation (i)]

Differentiating both sides with respect to x, we get,

⇒y’ = ex(a cos x + b sin x) + ex(-a sin x + b cos x)

⇒ y’ = ex[(a + b)cos x – (a – b) sin x)] … [equation (ii)]

Now, differentiating equation (ii) both sides,

We have,

y” = ex[(a + b) cos x – (a – b)sin x)] + ex[-(a + b)sin x – (a – b) cos x)]

On simplifying, we get,

⇒ y” = ex[2bcosx – 2asinx]

⇒ y” = 2ex(b cos x – a sin x) … [equation (iii)]

Now, adding equation (i) and (iii), we get,

6. Form the differential equation of the family of circles touching the y-axis at origin.

Solution:

By looking at the figure we can say that the center of the circle touching the y- axis at origin lies on the x – axis.

Let us assume (p, 0) be the centre of the circle.

Hence, it touches the y – axis at origin, its radius is p.

Now, the equation of the circle with centre (p, 0) and radius (p) is

⇒ (x – p)2 + y2 = p2

⇒ x2 + p2 – 2xp + y2 = p2

Transposing p2 and – 2xp to RHS then it becomes – p2 and 2xp

⇒ x2 + y2 = p2 – p2 + 2px

⇒ x2 + y2 = 2px … [equation (i)]

Now, differentiating equation (i) both sides,

We have,

⇒ 2x + 2yy’ = 2p

⇒ x + yy’ = p

Now, on substituting the value of ‘p’ in the equation, we get,

⇒ x2 + y2 = 2(x + yy’)x

⇒ 2xyy’ + x2 = y2

Hence, 2xyy’ + x2 = y2 is the required differential equation.

7. Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Solution:

The parabola having the vertex at origin and the axis along the positive y- axis is

x2 = 4ay … [equation (i)

8. Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Solution:

On simplifying,

⇒ -x (y’)2 – xyy” + yy’ = 0

⇒ xyy” + x (y’)2 – yy’ = 0

Hence, xyy” + x (y’)2 – yy’ = 0 is the required differential equation.

9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Solution:

⇒ x (y’)2 + xyy” – yy’ = 0

⇒ xyy” + x(y’)2 – yy’ = 0

Hence, xyy” + x (y’)2 – yy’ = 0is the required differential equation.

10. Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Solution:

Let us assume the centre of the circle on y – axis be (0, a).

We know that the differential equation of the family of circles with centre at (0, a) and radius 3 is: x2 + (y- a)2 = 32

⇒ x2 + (y- a)2 = 9 … [equation (i)]

Now, differentiating equation (i) both sides with respect to x,

⇒ 2x + 2(y – a) × y’ = 0 … [dividing both side by 2]

⇒ x + (y – a) × y’ = 0

Transposing x to the RHS it becomes – x.

⇒ (y – a) × y’ = x

11. Which of the following differential equations has y = c1 ex + c2 e-x as the general solution?

Solution:

Explanation:

12. Which of the following differential equations has y = x as one of its particular solution?

Solution:

Explanation:

= 0 – (x2 × 1) + (x × x)

= -x2 + x2

= 0

Exercise 9.4 Page No: 395

For each of the differential equations in Exercises 1 to 10, find the general solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

For each of the differential equations in Exercises 11 to 14, find a particular solution

Satisfying the given condition:

Solution:

Solution:

Solution:

Solution:

⇒ c = 1

Putting the value of c in 1

⇒ y = sec x

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x

Solution:

Find the solution curve passing through the point (1, –1).

Solution:

17. Find the equation of a curve passing through the point (0, –2) given that at any

point (x, y) on the curve, the product of the slope of its tangent and y coordinate

of the point is equal to the x coordinate of the point.

Solution:

18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the

line segment joining the point of contact to the point (– 4, –3). Find the equation

of the curve given that it passes through (–2, 1).

Solution:

19. The volume of spherical balloon being inflated changes at a constant rate. If

initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of

balloon after t seconds.

Solution:

20. In a bank, principal increases continuously at the rate of r% per year. Find the

value of r if Rs 100 double itself in 10 years (loge 2 = 0.6931).

Solution:

21. In a bank, principal increases continuously at the rate of 5% per year. An amount

of Rs 1000 is deposited with this bank, how much will it worth after 10 years

(e0.5 = 1.648).

Solution:

22. In a culture, the bacteria count is 1, 00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2, 00,000, if the rate of growth of bacteria is proportional to the number present?

Solution:

Solution:

(A) ex + e-y = C

Explanation:

Exercise 9.5 Page No: 406

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

1. (x2 + x y) dy = (x2 + y2) dx

Solution:

Solution:

3. (x – y) dy – (x + y) dx = 0

Solution:

4. (x2 – y2)dx + 2xy dy = 0

Solution:

On simplification

x2 + y2 = Cx

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

For each of the differential equations in Exercises from 11 to 15, find the particular

solution satisfying the given condition:

11. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

Solution:

12. x2dy + (x y + y2)dx = 0; y = 1 when x = 1

Solution:

Solution:

Solution:

Solution:

The required solution of the differential equation.

16. A homogeneous differential equation of the from can be solved by making the substitution.

(A) y = v x (B) v = y x (C) x = v y (D) x = v

Solution:

(C) x = v y

Explanation:

17. Which of the following is a homogeneous differential equation?

A. (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
B. (x y) dx – (x3 + y3) dy = 0
C. (x3 + 2y2) dx + 2xy dy = 0
D. y2dx + (x2 – x y – y2) dy = 0

Solution:

D. y2dx + (x2 – x y – y2) dy = 0

Explanation:

Exercise 9.6 Page No: 413

For each of the differential equations given in question, find the general solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

8. (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)

Solution:

Solution:

Solution:

11. y dx + (x – y2)dy = 0

Solution:

Solution:

⇒ x = 3y2 + Cy

Therefore, the required general solution of the given differential equation is x = 3y2 + Cy.

For each of the differential equations given in Exercises 13 to 15, find a particular

solution satisfying the given condition:

Solution:

Solution:

Solution:

16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Solution:

Now, it is given that curve passes through origin.

Thus, equation 2 becomes

1 = C

⇒ C = 1

Substituting C = 1 in equation 2, we get,

x + y + 1 = ex

Therefore, the required general solution of the given differential equation is

x + y + 1 = ex

17. Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Solution:

Thus, equation (2) becomes:

0 + 2 – 4 = C e0

⇒ – 2 = C

⇒ C = -2

Substituting C = -2 in equation (2), we get,

x + y – 4 =-2ex

⇒ y = 4 – x – 2ex

Therefore, the required general solution of the given differential equation is

y = 4 – x – 2ex

18. The Integrating Factor of the differential equation is

A. e–x B. e–y C. 1/x D. x

Solution:

C. 1/x

Explanation:

19. The Integrating Factor of the differential equation

Solution:

Explanation:

Miscellaneous Exercise Page No: 419

1. For each of the differential equations given below, indicate its order and degree (if defined).

Solution:

Therefore, its degree is three.

2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

Solution:

3. Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant.

Solution:

4. Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3–3xy2) dx = (y3–3x2y) dy, where c is a parameter.

Solution:

5. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Solution:

We know that the equation of a circle in the first quadrant with centre (a, a) and radius a which touches the coordinate axes is (x -a)2 + (y –a)2 = a2 …………1

Now differentiating above equation with respect to x, we get,

2(x-a) + 2(y-a) dy/dx = 0

⇒ (x – a) + (y – a) y’ = 0

On multiplying we get

⇒ x – a +yy’ – ay’ = 0

⇒ x + yy’ –a (1+y’) = 0

Therefore from above equation we have

6. Find the general solution of the differential equation

Solution:

On integrating, we get,

⇒ sin-1x + sin-1y = C

7. Show that the general solution of the differential equation

is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.

Solution:

8. Find the equation of the curve passing through the point (0, π/4) whose differential equation is sin x cos y dx + cos x sin y dy = 0.

Solution:

9. Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.

Solution:

10. Solve the differential equation

Solution:

11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

Solution:

12. Solve the differential equation

Solution:

13. Find a particular solution of the differential equation

(x ≠ 0), given that y = 0 when x = π/2

Solution:

14. Find a particular solution of the differential equation,

given that y = 0 when x = 0.

Solution:

15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Solution:

16. The general solution of the differential equation   is
A. xy = C B. x = Cy2 C. y = Cx D. y = Cx2

Solution:

C. y = Cx

Explanation:

Given question is

17. The general solution of a differential equation of the type is

Solution:

Explanation:

18. The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
A. x ey + x2 = C B. x ey + y2 = C C. y ex + x2 = C D. y ey + x2 = C

Solution:

C. y ex + x2 = C

Explanation:

 Also Access NCERT Exemplar for Class 12 Maths Chapter 9 CBSE Notes for Class 12 Maths Chapter 9

#### NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations

The major concepts of Maths covered in Chapter 9- Differential Equations of NCERT Solutions for Class 12 includes:

9.1 Introduction

9.2 Basic Concepts

9.2.1 Order of a differential equation

9.2.2 Degree of a differential equation

9.3 General and Particular Solutions of a Differential Equation

9.4 Formation of a Differential Equation whose General Solution is given

9.4.1 Procedure to form a differential equation that will represent a given family of curves

9.5 Methods of Solving First Order, First Degree Differential Equations

9.5.1 Differential equations with variables separable

9.5.2 Homogeneous differential equations

9.5.3 Linear differential equations

Exercise 9.1 Solutions 12 Questions

Exercise 9.2 Solutions 12 Questions

Exercise 9.3 Solutions 12 Questions

Exercise 9.4 Solutions 23 Questions

Exercise 9.5 Solutions 17 Questions

Exercise 9.6 Solutions 19 Questions

Miscellaneous Exercise On Chapter 9 Solutions 18 Questions

## NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations

The chapter Differential Equations belongs to the unit Calculus, that adds up to 35 marks of the total marks. There are 6 exercises along with a miscellaneous exercise in this chapter to help students understand the concepts of Differential Equations clearly. The Chapter 9 of NCERT Solutions for Class 12 Maths discusses the following:

1. An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation.
2. Order of a differential equation is the order of the highest order derivative occurring in the differential equation.
3. Degree of a differential equation is defined if it is a polynomial equation in its derivatives.
4. Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it.
5. A function which satisfies the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called a particular solution.
6. To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants.
7. Variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing y should remain with dy and terms containing x should remain with dx.

### Key Features of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations

Learning the chapter Differential Equations enables the students to understand the following:

Definition, order and degree, general and particular solutions of a differential equation. Formation of differential equation whose general solution is given. Solutions of differential equations by the method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type: