NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 – Free PDF Download

The NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations are provided here with the best possible explanations for every question available in the chapter. This chapter is part of the CBSE Syllabus 2023-24. Students learn about the order and degree of differential equations, the method of solving a differential equation, their properties and much more in this chapter. Solving the problems in the different exercises in this chapter using the NCERT Solutions for Class 12 Maths can help the students create a strong grasp of the concepts of Differential Equations.

Chapter 9 of NCERT Solutions for Class 12 Maths covers the concepts of Differential Equations in easy-to-understand methods, which favours the students in their board exam preparation. The answers for each question provided at BYJU’S can help the students understand how the questions have to be actually answered. Students can download these NCERT Solutions for Class 12 Maths Chapter 9 for free and practise them whenever they wish to.

NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

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

Exercise 9.1 Page: 382

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 1

Solution:

The given differential equation is,

NCERT Solutions for Class 12 Maths Chapter 9 - Image 2

⇒ 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, 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.

NCERT Solutions for Class 12 Maths Chapter 9 - Image 3

So, its degree is one.

NCERT Solutions for Class 12 Maths Chapter 9 - Image 4

So, its degree is not defined.

NCERT Solutions for Class 12 Maths Chapter 9 - Image 5

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.

NCERT Solutions for Class 12 Maths Chapter 9 - Image 6

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

Solution:-

(D) not defined

The given differential equation is,

NCERT Solutions for Class 12 Maths Chapter 9 - Image 8

The highest order derivative present in the differential equation is

\(\begin{array}{l}\frac{d^{2}y}{dx^{2}}\end{array} \)
.

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 9

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

Solution:-

(A) 2

The given differential equation is,

\(\begin{array}{l}2x^{2}\frac{d^{2}y}{dx^{2}}-3\frac{dy}{dx}+y=0\end{array} \)

The highest order derivative present in the differential equation is

\(\begin{array}{l}\frac{d^{2}y}{dx^{2}}\end{array} \)
.

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) are 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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 12

⇒ 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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 13

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 14

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:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 15

NCERT Solutions for Class 12 Maths Chapter 9 - Image 16

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 17

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:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 18

NCERT Solutions for Class 12 Maths Chapter 9 - Image 19

NCERT Solutions for Class 12 Maths Chapter 9 - Image 20

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 21

NCERT Solutions for Class 12 Maths Chapter 9 - Image 22

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

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 23

NCERT Solutions for Class 12 Maths Chapter 9 - Image 24

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

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 25

NCERT Solutions for Class 12 Maths Chapter 9 - Image 26

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Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 28

NCERT Solutions for Class 12 Maths Chapter 9 - Image 29

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

(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 is:

(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.

NCERT Solutions for Class 12 Maths Chapter 9 - Image 30

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 31

2. y2 = a (b2 – x2)

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 32

NCERT Solutions for Class 12 Maths Chapter 9 - Image 33

3. y = ae3x + be-2x

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 34

NCERT Solutions for Class 12 Maths Chapter 9 - Image 35

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 from 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,

NCERT Solutions for Class 12 Maths Chapter 9 - Image 36

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 equations (i) and (iii), we get,

NCERT Solutions for Class 12 Maths Chapter 9 - Image 37

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 38

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

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

Hence, it touches the y-axis at the origin, and 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 a vertex at origin and axis along positive y-axis.

Solution:

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

x2 = 4ay … [equation (i)

NCERT Solutions for Class 12 Maths Chapter 9 - Image 39

NCERT Solutions for Class 12 Maths Chapter 9 - Image 40

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 41

NCERT Solutions for Class 12 Maths Chapter 9 - Image 42

NCERT Solutions for Class 12 Maths Chapter 9 - Image 43

NCERT Solutions for Class 12 Maths Chapter 9 - Image 44

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 the origin.

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 45

NCERT Solutions for Class 12 Maths Chapter 9 - Image 46

NCERT Solutions for Class 12 Maths Chapter 9 - Image 47

⇒ 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 a centre on the y-axis and a radius of 3 units.

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 48

Let us assume the centre of the circle on the 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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 49

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 50

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 51

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 52

NCERT Solutions for Class 12 Maths Chapter 9 - Image 53

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 54

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 55

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 56

Exercise 9.4 Page No: 395

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 57

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 58

NCERT Solutions for Class 12 Maths Chapter 9 - Image 59

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 60

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 81

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

Satisfying the given condition:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 82

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 83

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Solution:

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Solution:

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 95

⇒ 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 96

NCERT Solutions for Class 12 Maths Chapter 9 - Image 97

NCERT Solutions for Class 12 Maths Chapter 9 - Image 98

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 99

NCERT Solutions for Class 12 Maths Chapter 9 - Image 100

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 101

NCERT Solutions for Class 12 Maths Chapter 9 - Image 102

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 103

NCERT Solutions for Class 12 Maths Chapter 9 - Image 104

19. The volume of a 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

the balloon after t seconds.

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 105

NCERT Solutions for Class 12 Maths Chapter 9 - Image 106

20. In a bank, the 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 107

NCERT Solutions for Class 12 Maths Chapter 9 - Image 108

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

of Rs 1000 is deposited with this bank. How much will it be worth after 10 years?

(e0.5 = 1.648).

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 109

 

22. In a culture, the bacteria count is 1,00,000. The number 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 111

NCERT Solutions for Class 12 Maths Chapter 9 - Image 112

NCERT Solutions for Class 12 Maths Chapter 9 - Image 113

NCERT Solutions for Class 12 Maths Chapter 9 - Image 114

Solution:

(A) ex + e-y = C

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 115

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 116

NCERT Solutions for Class 12 Maths Chapter 9 - Image 117

NCERT Solutions for Class 12 Maths Chapter 9 - Image 119

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 120

NCERT Solutions for Class 12 Maths Chapter 9 - Image 121

NCERT Solutions for Class 12 Maths Chapter 9 - Image 122

3. (x – y) dy – (x + y) dx = 0

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 123

NCERT Solutions for Class 12 Maths Chapter 9 - Image 124

NCERT Solutions for Class 12 Maths Chapter 9 - Image 125

NCERT Solutions for Class 12 Maths Chapter 9 - Image 126

4. (x2 – y2)dx + 2xy dy = 0

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 127

NCERT Solutions for Class 12 Maths Chapter 9 - Image 128

NCERT Solutions for Class 12 Maths Chapter 9 - Image 129

NCERT Solutions for Class 12 Maths Chapter 9 - Image 130

On simplification

x2 + y2 = Cx

NCERT Solutions for Class 12 Maths Chapter 9 - Image 131

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 132

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 136

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Solution:

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 147

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 152

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 158

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NCERT Solutions for Class 12 Maths Chapter 9 - Image 160

NCERT Solutions for Class 12 Maths Chapter 9 - Image 161

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 162

NCERT Solutions for Class 12 Maths Chapter 9 - Image 163

NCERT Solutions for Class 12 Maths Chapter 9 - Image 164

NCERT Solutions for Class 12 Maths Chapter 9 - Image 165

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 167

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NCERT Solutions for Class 12 Maths Chapter 9 - Image 169

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 173

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 178

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 182

NCERT Solutions for Class 12 Maths Chapter 9 - Image 183

NCERT Solutions for Class 12 Maths Chapter 9 - Image 184

The required solution of the differential equation.

16. A homogeneous differential equation of the from NCERT Solutions for Class 12 Maths Chapter 9 - Image 185can 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 186

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 187

Exercise 9.6 Page No: 413

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 188

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 189

NCERT Solutions for Class 12 Maths Chapter 9 - Image 190

NCERT Solutions for Class 12 Maths Chapter 9 - Image 191

Solution:

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 205

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 207

NCERT Solutions for Class 12 Maths Chapter 9 - Image 208

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 210

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Solution:

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11. y dx + (x – y2)dy = 0

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 215

NCERT Solutions for Class 12 Maths Chapter 9 - Image 216

NCERT Solutions for Class 12 Maths Chapter 9 - Image 217

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 218

⇒ 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 219

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 220

NCERT Solutions for Class 12 Maths Chapter 9 - Image 221

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 222

NCERT Solutions for Class 12 Maths Chapter 9 - Image 223

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 224

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 225

Now, it is given that the curve passes through the 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 226

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 227

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 228

19. The integrating factor of the differential equation

NCERT Solutions for Class 12 Maths Chapter 9 - Image 229

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 230

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 231

Miscellaneous Exercise Page No: 419

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

(i) (d2y/dx2) + 5x(dy/dx)2 – 6y = log x

(ii) (dy/dx)3 – 4 (dy/dx)2 + 7y = sin x

(iii) (d4y/dx4) – sin (d3y/dx3) = 0

Solution:

(i) (d2y/dx2) + 5x(dy/dx)2 – 6y = log x

Rearranging the given equation, we get

(d2y/dx2) + 5x(dy/dx)2 – 6y – log x = 0

Hence, the highest order derivative present in the given differential equation is d2y/dx2.

Therefore, the order is 2.

Also, the highest power raised to d2y/dx2 is 1.

Hence, the degree is 1.

(ii) (dy/dx)3 – 4 (dy/dx)2 + 7y = sin x

Rearranging the given equation, we get

(dy/dx)3 – 4 (dy/dx)2 + 7y – sin x = 0

Hence, the highest order derivative present in the given differential equation is dy/dx.

Therefore, the order is 1.

And the highest power raised to dy/dx is 3.

Hence, the degree is 3.

(iii) (d4y/dx4) – sin (d3y/dx3) = 0

The highest order derivative present in the given differential equation is d4y/dx4. Hence, the order is 4.

Since the given differential equation is not a polynomial equation, the degree of the equation is not defined.

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

(i) y = aex + be-x + x2: x (d2y/dx2) + 2(dy/dx) – xy + x2 – 2 = 0

(ii) y = ex (a cos x + b sin x): (d2y/dx2) – 2(dy/dx) + 2y = 0

(iii) y = x sin 3x: (d2y/dx2) + 9y – 6 cos 3x = 0

(iv) x2 = 2y2 log y: (x2 + y2)(dy/dx) – xy = 0

Solution:

(i) y = aex + be-x + x2 : x (d2y/dx2) + 2(dy/dx) – xy + x2 – 2 = 0

Given: y = aex + be-x + x2

Differentiate the function with respect to x, and we get

dy/dx = aex – be-x + 2x … (1)

Now, again differentiate with respect to x, and we get

d2y/dx2 = aex + be-x + 2 …(2)

To check whether the given function is the solution of the given differential equation, substitute (1) and (2) in the given differential equation.

L.H.S of the given differential equation

= x (d2y/dx2) + 2(dy/dx) – xy + x2

Now, substituting the values, we get

= x (aex + be-x + 2) + 2 (aex – be-x + 2x) – x (aex + be-x + x2) + x2 – 2

= (xaex + x be-x + 2x) + (2aex – 2be-x + 4x) – (xaex + xbe-x + x3) + x2 – 2

On simplifying the above equation, we get

= 2aex – 2be-x – x3 + x2 + 6x – 2 ≠ 0

Hence, L.H.S ≠ R.H.S.

Therefore, the given function is not a solution to the corresponding differential equation.

(ii) y = ex (a cos x + b sin x): (d2y/dx2) – 2(dy/dx) + 2y = 0

Given: y = ex (a cos x + b sin x)

The given function can be written as follows:

y = ex a cos x + ex b sin x

Differentiating the function on both sides, we get

dy/dx = (a + b)ex cos x + (b – a)ex sin x …(1)

Again, differentiate the above equation on both sides with respect to x, and we get

d2y/dx2 = [(a + b) (d/dx) (ex cos x ] + [(b-a) (d/dx) (ex sin x]

d2y/dx2 = [(a+ b) (ex cos x – ex sin x) ] + [(b- a)(ex sin x + ex cos x)]

d2y/dx2 = ex [(a+ b) (cos x – sin x) + (b- a)(sin x + cos x)]

On simplifying the above equation, we get

d2y/dx2 = 2ex (b cos x – a sin x) …(2)

Now, substitute (1) and (2) in the given differential equation.

L.H.S = (d2y/dx2) – 2(dy/dx) + 2y

= [2ex (b cos x – a sin x)] – 2[(a + b)ex cos x + (b – a)ex sin x] + 2 ex (a cos x + b sin x)

= ex [(2b cos x – 2a sin x)- (2a cos x + 2b cos x) – (2b sin x – 2a sin x) + (2a cos x + 2b sin x)]

= ex [2b cos x – 2a sin x- 2a cos x – 2b cos x – 2b sin x + 2a sin x + 2a cos x + 2b sin x]

= ex [0]

= 0 = R.H.S

As L.H.S = R.H.S, the given function is the solution of the corresponding differential equation.

(iii) y = x sin 3x: (d2y/dx2) + 9y – 6 cos 3x = 0

Given: y = x sin 3x

Now, differentiating the given function with respect to x, and we get

dy/dx = sin 3x + x. cos 3x. 3

dy/dx = sin 3x + 3x cos 3x …(1)

Again differentiate (1) with respect to x, we get

d2y/dx2 = (d/dx) (sin 3x) + 3 (d/dx) (x cos 3x)

d2y/dx2 = 3 cos 3x + 3 [cos 3x + x (- sin 3x). 3]

On simplifying the above equation, we get

d2y/dx2 = 6 cos 3x – 9x sin 3x …(2)

Now, substitute (1) and (2) in the given differential equation, and we get the following:

L.H.S = (d2y/dx2) + 9y – 6 cos 3x

= (6 cos 3x – 9x sin 3x) + 9(x sin 3x) – 6 cos 3x

= 6 cos 3x – 9x sin 3x + 9x sin 3x – 6 cos 3x

= 0 = R.H.S

As L.H.S = R.H.S, the given function is the solution of the corresponding differential equation.

(iv) x2 = 2y2 log y: (x2 + y2)(dy/dx) – xy = 0

Given: x2 = 2y2 log y

Now, differentiate the function with respect to x, and we get

2x = 2 (d/dx) (y2 log y)

On simplifying the above equation, we get

x = (d/dx) (y2 log y)

x = [2y log y .(dy/dx) + y2. (1/y). (dy/dx)]

x = (dy/dx)[2y log y + y]

Hence, we get

dy/dx = x / [y(1 + 2 log y)] …(1)

Now, substitute (1) in the given differential equation.

L.H.S = (x2 + y2)(dy/dx) – xy

= [ 2y2 log y + y2] . [ x / [y(1 + 2 log y)] ] – xy

= [y2(2 log y + 1)] . [ x / [y(1 + 2 log y)] ] – xy

= xy – xy

= 0 = R.H.S

As L.H.S = R.H.S, the given function is the solution of the corresponding differential equation.

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

Solution:

Given equation: (x- a)2 + 2y2 = a2

The given equation can be written as:

⇒ x2 + a2 – 2ax + 2y2 = a2

On rearranging the above equation, we get

⇒ 2y2 = 2ax – x2 …(1)

Now, differentiate equation (1) with respect to x,

⇒ 2 . 2y (dy/dx) = 2a – 2x

⇒ 2y(dy/dx) = (2a – 2x) /2

⇒ dy/dx = (a-x)/2y

⇒ dy/dx = (2ax – 2x2) / 4xy … (2)

From equation (1), we get

2ax = 2y2 + x2

Substitute the value in equation (2), and we get

dy/dx = [2y2 + x2 – 2x2]/4xy

dy/dx = (2y2 – x2) / 4xy

Therefore, the differential equation representing the family of curves given by (x- a)2 + 2y2 = a2 is (2y2 – x2) / 4xy.

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

Solution:

Given differential equation: (x3 – 3xy2 ) dx = (y3 – 3x2y) dy

The equation can be rewritten as:

dy/dx = (x3 – 3xy2) / (y3 – 3x2y) …(1)

The above equation is a homogeneous equation.

To simplify the equation, let us assume y = vx.

⇒ (d/dx) y = (d/dx) (vx)

⇒ dy/dx = v + x(dv/dx) …(2)

Using equations (1) and (2), we get

⇒v + x(dv/dx) = (x3 – 3xy2) / (y3 – 3x2y)

⇒v + x(dv/dx) = (x3 – 3x(vx)2) / ((vx)3 – 3x2(vx))

⇒v + x(dv/dx) = [(1-3v2)/(v3-3v)]

⇒x(dv/dx) = [(1-3v2)/(v3-3v)] – v

On simplifying the above equation, we get

⇒x(dv/dx) = (1-v4)/(v3 – 3v)

Rearranging the above equation,

⇒ [(v3 – 3v)/(1-v4)]dv = (dx/x).

Integrate both sides, and we get

\(\begin{array}{l}\Rightarrow \int \left ( \frac{v^3 – 3v}{1-v^4} \right )dv = log x + log C’ …(3)\end{array} \)
\(\begin{array}{l}\Rightarrow \int \left ( \frac{v^3 – 3v}{1-v^4} \right )dv = \int \frac{v^3dv}{1-v^4} – 3 \int \frac{vdv}{1-v^4}\end{array} \)
\(\begin{array}{l}\Rightarrow \int \left ( \frac{v^3 – 3v}{1-v^4} \right )dv = I_{1} – 3I_{2}…(4)\end{array} \)

Where I1 = ∫ [(v3dv)/(1-v4)] and I2 = ∫ [(vdv)/(1-v4)]

Now, let us assume 1- v4 = t

Hence, we get

⇒ (d/dv) (1-v4) = (dt/dv)

⇒ -4v3 = dt/dv

⇒v3 dv = -dt/4

Now,

I1 = ∫ -(dt/4t) = (-¼) log t = -(¼) log(1-v4) …(5)

Similarly,

I2 = ∫ [(vdv)/(1-v4)] = ∫ [(vdv)/(1-(v2)2)]

Assume that v2 = p

Hence, we get

⇒(d/dv)v2 = dp/dv

⇒ 2v = dp/dv

⇒vdv = dp/2

Now,

I2 = (½) ∫ [dp/(1-p2)]

\(\begin{array}{l}I_{2} = \frac{1}{2 \times 2} log \left|\frac{1 + p}{1 – p} \right|\end{array} \)
\(\begin{array}{l}I_{2} = \frac{1}{4} log \left|\frac{1 + v^2}{1 – v^2} \right| …(6)\end{array} \)

Using equations (4), (5) and (6), we get

\(\begin{array}{l}\int\left ( \frac{v^3 – 3v}{1-v^4} \right )dv = -\frac{1}{4} log (1-v^4) – \frac{3}{4}log \left| \frac{1+v^2}{1-v^2} \right| …(7)\end{array} \)

Now, using equations (2) and (7)

\(\begin{array}{l}-\frac{1}{4} log (1-v^4) – \frac{3}{4}log \left| \frac{1+v^2}{1-v^2} \right| = log x + logc'\end{array} \)
\(\begin{array}{l}-\frac{1}{4} log \left [ (1-v^4)\left| \frac{1+v^2}{1-v^2} \right|^3 \right ] = log C’x\end{array} \)
\(\begin{array}{l}-\frac{1}{4} log \left [ (1-v^2)(1+v^2)\left| \frac{1+v^2}{1-v^2} \right|^3 \right ] = log C’x\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{(1+v^2)^4}{(1-v^2)^2} = (C’x)^{-4}\end{array} \)

Now, replace v with y/x in the above equation and simplify it.

Hence, we get

⇒ (x2 – y2)2 = C’4 (x2 + y2 )4

Now, take the square root on both sides of the above equation, and we get

⇒ (x2 – y2) = C’2 (x2 + y2 )2

⇒ (x2 – y2) = C’ (x2 + y2 )2 , where C = C’2

Hence, x2 – y2 = C (x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2 ) dx = (y3 – 3x2y) dy, it is proved.

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

Solution:

We know that the equation of a circle with centre (a, a) and radius “a” in the first quadrant, which touches the coordinate axes, is:

(x-a)2 + (y-a)2 = a2 …(1)

Circle that touches first quadrant

Differentiating the above equation of both sides with respect to x, we get

⇒ 2(x-a) + 2(y-a)(dy/dx) = 0

Now, the equation can be written as

⇒ (x-a) + (y-a)y’ = 0

⇒ (x-a) + yy’ – ay’ = 0

⇒ x + yy’ -a(1+ y’) = 0

⇒ x + yy’ = a (1 + y’)

Rearranging the above equation, we get

⇒ a = (x + yy’) / (1+y’)

Now, substitute the value of “a” in equation (1), and we get

\(\begin{array}{l}\Rightarrow \left [ x – \left ( \frac{x + yy’}{1+y’} \right ) \right ]^2 + \left [ y – \left ( \frac{x + yy’}{1+y’} \right ) \right ]^2 =\left [ \frac{x + yy’}{1+y’} \right ]^2\end{array} \)
\(\begin{array}{l}\Rightarrow \left [ \frac{(x-y)y’}{(1+y’)} \right ]^2 + \left [ \frac{y-x}{1+y’} \right ]^2 = \left [ \frac{x+yy’}{1+y’} \right]^2\end{array} \)

On simplifying the above equation, we get

⇒ (x-y)2.y’2 + (y-x)2 = (x+ yy’)2

Therefore, the differential equation of the family of circles in the first quadrant, which touches the coordinate axes, is (x-y)2.y’2 + (y-x)2 = (x+ yy’)2.

6. Find the general solution of the differential equation (dy/dx) + √[(1-y2)/(1-x2)] = 0.

Solution:

Given: (dy/dx) + √[(1-y2)/(1-x2)] = 0.

The given differential equation can be written as dy/dx = -√[(1-y2)/(1-x2)]

⇒ dy /√(1-y2) = -dx/√(1-x2)

Now, integrate both sides, and we get

⇒ sin-1 y = – sin-1 x + C

Now, rearrange the equation, and we get

⇒ sin-1 x + sin-1 y = C.

7. Show that the general solution of the differential equation (dy/dx) + [(y2 + y + 1) / (x2 + x + 1)] = 0 is given by (x + y + 1) = A (1 – x – y – 2xy), where A is the parameter.

Solution:

Given differential equation: (dy/dx) + [(y2 + y + 1) / (x2 + x + 1)] = 0

Rearranging the given equation, we get

dy/dx = – [(y2 + y + 1) / (x2 + x + 1)]

⇒ dy/(y2 + y + 1) = – dx/ (x2 + x + 1)

Now, integrate both sides, and we get

⇒ ∫[ dy/(y2 + y + 1)] = -∫ [dx/ (x2 + x + 1)]

\(\begin{array}{l}\Rightarrow \int \frac{dy}{(y + \frac{1}{2})^2 + \left ( \frac{\sqrt{3}}{2} \right )^2} = – \int \frac{dx}{(x + \frac{1}{2})^2 + \left ( \frac{\sqrt{3}}{2} \right )^2}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{2}{\sqrt{3}} tan^{-1}\left [ \frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right ] = -\frac{2}{\sqrt{3}} tan^{-1}\left [ \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right ] + C\end{array} \)
\(\begin{array}{l}\Rightarrow tan^{-1}\left [ \frac{2y + 1}{\sqrt{3}} \right ] + tan^{-1}\left [ \frac{2x + 1}{\sqrt{3}} \right ] = \frac{\sqrt{3}C}{2}\end{array} \)

On simplifying the above equation, we get

\(\begin{array}{l}\Rightarrow \left ( \frac{\frac{2x + 2y + 2}{\sqrt{3}}}{1 – \frac{4xy + 2x + 2y + 1}{3}} \right )= tan \left [ \frac{\sqrt{3}C}{2} \right ]\end{array} \)
\(\begin{array}{l}\Rightarrow \left ( \frac{\frac{2x + 2y + 2}{\sqrt{3}}}{\frac{3 – (4xy + 2x + 2y + 1)}{3}} \right )=C_{1} \ \ \text{Where, }C_{1} =tan \left [ \frac{\sqrt{3}C}{2} \right ]\end{array} \)
[la
\(\begin{array}{l}\Rightarrow \frac{\sqrt{3}(2x + 2y +2)}{3 – (4xy + 2x + 2y + 1)} = C_{1}\end{array} \)

⇒ 2√3 (x + y + 1) = C1 (3 – 4xy + 2x + 2y + 1)

⇒ 2√3 (x + y + 1) = C1 (2 – 4xy – 2x – 2y)

⇒ 2√3 (x + y + 1) = 2C1 (1 – 2xy – x – y)

⇒√3 (x + y + 1) = C1 (1 – 2xy – x – y)

⇒√3 (x + y + 1) = C1 (1 – x – y – 2xy)

⇒ (x + y + 1) = (C1/√3)(1 – x – y – 2xy)

⇒ (x + y + 1) = A (1 – x – y – 2xy), where A = (C1/√3)

Hence, proved.

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:

The given differential equation is sin x cos y dx + cos x sin y dy = 0.

It can also be written as:

⇒ (sin x cos y dx + cos x sin y dy) / cos x cos y = 0.

We know that sin x / cos = tan x,

And simplify the above equation

⇒ tan x dx + tan y dy = 0

⇒ log (sec x ) + log (sec y) = log C

⇒ log (sec x. sec y) = log C

On simplification, we get

sec x sec y = C

It is given that the curve passes through the point (0, π/4).

⇒ 1 × √2 = C

⇒ C = √2

Hence,

sec x × sec y = √2

⇒ sec x × (1/cos y) = √2

⇒ cos y = sec x / √2

Hence, 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 is cos y = sec x /√2.

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:

Given differential equation: (1 + e2x) dy + (1 + y2)ex dx = 0

Rearranging the equation, we get

⇒ [dy/(1+y2)] + [(ex dx)/(1 + e2x)] = 0

Integrating both sides of the equation, we get

tan-1 y + ∫ [(ex dx) / (1+e2x)] = C …(1)

Let ex = t, and hence, e2x = t2

(d/dx) (ex) = (dt/dx)

⇒ ex = dt/dx

⇒ ex dx = dt

Substituting the value in equation (1), we get

tan-1 y + ∫ [(dt) / (1+t2)] = C

⇒ tan-1 y + tan-1t = C

⇒ tan-1 y + tan-1 (ex) = C

If x = 0 and y = 1, we get

⇒ tan-1 1 + tan-1 (e0) = C

⇒ tan-1 1 + tan-1 1 = C

⇒ (π/4) + (π/4) = C

⇒ C = π/2

Hence, tan-1 y + tan-1 (ex) = π/2, which is the particular solution of the given differential equation.

10. Solve the differential equation y ex/y dx = (xex/y + y2)dy ( y ≠ 0).

Solution:

Given: y ex/y dx = (xex/y + y2)dy

Rearranging the given equation, we get

y ex/y (dx/dx) = xex/y + y2

⇒ex/y [y(dx/dy) – x ] = y2

⇒ [ex/y [y(dx/dy) – x ]]/y2 = 1 ….(1)

Assume that ex/y = z

Differentiate with respect to y, and we get

(d/dy)(ex/y) = dz/dy

⇒ex/y [(y(dx/dy) – x)/y2] = dz/dy …(2)

Comparing equations (1) and (2), we get

⇒dz/dy = 1

⇒ dz = dy

Now, integrating both sides, we get

⇒ z = y + C

⇒ ex/y = y + C, which is the solution of the given differential equation.

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:

Given differential equation: (x – y) (dx + dy) = dx – dy

On simplifying the above equation, we get

⇒ (dy/dx) = (1-x+y)/ (x -y +1)

⇒ (dy/dx) = [1 – (x-y)] / [1 + (x-y)] …(1)

Given: x – y = t …(2)

(d/dx) (x – y) = dt/dx

⇒ 1 – (dy/dx) = dt/dx

⇒ 1 – (dt/dx) = dy/dx …(3)

Using the equations (1), (2) and (3), we get

⇒ 1 – (dt/dx) =(1-t)/(1+t)

⇒ dt/dx = 1 – [(1-t)/(1+t)]

On simplification, we get

⇒dt/dx = 2t / (1+t)

⇒ [(1+t)/t]dt = 2 dx

⇒ [ 1 + (1/t)]dt = 2dx

Now, integrating both sides, we get

⇒ t + log |t| = 2x + C

⇒ (x-y) + log |x-y| = 2x + C

⇒ log |x-y| = x + y + C

When x = 0 and y = -1, we get

⇒ log 1 = 0 – 1 + C

⇒ C = 1

Hence, log |x-y| = x + y + 1.

Therefore, log |x-y| = x + y +1 is a particular solution of the differential equation (x – y) (dx + dy) = dx – dy.

12. Solve the differential equation [ (e-2√x /√x ) – (y/√x) ](dx/dy) = 1 (x ≠ 0).

Solution:

Given: [ (e-2√x /√x ) – (y/√x) ](dx/dy) = 1

Rearranging the given equation, we get

⇒ dy/dx = (e-2√x /√x ) – (y/√x)

⇒ (dy/dx) + (y/√x) = (e-2√x /√x )

The above equation is a linear equation of the form (dy/dx) + Py = Q

Where, P = 1/√x and Q = e-2√x/√x

Now, I.F = e∫ P dx = e∫1/√x dx = e2√x

Hence, the general solution of the given differential equation is:

y. (I.F) = ∫ (Q × I.F) dx + C

Now, substituting the values, we get

⇒ ye2√x = ∫ [ (e-2√x/√x) ×e2√x ]dx + C

⇒ye2√x = ∫ (1/√x) dx + C

⇒ye2√x = 2√x +C, which is the solution of the given differential equation.

13. Find a particular solution of the differential equation (dy/dx) + y cot x = 4x cosec x, (x≠ 0), given that y = 0 when x = π/2.

Solution:

Given: (dy/dx) + y cot x = 4x cosec x

The given equation is a linear differential equation of the form (dy/dx) + Py = Q

Where

P = cot x and Q = 4x cosec x

Now, I.F = e∫ P dx = e∫ cot x dx = elog |sin x| = sin x

Hence, the general solution of the given differential equation is:

y. (I.F) = ∫ (Q × I.F) dx + C

⇒ y sin x = ∫ (4x cosec x × sin x) dx + C

⇒ y sin x = 4 ∫x dx + C

⇒ y sin x = 4 (x2/2) + C

⇒ y sin x =2x2 + C

when x = π/2 and y = 0,

Substituting the values in the above equation, we get

⇒ 0 =2(π/2)2 + C

⇒ 0 = 2(π2/4) + C

⇒ 0 = π2/2 + C

⇒ C = – π2/2

Hence, y sin x =2x2 – (π2/2)

Therefore, the particular solution of the differential equation (dy/dx) + y cot x = 4x cosec x is y sin x =2x2 – (π2/2).

14. Find a particular solution of the differential equation (x + 1) (dy/dx) = 2e-y – 1. Given that y = 0 when x = 0.

Solution:

Given differential equation: (x + 1) (dy/dx) = 2e-y – 1

Rearranging the equation, we get

⇒ dy/2e-y – 1 = dx/(x + 1)

⇒ (ey dy)/(2-ey) = dx/(x+1)

Integrate on both sides, we get

∫ [(ey dy)/(2-ey)] = log |x+1| + log C …(1)

Assume that 2-ey = t

⇒ (d/dy) (2-ey) = dt/dy

⇒ -ey = dt/dy

⇒ -ey dy = dt

Substituting the value in equation (1), we get

⇒∫ [(dt)/(t)] = log |x+1| + log C

⇒ – log |t| = log |C (x+1)|

⇒ – log |2-ey| = log |C (x+1)|

⇒ 1/ (2-ey) = C (x+1)

⇒2-ey = 1/[C(x+1)]

When x = 0 and y = 0, we get

⇒ 2 – 1 = 1/C

We get C = 1.

Therefore, 2-ey = 1/[1(x+1)]

⇒2-ey = 1/(x+1)

⇒ ey = 2 – [1/(x+1)]

On simplification, we get

ey = (2x +1) / (x+1)

⇒ y = log |(2x +1) / (x+1)|, where x ≠ -1.

Therefore, the particular solution of the differential equation (x + 1) (dy/dx) = 2e-y -1 is y = log |(2x +1) / (x+1)|, where x ≠ -1.

15. The population of a village increases continuously at a 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:

Let us assume that the population at any instant (t) be y.

Also, given that the rate of Increase in population is proportional to the number of inhabitants at any instant.

(dy/dx) ∝ y

⇒ (dy/dx) = ky

⇒ dy/y = kdt (Where k is a constant)

Now, integrating both sides of the above equation, we get

log y = kt + C … (1)

In 1999, t = 0 and y = 20000, we get log 20000 = C … (2)

In 2004, t = 5 and y = 25000, we get 1og 25000=k.5 + C

⇒ log 25000 = 5k + Iog 20000

⇒ 5k = log(25000/20000) = log(5/4)

⇒ k = (⅕) log(5/4) …(3)

In 2009, t = 10 years.

Substitute the values of k, t and C in (1), and we get

log y = 10 [ (⅕) log(5/4)] + log 20000

On simplification, we get

y = (20000)(5/4)(5/4)

y = 31250

Therefore, the population of the village in 2009 was 31250.

16. The general solution of the differential equation [(y dx – x dy)/y] = 0 is:

  1. xy = C (B) x = Cy2 (C) y = Cx (D) y = Cx2

Solution:

The differential equation is [(y dx – x dy)/y] = 0.

The given equation can be written as:

(ydx / y ) – (xdy/y) = 0

Thus, we get

dx = xdy /y

dx/x = dy/y

(1/x)dx – (1/y)dy = 0

Now, integrating the above equation on both sides, we get

log |x| – log |y| = log k

⇒ log |x/y| = log k

⇒ x/y = k

⇒ y = (1/k) x

⇒ y = Cx [Where C = 1/k]

Hence, the correct answer is option (C) y = Cx.

17. The general solution of a differential equation of the type (dx/dy) + P1x = Q1 is:

  1. \(\begin{array}{l}ye^{\int P_{1}dy} = \int (Q_{1}e^{\int P_{1}dy})dy + C\end{array} \)
  2. \(\begin{array}{l}ye^{\int P_{1}dx} = \int (Q_{1}e^{\int P_{1}dx})dx + C\end{array} \)
  3. \(\begin{array}{l}xe^{\int P_{1}dy} = \int (Q_{1}e^{\int P_{1}dy})dy + C\end{array} \)
  4. \(\begin{array}{l}xe^{\int P_{1}dx} = \int (Q_{1}e^{\int P_{1}dx})dx + C\end{array} \)

Solution:

As we know, the integrating factor of the differential equation (dx/dy) + P1x = Q1 is e∫ P1 dy.

Hence,

⇒ x. (I.F) = ∫ (Q1 × I.F) dy + C

⇒ x. e∫ P1 dy = ∫ (Q1 × e∫ P1 dy) dy + C

Hence, the correct answer is option (C).

\(\begin{array}{l}xe^{\int P_{1}dy} = \int (Q_{1}e^{\int P_{1}dy})dy + C\end{array} \)

18. The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is:

  1. x ey + x2 = C
  2. x ey + y2 = C
  3. y ex + x2 = C
  4. y ey + x2 = C

Solution:

The correct answer is option (C) y ex + x2 = C

Explanation:

The given differential equation is ex dy + (y ex + 2x) dx = 0

⇒ ex (dy/dx) + yex + 2x = 0

Hence, we get

⇒ (dy/dx) + y = -2xe-x.

The above equation is the linear differential equation of the form (dy/dx) + Py = Q, where P = 1 and Q = -2xe-x.

Now,

I.F = e∫ P dx = e∫ dx = ex.

⇒ y. (I.F) = ∫ (Q × I.F) dx + C

⇒ y. ex = ∫ (-2xe-x × ex) dx + C

⇒ yex = ∫-2x dx + C

⇒ yex = -x2 + C

On rearranging the above equation, we get

⇒ yex + x2 = C

Hence, option (C) is the general solution of the given differential equation.


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 include:

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, which 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. Chapter 9 of NCERT Solutions for Class 12 Maths discusses the following:

  1. An equation involving derivatives of the dependent variable with respect to the independent variable (variables) is known as a differential equation.
  2. The order of a differential equation is the order of the highest order derivative occurring in the differential equation.
  3. The degree of a differential equation is defined if it is a polynomial equation in its derivatives.
  4. The 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. The 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 using the NCERT Solutions 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, solutions of differential equations by the method of separation of variables, solutions of homogeneous differential equations of the first order and first degree, and solutions of linear differential equation of the type:

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

Disclaimer – 

Dropped topics – 9.4 Formation of Differential Equations Whose General Solution is Given, Page 415-416 Example 25, Ques. 3, 5 and 15 (Miscellaneous Exercise), Point Six of the Summary

Frequently Asked Questions on NCERT Solutions for Class 12 Maths Chapter 9

Q1

What is a differential equation in Chapter 9 of NCERT Solutions for Class 12 Maths?

The equation which involves derivatives of dependent variables with respect to independent variables is called a differential equation. This is one of the important chapters in Class 12, as these concepts have numerous applications in our daily life. Students are advised to solve the questions in this chapter using the solutions designed by the experts at BYJU’S to score well in Class 12 board exams.
Q2

What is the variable separable method in Chapter 9 of NCERT Solutions for Class 12 Maths?

The variable separable method is used in equations in which the variables can be separated completely. This means the terms having y should remain with dy, and the terms having x should remain with dx. Students can understand this concept more efficiently by practising the solutions framed by the expert faculty at BYJU’S.
Q3

Do NCERT Solutions for Class 12 Maths Chapter 9 have answers for all the textbook questions?

Yes, the NCERT Solutions for Class 12 Maths Chapter 9, available in PDF format, is designed by the subject experts as per the textbook questions. These solutions are completely based on the latest CBSE Syllabus 2023-24 and cover all the important concepts for the board exam. The textbook problems are solved in a stepwise manner as per the marks weightage in the CBSE Board exams. Both chapter-wise and exercise-wise PDF links are available on BYJU’S website, which can be accessed by the students to get their doubts clarified instantly.

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