NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations – PDF Download

Q: For each of the differential equations in Question 11 to 12, find a particular solution satisfying the given condition: 47. (x3+x2+x+1)dydx=2x2+x

The given D.E is (x3+x2+x+1)dydx=2x2+x ⇒dy=(2x2+xx3+x2+x+1)dx⇒dy=2x2+xx2(x+1)+(x+1)dx=2x2+x(x+1)(x2+1)dx Integrating both sides we get, ∫dy=∫2x2+x(x+1)(x2+1)dx Let, 2x2+x(x+1)(x2+1)=Ax+1+Bx+cx2+1 ⇒2x2+2=A(x2+1)+(Bx+c)(x+1)=Ax2+A+Bx2+Bx+Cx+C=(A+B)x2+(B+C)x2+(A+C) Comparing the co-efficient we get, A+B=2−−−−−−(1)B+C=1−−−−−−(2)A+C=0−−−−−−(3) Subtracting equation (1) – (2), we get A+B−(B+C)=2−1→A−C=1 But from equation (3) A=−C so, we get, A(−C)−C=1→−2C=1→C=−12&A=−(−12)=12 And putting value of A in equation (1), 12+B=2⇒B=2−12=4−12=32 Putting value of A,B and C in 2x2+x(x+1)(x2+1)=12x+1+32x−12x2+1=12(x+1)+32(xx2+1)−12(1x2+1) Hence, the integration becomes ∫dy=∫12(x+1)dx+∫34(2xx2+1)dx−∫12(1x2+1)dx⇒y=12log(x+1)+34log(x2+1)−12tan−1x1+c Given, At x=0,y=1 Then, 1=12log1+34log1−12tan−1(0)+C ⇒1=0+0−0+C{?log1=0tan−10−0}⇒c=1 ∴ The required particular solution is: y=12log(x+1)+34log(x2+1)−12tan−1x+1=14[2log(x+1)+3log(x2+1)]−12tan−1x+1=14[log(x+1)2+log(x2+1)3]−12tan−1x+1=14[log(x+1)2(x2+1)3]−12tan−1x+1

Q: 41. (ex+e−x)dy−(ex−e−x)dx=0

Given, (ex+e−x)dy− (ex−e−x)dx=0 ⇒ (ex+e−x)dy= (ex−e−x)dx⇒dy=ex−e−xex+e−xdx Integrating both sides ⇒∫dy=ex−e−xex+e−xdx {? ∫f| (x)f (x)dx=log|x|} ⇒y=log|ex+e−x|+c is the required general solution.

Q: 58. 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.

Let ‘x’ be the number of bacteria present in instantaneous time t. Then, dxdt∝x ⇒dxdt=kx,where,k= constant of proportionality. ⇒dxx=kdt Integrating both sides, ∫dxx=∫kdt⇒logx=kt+c Given, at t=0,x=x0(say)then, logx0=c(Initial,x0=100000) So, the differential equation is logx=kt+logx0⇒logx−logx0=kt⇒logxx0=kt As the bacteria number increased by 10% in 2 hours. The number of bacteria increased in 2hours =10%×100000=10000 Hence, at t=2, x=100000+10000=110000 So, log110000100000=2k⇒k=12log(1110) Hence, logxx0=[12log1110]×t when,x=200000, then we get, log200000100000=12log1110×t ⇒2log2=log(1110)×t⇒t=2log2log(1110)hours

Updated on Aug 6, 2025 12:15 IST

By Pallavi Pathak, Assistant Manager Content

Differential Equation Class 12 covers basic concepts of the chapter, general and particular solutions of a differential equation, methods to solve a first-order, first-degree differential equation, formation of differential equations, and applications of differential equations in different areas. It is an important chapter as an in-depth study of differential equations is extremely significant in all modern scientific investigations.
Differential Equations Class 12 NCERT Solutions is an accurate and reliable study material for Class 12 students preparing for the CBSE Board exam and entrance tests like JEE Mains. The student can also download the Differential Equations Class 12 PDF from this page to get well-structured solutions of all the NCERT textbook questions of this chapter. To get access to the NCERT solutions of all the Maths, Physics, and Chemistry of Class 11 and Class 12, check - NCERT Solutions Class 11 and 12.

Table of content

Quick Summary of Differential Equations – Class 12 Maths
Class 12 Math Chapter 9 Differential Equations NCERT Solution PDF
Class 12 Math Chapter 9 Differential Equations: Key Topics, Weightage
Important Formulas of Differential Equation Class 12
Class 12 Differential Equations Exercise-wise Solution
Differential Equations Exercise 9.1 Solutions

Quick Summary of Differential Equations – Class 12 Maths

Here is an overview of the Class 12 Differential Equations:

The chapter states that a differential equation is an equation involving derivatives of the dependent variable with respect to the independent variable.
A polynomial equation in its derivatives is the degree of a differential equation.
The order of the highest order derivative occurring in the differential equation is the order of the differential equation.
Homogeneous differential equation - a differential equation that can be expressed as math ml code are homogenous function of degree zero.

To read the chapter-wise important topics and get free PDFs, check Class 12 Maths NCERT Solutions.

Class 12 Math Chapter 9 Differential Equations NCERT Solution PDF

To get access to the step-by-step solutions of all the NCERT questions of the Differential Equations Class 12, download the PDF from the link given here. The solutions are ideal for CBSE Board exam preparation and other competitive exams like JEE Mains.

Class 12 Math Differential Equations NCERT Solutions: Free PDF Download

Class 12 Math Chapter 9 Differential Equations: Key Topics, Weightage

Differential Equation Class 12 is an important chapter as the concepts are used in all modern scientific investigations. Students should master the concepts to score high in the CBSE Board exam and in other competitive exams. Find below the topics covered in this chapter:

Exercise	Topics Covered
9.1	Introduction
9.2	Basic Concepts
9.3	General and Particular Solutions of a Differential Equation
9.4	Methods of Solving First Order, First Degree Differential Equations

Differential Equation Class 12 Weightage in JEE Mains

Exam	Number of Questions	Weightage
JEE Main	1-2 questions	3.3% to 7%

Important Formulas of Differential Equation Class 12

Differential Equations Important Formulae for CBSE and Competitive Exams

Students can check the important formulae for CBSE and other engineering entrance exams such as JEE Main, CUSAT CAT, and more... below;

General Solution:
$y = \int f(x)dx + C$
Variable Separation:
$\int \frac{1}{g(y)}dy = \int f(x)dx$
Linear Differential Equations:
- General form: $\frac{dy}{dx} + Py = Q$
- Solution form: $y \cdot e^{\int P d x} = \int Q \cdot e^{\int P d x} d x + C$
Homogeneous Equations:
- Substitution: $y = vx$ or $x = vy$
- Simplified separable form: $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$
Exact Differential Equations:
- General form: $M(x, y)dx + N(x, y)dy = 0$
- Solution: $\int Mdx + \int \left(N - \frac{\partial}{\partial y} \left( \int Mdx \right)\right)dy = C$
Exponential Growth/Decay Differential Equations: $\frac{d y}{d t} = k y Solution: y = y_{0} e^{k t}$

Class 12 Differential Equations Exercise-wise Solution

Class 12 Differential Equations is an important chapter, which consists of essential concepts of Calculus. Various exercises of Differential Equations deal with key concepts such as order and degree of a differential equation, formation of differential equations by eliminating arbitrary constants, and methods to solve them, including variable separation, homogeneous equations, and linear differential equations. Shiksha provides exercise-wise solutions with step-by-step explanations, Students can take the help of solutions to master these concepts. Students can check the Exercise-wise Chapter 9 Differential Equation Class 12 math solutions here.

Differential Equations Exercise 9.1 Solutions

Class 12 Differential Equations Exercise 9.1 focuses on Fundamental concepts such as understanding the order and degree of a differential equation, identifying differential equations from given relationships, and forming differential equations by eliminating arbitrary constants from given functions. Differential Equation Exercise 9.1 Solutions consists of 12 Questions. Students can check the complete Exercise 9.1 Solutions below;

Class 12 Chapter 9 Differential Equations Exercise 9.1 NCERT Solution

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

Q1. $\frac{d^{4} y}{d x^{4}} + s i n (y') = 0$

A.1. The highest order derivation present in the differential equation (D.E.) is $\frac{d^{4} y}{d x^{4}}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative ,its degree is not defined.

Q2. y' + 5y = 0

A.2. The highest order derivation present in the D.E. is y , so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

Q3.

{(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0

A.3. The highest order derivation present in the D.E. is $\frac{d^{2} s}{d t^{2}}$ so its order is 2 .

As the given D.E. is a polynomial equation in its derivative its degree is 1.

Q4.

\frac{d^{2} y}{{(d x^{2})}^{2}} + c o s (\frac{d y}{d x}) = 0

A.4.

\frac{d^{4} y}{d x^{4}}

As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

Commonly asked questions

For each of the differential equations in Question 11 to 12, find a particular solution satisfying the given condition:

47. $(x^{3} + x^{2} + x + 1) \frac{d y}{d x} = 2 x^{2} + x$

41. $(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0$

58. 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.

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

1. $\frac{d^{4} y}{d x^{4}} + s i n (y') = 0$

22. Kindly Consider the following

37. $\frac{d y}{d x} = \frac{1 - c o s x}{1 + c o s x}$

39. $\frac{d y}{d x} + y = 1 (y \neq 1)$

42. $\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$

65. Kindly Consider the following

18. Kindly Consider the following

Kindly Consider the following

25. $\frac{x}{a} + \frac{y}{b} = 1$

21. x + y = tan^-1 y : y²y' + y²+ 1 = 0

40. $s e c^{2} x t a n y d x + s e c^{2} y t a n x d y = 0$

43. $y l o g y d x - x d y = 0$

44. $x^{5} \frac{d y}{d x} = - y^{5}$

45. $\frac{d y}{d x} = s i n^{- 1} x$

46. $e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$

53 Find the equation of the 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.

56. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).

57. 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

(e^0.5 = 1.645).

62. $(x - y) d y - (x + y) d x = 0$

38. Kindly Consider the following

54. 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).

55. The volume of the 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.

59. The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is:

(A) e^x + e^-y = C

(B) e^x + e^y = C

(C) e^-x + e^y = C

(D) e^x + e^-y = C

60. $(x^{2} + x y) d y = (x^{2} + y^{2}) d x$

61. Kindly Consider the following

$y' = \frac{x + y}{x}$

63. $(x^{2} - y^{2}) d x + 2 x y d y = 0$

64. $x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + x y$

66. ${x c o s (\frac{y}{x}) + y s i n (\frac{y}{x})} y d x = {y s i n (\frac{y}{x}) + x c o s (\frac{y}{x})} x d y$

67. $x \frac{d y}{d x} - y = x s i n (\frac{y}{x}) = 0$

68. $y d x = x l o g (\frac{y}{x}) d y - 2 x d y = 0$

69. $(1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0$

For each of the differential equations in Questions from 11 to 15, find the particular solution satisfying the given condition

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

6. (y′′′) + (y′′) + (y′)⁴ + y⁵ =0

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{2} + s i n (\frac{d y}{d x}) + 1 = 0$ is:

(A) 3

(B) 2

(C) 1

(D) Not defined

12. The order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + y = 0$ is:

(A) 2

(B) 1

(C) 0

(D) Not defined

Kindly Consider the following

2. y' + 5y = 0

3. ${(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0$

4. $\frac{d^{2} y}{{(d x^{2})}^{2}} + c o s (\frac{d y}{d x}) = 0$

$\frac{d^{2} y}{d x^{2}} = c o s 3 x + s i n 3 x$

7. y''' + 2y'' + y' = 0

8. y′ + y = e^x

9. y'′ + (y')² + 2y = 0

10. y'′ + 2y' + sin y = 0

13. y = e^x + 1 : y″ – y′ = 0

14. y = x² + 2x + C : y′ – 2x – 2 = 0

15. y = cos x + C : y′ + sin x = 0

16. y = √1 + x² ; y/= $\frac{x y}{1 + x^{2}}$

17. y = Ax : xy′ = y (x ≠ 0)

19. xy = log y + C :

$y' = \frac{y^{2}}{1 - x y} (x y \neq 1)$

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

23. 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

24. 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

26. $y^{2} = a (b^{2} - x^{2})$

27. $y = a e^{3 x} + b e^{- 2 x}$

Kindly Consider the following

28. $y = e^{2 x} (a + b x)$

29. $y = e^{x} (a c o s x + b s i n x)$

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

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

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

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

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

35. Which of the following differential equation has $y = c_{1} e^{x} + c_{2} e^{- x}$ as the general solution:

$\begin{array}{l} (A) \frac{d^{2} y}{d x^{2}} + y = 0 \\ (B) \frac{d^{2} y}{d x^{2}} - y = 0 \\ (C) \frac{d^{2} y}{d x^{2}} + 1 = 0 \\ (D) \frac{d^{2} y}{d x^{2}} - 1 = 0 \end{array}$

36. Which of the following differential equations has y = x as one of its particular solutions:

$\begin{array}{l} (A) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = x \\ (B) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = x \\ (C) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0 \\ (D) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = 0 \end{array}$

48. $x (x^{2} - 1) \frac{d y}{d x} = 1; y = 0 w h e n x = 2$

49. $c o s (\frac{d x}{d y}) = a (a \in R); y = 1 w h e n x = 0$

50. $\frac{d y}{d x} = y t a n x; y = 1 w h e n x = 0$

51. Find the equation of the curve passing through the point (0, 0) and whose differential equation is y' = e^x sin x

52. For the differential equation $x y \frac{d y}{d x} = (x + 2) (y + 2)$ find the solution curve passing through the point (1,-1)

71. x² dy + (xy + y² ) dx = 0; y = 1 when x = 1

72. $[x s i n^{2} (\frac{y}{x} - y)] d x + x d y = 0; y = \frac{π}{4} w h e n x = 1$

73. $\frac{d y}{d x} - \frac{y}{x} + c o s e c (\frac{y}{x}) = 0; y = 0 w h e n x = 1$

74. $2 x y + y^{2} - 2 x^{2} \frac{d y}{d x} = 0; y = 2 w h e n x = 1$

75. A homogeneous differential equation of the form $\frac{d x}{d y} = h (\frac{x}{y})$ can be solved by making the substitution:

(A) y = vx

(B) v = yx

(C) x = vy

(D) x = v

76. Which of the following is a homogeneous differential equation:

(A) (4x +6y +5) dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – x³ + y³) dy = 0

(C) (x² + 2y²) dx + (2xy + dy) = 0

(D) y² dx + (x²– xy + y²) dy = 0

77. Kindly Consider the following

$\frac{d y}{d x} + 2 y = 2 s i n x$

78. $\frac{d y}{d x} + 3 y = e^{- 2 x}$

79. Kindly Consider the following

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

80. $\frac{d y}{d x} + s e c x y = t a n x (0 \leq x \leq \frac{π}{2})$

81. $c o s^{2} x \frac{d y}{d x} + y = t a n x (0 \leq x \leq \frac{π}{2})$

82. Kindly Consider the following

$x \frac{d y}{d x} + 2 y = x^{2} l o g x$

83. $x l o g x \frac{d y}{d x} + y = \frac{2}{x} l o g x$

84. $(1 + x^{2}) d y + 2 x y d x = c o t x d x (x \neq 0)$

85. $x \frac{d y}{d x} + y - x + x y c o t x = 0 (x \neq 0)$

86. $(x + y) \frac{d y}{d x} = 1$

87. $y d x + (x - y^{2}) d y = 0$

88. $(x + 3 y^{2}) \frac{d y}{d x} = y (y > 0)$

For each of the differential equations given in Questions 13 to 15, find a particular solution satisfying the given condition:

89. $\frac{d y}{d x} + 2 y t a n x = s i n x; y = 0 w h e n x = \frac{π}{3}$

90. $(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x^{2}}; y = 0 w h e n x = 1$

91. $\frac{d y}{d x} - 3 y c o t x = s i n 2 x; y = 2 w h e n x = \frac{π}{2}$

92. Find the equation of the 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 coordinates of that point.

93. Find the equation of the 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 tangents to the curve at that point by 5.

94. Choose the correct answer:

The integrating factor of the differential equation $\frac{d y}{d x} - y = 2 x^{2}$ is:

(A) e^–x

(B) e^–y

(C) $\frac{1}{x}$

(D) x

95. Choose the correct answer:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < < 1)$ .

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

$\begin{array}{l} (i) \frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x \\ (i i) {(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x \\ (i i i) \frac{d^{4} y}{d x^{4}} - s i n \frac{d^{3} y}{d x^{3}} = 0 \end{array}$

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

$\begin{array}{l} (i) y a e^{2} + b e^{- x} + x^{2} : x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0 \\ (i i) y = e^{x} (a c o s x + b s i n x) : \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0 \\ (i i i) y = x s i n 3 x : \frac{d^{2} y}{d x^{2}} + 9 y - 6 c o s 3 x = 0 \\ (i v) x^{2} = 2 y^{2} l o g y : (x^{2} + y^{2}) \frac{d y}{d x} - x y = 0 \end{array}$

98. Form the differential equation representing the family of curves ${(x - a)}^{2} + 2 y^{2} = a^{2}$ where a an arbitrary constant.

99. Prove that $x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2}$ is the general equation of the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ where c is a parameter.

100. For the differential equation of the family of the circles in the first quadrant which touch the coordinate axes.

101. Kindly Consider the following

102. Show that the general solution of the differential equation

$\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$ is given by $(x + y + 1) A (1 - x - y - 2 x y),$ where A is parameter.

103. 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.

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

105. Solve the differential equation: $y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y (y \neq 0)$

106. 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)

107. Kindly Consider the following

108. Find the particular solution of the differential equation $\frac{d y}{d x} + y c o t x = 4 x c o s e c x (x \neq 0)$

given that y = 0 when x = π/2

109. Find the particular solution of the differential equation $(x + 1) \frac{d y}{d x} = 2 e^{- y} - 1$ given that y = 0 when x = 0

110. 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 25,000 in the year 2004, what will be the population of the village in 2009?

Choose the correct answer:

111. The general solution of the differential equation $\frac{y d x - x d y}{y} = 0$ is:

$\begin{array}{l} (A) x y = c \\ (B) x = c y^{2} \\ (C) y = c x \\ (D) y = c x^{2} \end{array}$

112. The general equation of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q, i s :$

$\begin{array}{l} (A) y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \\ (B) y . e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C \\ (C) x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \\ (D) x . e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C \end{array}$

113. The general solution of the differential equation $e^{x} d y + (y e^{x} + 2 x) d x = 0$ is:

$\begin{array}{l} (A) x e^{y} + x^{2} = C \\ (B) x e^{y} + y^{2} = C \\ (C) y e^{x} + x^{2} = C \\ (D) y e^{y} + x^{2} = C \end{array}$

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