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 explanation 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 have a strong grasp over the concept of Differential Equations.

The chapter covers the concepts of Differential Equations in an easy to understand the technique to favour the students in their exam preparation. Students can download these NCERT solutions for free to practice them whenever they wish to. The answers for each question provided at BYJU’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

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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

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 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.

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
NCERT Solutions for Class 12 Maths Chapter 9 - Image 7

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

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,

NCERT Solutions for Class 12 Maths Chapter 9 - Image 11NCERT Solutions for Class 12 Maths Chapter 9 - Image 10

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,

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:-

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NCERT Solutions for Class 12 Maths Chapter 9 - Image 20

Solution:-

NCERT Solutions for Class 12 Maths Chapter 9 - Image 21

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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

Therefore, the given function is the solution of the corresponding differential equation.
NCERT Solutions for Class 12 Maths Chapter 9 - Image 27

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 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.

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:-

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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 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,

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 equation (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 origin.

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 38

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)

NCERT Solutions for Class 12 Maths Chapter 9 - Image 39

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 41

NCERT Solutions for Class 12 Maths Chapter 9 - Image 42

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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:

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 centre on y-axis and radius 3 units.

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 48

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

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:

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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?

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 55

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 56

= 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 57

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 58

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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:

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Solution:

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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:

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⇒ 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

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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 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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 105

NCERT Solutions for Class 12 Maths Chapter 9 - Image 106

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 107

NCERT Solutions for Class 12 Maths Chapter 9 - Image 108

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 109

NCERT Solutions for Class 12 Maths Chapter 9 - Image 110

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:

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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:

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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

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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

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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

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On simplification

x2 + y2 = Cx

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Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 132

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Solution:

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Solution:

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Solution:

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Solution:

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NCERT Solutions for Class 12 Maths Chapter 9 - Image 155

NCERT Solutions for Class 12 Maths Chapter 9 - Image 156

NCERT Solutions for Class 12 Maths Chapter 9 - Image 157

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 158

NCERT Solutions for Class 12 Maths Chapter 9 - Image 159

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 166

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 167

NCERT Solutions for Class 12 Maths Chapter 9 - Image 168

NCERT Solutions for Class 12 Maths Chapter 9 - Image 169

NCERT Solutions for Class 12 Maths Chapter 9 - Image 170

NCERT Solutions for Class 12 Maths Chapter 9 - Image 171

NCERT Solutions for Class 12 Maths Chapter 9 - Image 172

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 173

NCERT Solutions for Class 12 Maths Chapter 9 - Image 174

NCERT Solutions for Class 12 Maths Chapter 9 - Image 175

NCERT Solutions for Class 12 Maths Chapter 9 - Image 176

NCERT Solutions for Class 12 Maths Chapter 9 - Image 177

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 178

NCERT Solutions for Class 12 Maths Chapter 9 - Image 179

NCERT Solutions for Class 12 Maths Chapter 9 - Image 180

NCERT Solutions for Class 12 Maths Chapter 9 - Image 181

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 192

NCERT Solutions for Class 12 Maths Chapter 9 - Image 193

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 194

NCERT Solutions for Class 12 Maths Chapter 9 - Image 195

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 196

NCERT Solutions for Class 12 Maths Chapter 9 - Image 197

NCERT Solutions for Class 12 Maths Chapter 9 - Image 198

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 199

NCERT Solutions for Class 12 Maths Chapter 9 - Image 200

NCERT Solutions for Class 12 Maths Chapter 9 - Image 201

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 202

NCERT Solutions for Class 12 Maths Chapter 9 - Image 203

NCERT Solutions for Class 12 Maths Chapter 9 - Image 204

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 205

NCERT Solutions for Class 12 Maths Chapter 9 - Image 206

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 209

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 210

NCERT Solutions for Class 12 Maths Chapter 9 - Image 211

NCERT Solutions for Class 12 Maths Chapter 9 - Image 212

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 213

NCERT Solutions for Class 12 Maths Chapter 9 - Image 214

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 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:

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).

NCERT Solutions for Class 12 Maths Chapter 9 - Image 232

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 233

NCERT Solutions for Class 12 Maths Chapter 9 - Image 234

Therefore, its degree is three.

NCERT Solutions for Class 12 Maths Chapter 9 - Image 235

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 236

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 237

NCERT Solutions for Class 12 Maths Chapter 9 - Image 238

NCERT Solutions for Class 12 Maths Chapter 9 - Image 239

NCERT Solutions for Class 12 Maths Chapter 9 - Image 240

NCERT Solutions for Class 12 Maths Chapter 9 - Image 241

NCERT Solutions for Class 12 Maths Chapter 9 - Image 242

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 243

NCERT Solutions for Class 12 Maths Chapter 9 - Image 244

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 245

NCERT Solutions for Class 12 Maths Chapter 9 - Image 246

NCERT Solutions for Class 12 Maths Chapter 9 - Image 247

NCERT Solutions for Class 12 Maths Chapter 9 - Image 248

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 249

Therefore from above equation we have

NCERT Solutions for Class 12 Maths Chapter 9 - Image 250

NCERT Solutions for Class 12 Maths Chapter 9 - Image 251

6. Find the general solution of the differential equation

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 252

On integrating, we get,

NCERT Solutions for Class 12 Maths Chapter 9 - Image 253

⇒ sin-1x + sin-1y = C

NCERT Solutions for Class 12 Maths Chapter 9 - Image 254

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 255

NCERT Solutions for Class 12 Maths Chapter 9 - Image 256

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 257

NCERT Solutions for Class 12 Maths Chapter 9 - Image 258

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 259

NCERT Solutions for Class 12 Maths Chapter 9 - Image 261

NCERT Solutions for Class 12 Maths Chapter 9 - Image 260

10. Solve the differential equation

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 262

NCERT Solutions for Class 12 Maths Chapter 9 - Image 263

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 264

NCERT Solutions for Class 12 Maths Chapter 9 - Image 266
NCERT Solutions for Class 12 Maths Chapter 9 - Image 265

12. Solve the differential equation

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 267

NCERT Solutions for Class 12 Maths Chapter 9 - Image 268

NCERT Solutions for Class 12 Maths Chapter 9 - Image 269

13. Find a particular solution of the differential equation 

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

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 270

NCERT Solutions for Class 12 Maths Chapter 9 - Image 271

NCERT Solutions for Class 12 Maths Chapter 9 - Image 272

14. Find a particular solution of the differential equation,

given that y = 0 when x = 0.

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 273

NCERT Solutions for Class 12 Maths Chapter 9 - Image 274

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 275

NCERT Solutions for Class 12 Maths Chapter 9 - Image 276

NCERT Solutions for Class 12 Maths Chapter 9 - Image 277

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 278

NCERT Solutions for Class 12 Maths Chapter 9 - Image 279

NCERT Solutions for Class 12 Maths Chapter 9 - Image 280

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

NCERT Solutions for Class 12 Maths Chapter 9 - Image 281

Solution:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 282

Explanation:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 283

NCERT Solutions for Class 12 Maths Chapter 9 - Image 284

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:

NCERT Solutions for Class 12 Maths Chapter 9 - Image 285


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. Solution 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:

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

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