class 12th maths soultion

1. Class XII Chapter 9 – Differential Equations Maths
Page 1 of 120
Exercise 9.1
Question 1:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is four.
The given differential equation is not a polynomial equation in its derivatives. Hence, its
degree is not defined.
Question 2:
Determine order and degree(if defined) of differential equation
Answer
The given differential equation is:
The highest order derivative present in the differential equation is . Therefore, its order
is one.
It is a polynomial equation in . The highest power raised to is 1. Hence, its degree is
one.
Question 3:
Determine order and degree(if defined) of differential equation
Answer

2. Class XII Chapter 9 – Differential Equations Maths
Page 2 of 120
The highest order derivative present in the given differential equation is . Therefore,
its order is two.
It is a polynomial equation in and . The power raised to is 1.
Hence, its degree is one.
Question 4:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the given differential equation is . Therefore,
its order is 2.
The given differential equation is not a polynomial equation in its derivatives. Hence, its
degree is not defined.
Question 5:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is two.

3. Class XII Chapter 9 – Differential Equations Maths
Page 3 of 120
It is a polynomial equation in and the power raised to is 1.
Hence, its degree is one.
Question 6:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is three.
The given differential equation is a polynomial equation in .
The highest power raised to is 2. Hence, its degree is 2.
Question 7:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is three.
It is a polynomial equation in . The highest power raised to is 1. Hence, its
degree is 1.
Question 8:
Determine order and degree(if defined) of differential equation
Answer

4. Class XII Chapter 9 – Differential Equations Maths
Page 4 of 120
The highest order derivative present in the differential equation is . Therefore, its order
is one.
The given differential equation is a polynomial equation in and the highest power
raised to is one. Hence, its degree is one.
Question 9:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is two.
The given differential equation is a polynomial equation in and and the highest
power raised to is one.
Hence, its degree is one.
Question 10:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is two.
This is a polynomial equation in and and the highest power raised to is one.
Hence, its degree is one.
Question 11:
The degree of the differential equation
is

5. Class XII Chapter 9 – Differential Equations Maths
Page 5 of 120
(A) 3 (B) 2 (C) 1 (D) not defined
Answer
The given differential equation is not a polynomial equation in its derivatives. Therefore,
its degree is not defined.
Hence, the correct answer is D.
Question 12:
The order of the differential equation
is
(A) 2 (B) 1 (C) 0 (D) not defined
Answer
The highest order derivative present in the given differential equation is . Therefore,
its order is two.
Hence, the correct answer is A.

6. Class XII Chapter 9 – Differential Equations Maths
Page 6 of 120
Exercise 9.2
Question 1:
Answer
Differentiating both sides of this equation with respect to x, we get:
Now, differentiating equation (1) with respect to x, we get:
Substituting the values of in the given differential equation, we get the L.H.S.
as:
Thus, the given function is the solution of the corresponding differential equation.
Question 2:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
L.H.S. = = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.

7. Class XII Chapter 9 – Differential Equations Maths
Page 7 of 120
Question 3:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
L.H.S. = = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 4:
Answer
Differentiating both sides of the equation with respect to x, we get:

8. Class XII Chapter 9 – Differential Equations Maths
Page 8 of 120
L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 5:
Answer
Differentiating both sides with respect to x, we get:
Substituting the value of in the given differential equation, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question 6:
Answer

9. Class XII Chapter 9 – Differential Equations Maths
Page 9 of 120
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question 7:
Answer
Differentiating both sides of this equation with respect to x, we get:

10. Class XII Chapter 9 – Differential Equations Maths
Page 10 of 120
L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 8:
Answer
Differentiating both sides of the equation with respect to x, we get:
Substituting the value of in equation (1), we get:

11. Class XII Chapter 9 – Differential Equations Maths
Page 11 of 120
Hence, the given function is the solution of the corresponding differential equation.
Question 9:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:

12. Class XII Chapter 9 – Differential Equations Maths
Page 12 of 120
Hence, the given function is the solution of the corresponding differential equation.
Question 10:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
Hence, the given function is the solution of the corresponding differential equation.

13. Class XII Chapter 9 – Differential Equations Maths
Page 13 of 120
Question 11:
The numbers of arbitrary constants in the general solution of a differential equation of
fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4
Answer
We know that the number of constants in the general solution of a differential equation
of order n is equal to its order.
Therefore, the number of constants in the general equation of fourth order differential
equation is four.
Hence, the correct answer is D.
Question 12:
The numbers of arbitrary constants in the particular solution of a differential equation of
third order are:
(A) 3 (B) 2 (C) 1 (D) 0
Answer
In a particular solution of a differential equation, there are no arbitrary constants.
Hence, the correct answer is D.

14. Class XII Chapter 9 – Differential Equations Maths
Page 14 of 120
Exercise 9.3
Question 1:
Answer
Differentiating both sides of the given equation with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Hence, the required differential equation of the given curve is
Question 2:
Answer
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:

15. Class XII Chapter 9 – Differential Equations Maths
Page 15 of 120
Dividing equation (2) by equation (1), we get:
This is the required differential equation of the given curve.
Question 3:
Answer
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Multiplying equation (1) with (2) and then adding it to equation (2), we get:
Now, multiplying equation (1) with equation (3) and subtracting equation (2) from it, we
get:
Substituting the values of in equation (3), we get:

16. Class XII Chapter 9 – Differential Equations Maths
Page 16 of 120
This is the required differential equation of the given curve.
Question 4:
Answer
Differentiating both sides with respect to x, we get:
Multiplying equation (1) with equation (2) and then subtracting it from equation (2), we
get:
Differentiating both sides with respect to x, we get:
Dividing equation (4) by equation (3), we get:
This is the required differential equation of the given curve.

17. Class XII Chapter 9 – Differential Equations Maths
Page 17 of 120
Question 5:
Answer
Differentiating both sides with respect to x, we get:
Again, differentiating with respect to x, we get:
Adding equations (1) and (3), we get:
This is the required differential equation of the given curve.
Question 6:
Form the differential equation of the family of circles touching the y-axis at the origin.
Answer
The centre of the circle touching the y-axis at origin lies on the x-axis.
Let (a, 0) be the centre of the circle.
Since it touches the y-axis at origin, its radius is a.
Now, the equation of the circle with centre (a, 0) and radius (a) is

18. Class XII Chapter 9 – Differential Equations Maths
Page 18 of 120
Differentiating equation (1) with respect to x, we get:
Now, on substituting the value of a in equation (1), we get:
This is the required differential equation.
Question 7:
Form the differential equation of the family of parabolas having vertex at origin and axis
along positive y-axis.
Answer
The equation of the parabola having the vertex at origin and the axis along the positive
y-axis is:

19. Class XII Chapter 9 – Differential Equations Maths
Page 19 of 120
Differentiating equation (1) with respect to x, we get:
Dividing equation (2) by equation (1), we get:
This is the required differential equation.
Question 8:
Form the differential equation of the family of ellipses having foci on y-axis and centre at
origin.
Answer
The equation of the family of ellipses having foci on the y-axis and the centre at origin is
as follows:

20. Class XII Chapter 9 – Differential Equations Maths
Page 20 of 120
Differentiating equation (1) with respect to x, we get:
Again, differentiating with respect to x, we get:
Substituting this value in equation (2), we get:
This is the required differential equation.

21. Class XII Chapter 9 – Differential Equations Maths
Page 21 of 120
Question 9:
Form the differential equation of the family of hyperbolas having foci on x-axis and
centre at origin.
Answer
The equation of the family of hyperbolas with the centre at origin and foci along the x-
axis is:
Differentiating both sides of equation (1) with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Substituting the value of in equation (2), we get:

22. Class XII Chapter 9 – Differential Equations Maths
Page 22 of 120
This is the required differential equation.
Question 10:
Form the differential equation of the family of circles having centre on y-axis and radius
3 units.
Answer
Let the centre of the circle on y-axis be (0, b).
The differential equation of the family of circles with centre at (0, b) and radius 3 is as
follows:
Differentiating equation (1) with respect to x, we get:
Substituting the value of (y – b) in equation (1), we get:

23. Class XII Chapter 9 – Differential Equations Maths
Page 23 of 120
This is the required differential equation.
Question 11:
Which of the following differential equations has as the general solution?
A.
B.
C.
D.
Answer
The given equation is:
Differentiating with respect to x, we get:
Again, differentiating with respect to x, we get:

24. Class XII Chapter 9 – Differential Equations Maths
Page 24 of 120
This is the required differential equation of the given equation of curve.
Hence, the correct answer is B.
Question 12:
Which of the following differential equation has as one of its particular solution?
A.
B.
C.
D.
Answer
The given equation of curve is y = x.
Differentiating with respect to x, we get:
Again, differentiating with respect to x, we get:
Now, on substituting the values of y, from equation (1) and (2) in each of
the given alternatives, we find that only the differential equation given in alternative C is
correct.

25. Class XII Chapter 9 – Differential Equations Maths
Page 25 of 120
Hence, the correct answer is C.

26. Class XII Chapter 9 – Differential Equations Maths
Page 26 of 120
Exercise 9.4
Question 1:
Answer
The given differential equation is:
Now, integrating both sides of this equation, we get:
This is the required general solution of the given differential equation.
Question 2:
Answer
The given differential equation is:

27. Class XII Chapter 9 – Differential Equations Maths
Page 27 of 120
Now, integrating both sides of this equation, we get:
This is the required general solution of the given differential equation.
Question 3:
Answer
The given differential equation is:
Now, integrating both sides, we get:

28. Class XII Chapter 9 – Differential Equations Maths
Page 28 of 120
This is the required general solution of the given differential equation.
Question 4:
Answer
The given differential equation is:
Integrating both sides of this equation, we get:

29. Class XII Chapter 9 – Differential Equations Maths
Page 29 of 120
Substituting these values in equation (1), we get:
This is the required general solution of the given differential equation.
Question 5:
Answer
The given differential equation is:
Integrating both sides of this equation, we get:

30. Class XII Chapter 9 – Differential Equations Maths
Page 30 of 120
Let (ex
+ e–x
) = t.
Differentiating both sides with respect to x, we get:
Substituting this value in equation (1), we get:
This is the required general solution of the given differential equation.
Question 6:
Answer
The given differential equation is:
Integrating both sides of this equation, we get:

31. Class XII Chapter 9 – Differential Equations Maths
Page 31 of 120
This is the required general solution of the given differential equation.
Question 7:
Answer
The given differential equation is:
Integrating both sides, we get:
Substituting this value in equation (1), we get:

32. Class XII Chapter 9 – Differential Equations Maths
Page 32 of 120
This is the required general solution of the given differential equation.
Question 8:
Answer
The given differential equation is:
Integrating both sides, we get:
This is the required general solution of the given differential equation.

33. Class XII Chapter 9 – Differential Equations Maths
Page 33 of 120
Question 9:
Answer
The given differential equation is:
Integrating both sides, we get:
Substituting this value in equation (1), we get:

34. Class XII Chapter 9 – Differential Equations Maths
Page 34 of 120
This is the required general solution of the given differential equation.
Question 10:
Answer
The given differential equation is:
Integrating both sides, we get:

35. Class XII Chapter 9 – Differential Equations Maths
Page 35 of 120
Substituting the values of in equation (1), we get:
This is the required general solution of the given differential equation.
Question 11:
Answer
The given differential equation is:
Integrating both sides, we get:

36. Class XII Chapter 9 – Differential Equations Maths
Page 36 of 120
Comparing the coefficients of x2
and x, we get:
A + B = 2
B + C = 1
A + C = 0
Solving these equations, we get:
Substituting the values of A, B, and C in equation (2), we get:
Therefore, equation (1) becomes:

37. Class XII Chapter 9 – Differential Equations Maths
Page 37 of 120
Substituting C = 1 in equation (3), we get:
Question 12:
Answer
Integrating both sides, we get:

38. Class XII Chapter 9 – Differential Equations Maths
Page 38 of 120
Comparing the coefficients of x2
, x, and constant, we get:
Solving these equations, we get
Substituting the values of A, B, and C in equation (2), we get:
Therefore, equation (1) becomes:

39. Class XII Chapter 9 – Differential Equations Maths
Page 39 of 120
Substituting the value of k2
in equation (3), we get:
Question 13:
Answer

40. Class XII Chapter 9 – Differential Equations Maths
Page 40 of 120
Integrating both sides, we get:
Substituting C = 1 in equation (1), we get:
Question 14:
Answer
Integrating both sides, we get:

41. Class XII Chapter 9 – Differential Equations Maths
Page 41 of 120
Substituting C = 1 in equation (1), we get:
y = sec x
Question 15:
Find the equation of a curve passing through the point (0, 0) and whose differential
equation is .
Answer
The differential equation of the curve is:
Integrating both sides, we get:

42. Class XII Chapter 9 – Differential Equations Maths
Page 42 of 120
Substituting this value in equation (1), we get:
Now, the curve passes through point (0, 0).
Substituting in equation (2), we get:
Hence, the required equation of the curve is
Question 16:
For the differential equation find the solution curve passing
through the point (1, –1).

43. Class XII Chapter 9 – Differential Equations Maths
Page 43 of 120
Answer
The differential equation of the given curve is:
Integrating both sides, we get:
Now, the curve passes through point (1, –1).
Substituting C = –2 in equation (1), we get:
This is the required solution of the given curve.
Question 17:
Find the equation of a curve passing through the point (0, –2) given that at any point
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.
Answer
Let x and y be the x-coordinate and y-coordinate of the curve respectively.

44. Class XII Chapter 9 – Differential Equations Maths
Page 44 of 120
We know that the slope of a tangent to the curve in the coordinate axis is given by the
relation,
According to the given information, we get:
Integrating both sides, we get:
Now, the curve passes through point (0, –2).
∴ (–2)2
– 02
= 2C
⇒ 2C = 4
Substituting 2C = 4 in equation (1), we get:
y2
– x2
= 4
This is the required equation of the curve.
Question 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).
Answer

45. Class XII Chapter 9 – Differential Equations Maths
Page 45 of 120
It is given that (x, y) is the point of contact of the curve and its tangent.
The slope (m1) of the line segment joining (x, y) and (–4, –3) is
We know that the slope of the tangent to the curve is given by the relation,
According to the given information:
Integrating both sides, we get:
This is the general equation of the curve.
It is given that it passes through point (–2, 1).
Substituting C = 1 in equation (1), we get:
y + 3 = (x + 4)2
This is the required equation of the curve.

46. Class XII Chapter 9 – Differential Equations Maths
Page 46 of 120
Question 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.
Answer
Let the rate of change of the volume of the balloon be k (where k is a constant).
Integrating both sides, we get:
⇒ 4π × 33
= 3 (k × 0 + C)
⇒ 108π = 3C
⇒ C = 36π

47. Class XII Chapter 9 – Differential Equations Maths
Page 47 of 120
At t = 3, r = 6:
⇒ 4π × 63
= 3 (k × 3 + C)
⇒ 864π = 3 (3k + 36π)
⇒ 3k = –288π – 36π = 252π
⇒ k = 84π
Substituting the values of k and C in equation (1), we get:
Thus, the radius of the balloon after t seconds is .
Question 20:
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).

48. Class XII Chapter 9 – Differential Equations Maths
Page 48 of 120
Answer
Let p, t, and r represent the principal, time, and rate of interest respectively.
It is given that the principal increases continuously at the rate of r% per year.
Integrating both sides, we get:
It is given that when t = 0, p = 100.
⇒ 100 = ek
… (2)
Now, if t = 10, then p = 2 × 100 = 200.
Therefore, equation (1) becomes:

49. Class XII Chapter 9 – Differential Equations Maths
Page 49 of 120
Hence, the value of r is 6.93%.
Question 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 .
Answer
Let p and t be the principal and time respectively.
It is given that the principal increases continuously at the rate of 5% per year.
Integrating both sides, we get:
Now, when t = 0, p = 1000.
⇒ 1000 = eC
… (2)
At t = 10, equation (1) becomes:

50. Class XII Chapter 9 – Differential Equations Maths
Page 50 of 120
Hence, after 10 years the amount will worth Rs 1648.
Question 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?
Answer
Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
Integrating both sides, we get:
Let y0 be the number of bacteria at t = 0.
⇒ log y0 = C
Substituting the value of C in equation (1), we get:
Also, it is given that the number of bacteria increases by 10% in 2 hours.

51. Class XII Chapter 9 – Differential Equations Maths
Page 51 of 120
Substituting this value in equation (2), we get:
Therefore, equation (2) becomes:
Now, let the time when the number of bacteria increases from 100000 to 200000 be t1.
⇒ y = 2y0 at t = t1
From equation (4), we get:
Hence, in hours the number of bacteria increases from 100000 to 200000.

52. Class XII Chapter 9 – Differential Equations Maths
Page 52 of 120
Question 23:
The general solution of the differential equation
A.
B.
C.
D.
Answer
Integrating both sides, we get:
Hence, the correct answer is A.

53. Class XII Chapter 9 – Differential Equations Maths
Page 53 of 120
Exercise 9.5
Question 1:
Answer
The given differential equation i.e., (x2
+ xy) dy = (x2
+ y2
) dx can be written as:
This shows that equation (1) is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Differentiating both sides with respect to x, we get:
Substituting the values of v and in equation (1), we get:

54. Class XII Chapter 9 – Differential Equations Maths
Page 54 of 120
Integrating both sides, we get:
This is the required solution of the given differential equation.

55. Class XII Chapter 9 – Differential Equations Maths
Page 55 of 120
Question 2:
Answer
The given differential equation is:
Thus, the given equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Differentiating both sides with respect to x, we get:
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:

56. Class XII Chapter 9 – Differential Equations Maths
Page 56 of 120
This is the required solution of the given differential equation.
Question 3:
Answer
The given differential equation is:
Thus, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:

57. Class XII Chapter 9 – Differential Equations Maths
Page 57 of 120
Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 4:
Answer
The given differential equation is:
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx

58. Class XII Chapter 9 – Differential Equations Maths
Page 58 of 120
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 5:
Answer
The given differential equation is:

59. Class XII Chapter 9 – Differential Equations Maths
Page 59 of 120
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:

60. Class XII Chapter 9 – Differential Equations Maths
Page 60 of 120
This is the required solution for the given differential equation.
Question 6:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx

61. Class XII Chapter 9 – Differential Equations Maths
Page 61 of 120
Substituting the values of v and in equation (1), we get:
Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 7:
Answer
The given differential equation is:

62. Class XII Chapter 9 – Differential Equations Maths
Page 62 of 120
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:

63. Class XII Chapter 9 – Differential Equations Maths
Page 63 of 120
Integrating both sides, we get:
This is the required solution of the given differential equation.

64. Class XII Chapter 9 – Differential Equations Maths
Page 64 of 120
Question 8:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:

65. Class XII Chapter 9 – Differential Equations Maths
Page 65 of 120
Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 9:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx

66. Class XII Chapter 9 – Differential Equations Maths
Page 66 of 120
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:

67. Class XII Chapter 9 – Differential Equations Maths
Page 67 of 120
Therefore, equation (1) becomes:
This is the required solution of the given differential equation.
Question 10:
Answer

68. Class XII Chapter 9 – Differential Equations Maths
Page 68 of 120
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
x = vy
Substituting the values of x and in equation (1), we get:
Integrating both sides, we get:

69. Class XII Chapter 9 – Differential Equations Maths
Page 69 of 120
This is the required solution of the given differential equation.
Question 11:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:

70. Class XII Chapter 9 – Differential Equations Maths
Page 70 of 120
Integrating both sides, we get:
Now, y = 1 at x = 1.
Substituting the value of 2k in equation (2), we get:
This is the required solution of the given differential equation.

71. Class XII Chapter 9 – Differential Equations Maths
Page 71 of 120
Question 12:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:

72. Class XII Chapter 9 – Differential Equations Maths
Page 72 of 120
Integrating both sides, we get:
Now, y = 1 at x = 1.
Substituting in equation (2), we get:
This is the required solution of the given differential equation.
Question 13:

73. Class XII Chapter 9 – Differential Equations Maths
Page 73 of 120
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve this differential equation, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:

74. Class XII Chapter 9 – Differential Equations Maths
Page 74 of 120
Now, .
Substituting C = e in equation (2), we get:
This is the required solution of the given differential equation.
Question 14:
Answer
Therefore, the given differential equation is a homogeneous equation.

75. Class XII Chapter 9 – Differential Equations Maths
Page 75 of 120
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
This is the required solution of the given differential equation.
Now, y = 0 at x = 1.
Substituting C = e in equation (2), we get:
This is the required solution of the given differential equation.

76. Class XII Chapter 9 – Differential Equations Maths
Page 76 of 120
Question 15:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the value of y and in equation (1), we get:
Integrating both sides, we get:

77. Class XII Chapter 9 – Differential Equations Maths
Page 77 of 120
Now, y = 2 at x = 1.
Substituting C = –1 in equation (2), we get:
This is the required solution of the given differential equation.
Question 16:
A homogeneous differential equation of the form can be solved by making the
substitution
A. y = vx
B. v = yx
C. x = vy
D. x = v
Answer

78. Class XII Chapter 9 – Differential Equations Maths
Page 78 of 120
For solving the homogeneous equation of the form , we need to make the
substitution as x = vy.
Hence, the correct answer is C.
Question 17:
Which of the following is a homogeneous differential equation?
A.
B.
C.
D.
Answer
Function F(x, y) is said to be the homogenous function of degree n, if
F(λx, λy) = λn
F(x, y) for any non-zero constant (λ).
Consider the equation given in alternativeD:
Hence, the differential equation given in alternative D is a homogenous equation.

79. Class XII Chapter 9 – Differential Equations Maths
Page 79 of 120
Exercise 9.6
Question 1:
Answer
The given differential equation is
This is in the form of
The solution of the given differential equation is given by the relation,

80. Class XII Chapter 9 – Differential Equations Maths
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Therefore, equation (1) becomes:
This is the required general solution of the given differential equation.
Question 2:
Answer
The given differential equation is
The solution of the given differential equation is given by the relation,
This is the required general solution of the given differential equation.
Question 3:
Answer
The given differential equation is:

81. Class XII Chapter 9 – Differential Equations Maths
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The solution of the given differential equation is given by the relation,
This is the required general solution of the given differential equation.
Question 4:
Answer
The given differential equation is:
The general solution of the given differential equation is given by the relation,
Question 5:
Answer

82. Class XII Chapter 9 – Differential Equations Maths
Page 82 of 120
By second fundamental theorem of calculus, we obtain
Question 6:
Answer
The given differential equation is:
This equation is in the form of a linear differential equation as:
The general solution of the given differential equation is given by the relation,

83. Class XII Chapter 9 – Differential Equations Maths
Page 83 of 120
Question 7:
Answer
The given differential equation is:
This equation is the form of a linear differential equation as:
The general solution of the given differential equation is given by the relation,

84. Class XII Chapter 9 – Differential Equations Maths
Page 84 of 120
Substituting the value of in equation (1), we get:
This is the required general solution of the given differential equation.
Question 8:
Answer
This equation is a linear differential equation of the form:

85. Class XII Chapter 9 – Differential Equations Maths
Page 85 of 120
The general solution of the given differential equation is given by the relation,
Question 9:
Answer
This equation is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

86. Class XII Chapter 9 – Differential Equations Maths
Page 86 of 120
Question 10:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

87. Class XII Chapter 9 – Differential Equations Maths
Page 87 of 120
Question 11:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

88. Class XII Chapter 9 – Differential Equations Maths
Page 88 of 120
Question 12:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

89. Class XII Chapter 9 – Differential Equations Maths
Page 89 of 120
Question 13:
Answer
The given differential equation is
This is a linear equation of the form:
The general solution of the given differential equation is given by the relation,
Now,
Therefore,
Substituting C = –2 in equation (1), we get:

90. Class XII Chapter 9 – Differential Equations Maths
Page 90 of 120
Hence, the required solution of the given differential equation is
Question 14:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
Now, y = 0 at x = 1.
Therefore,

91. Class XII Chapter 9 – Differential Equations Maths
Page 91 of 120
Substituting in equation (1), we get:
This is the required general solution of the given differential equation.
Question 15:
Answer
The given differential equation is
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
Now,
Therefore, we get:

92. Class XII Chapter 9 – Differential Equations Maths
Page 92 of 120
Substituting C = 4 in equation (1), we get:
This is the required particular solution of the given differential equation.
Question 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.
Answer
Let F (x, y) be the curve passing through the origin.
At point (x, y), the slope of the curve will be
According to the given information:
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,

93. Class XII Chapter 9 – Differential Equations Maths
Page 93 of 120
Substituting in equation (1), we get:
The curve passes through the origin.
Therefore, equation (2) becomes:
1 = C
⇒ C = 1
Substituting C = 1 in equation (2), we get:
Hence, the required equation of curve passing through the origin is
Question 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.
Answer
Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent
to the curve at (x, y) is
According to the given information:
This is a linear differential equation of the form:

94. Class XII Chapter 9 – Differential Equations Maths
Page 94 of 120
The general equation of the curve is given by the relation,
Therefore, equation (1) becomes:
The curve passes through point (0, 2).
Therefore, equation (2) becomes:
0 + 2 – 4 = Ce0
⇒ – 2 = C
⇒ C = – 2
Substituting C = –2 in equation (2), we get:

95. Class XII Chapter 9 – Differential Equations Maths
Page 95 of 120
This is the required equation of the curve.
Question 18:
The integrating factor of the differential equation is
A. e–x
B. e–y
C.
D. x
Answer
The given differential equation is:
This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,
Hence, the correct answer is C.
Question 19:
The integrating factor of the differential equation.
is
A.

96. Class XII Chapter 9 – Differential Equations Maths
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B.
C.
D.
Answer
The given differential equation is:
This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,
Hence, the correct answer is D.

97. Class XII Chapter 9 – Differential Equations Maths
Page 97 of 120
Miscellaneous Solutions
Question 1:
For each of the differential equations given below, indicate its order and degree (if
defined).
(i)
(ii)
(iii)
Answer
(i) The differential equation is given as:
The highest order derivative present in the differential equation is . Thus, its order is
two. The highest power raised to is one. Hence, its degree is one.
(ii) The differential equation is given as:

98. Class XII Chapter 9 – Differential Equations Maths
Page 98 of 120
The highest order derivative present in the differential equation is . Thus, its order is
one. The highest power raised to is three. Hence, its degree is three.
(iii) The differential equation is given as:
The highest order derivative present in the differential equation is . Thus, its order is
four.
However, the given differential equation is not a polynomial equation. Hence, its degree
is not defined.
Question 2:
For each of the exercises given below, verify that the given function (implicit or explicit)
is a solution of the corresponding differential equation.
(i)
(ii)
(iii)
(iv)
Answer
(i)
Differentiating both sides with respect to x, we get:

99. Class XII Chapter 9 – Differential Equations Maths
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Again, differentiating both sides with respect to x, we get:
Now, on substituting the values of and in the differential equation, we get:
⇒ L.H.S. ≠ R.H.S.
Hence, the given function is not a solution of the corresponding differential equation.
(ii)
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:

100. Class XII Chapter 9 – Differential Equations Maths
Page 100 of 120
Now, on substituting the values of and in the L.H.S. of the given differential
equation, we get:
Hence, the given function is a solution of the corresponding differential equation.
(iii)
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:

101. Class XII Chapter 9 – Differential Equations Maths
Page 101 of 120
Substituting the value of in the L.H.S. of the given differential equation, we get:
Hence, the given function is a solution of the corresponding differential equation.
(iv)
Differentiating both sides with respect to x, we get:
Substituting the value of in the L.H.S. of the given differential equation, we get:

102. Class XII Chapter 9 – Differential Equations Maths
Page 102 of 120
Hence, the given function is a solution of the corresponding differential equation.
Question 3:
Form the differential equation representing the family of curves given by
where a is an arbitrary constant.
Answer
Differentiating with respect to x, we get:
From equation (1), we get:
On substituting this value in equation (3), we get:

103. Class XII Chapter 9 – Differential Equations Maths
Page 103 of 120
Hence, the differential equation of the family of curves is given as
Question 4:
Prove that is the general solution of differential
equation , where c is a parameter.
Answer
This is a homogeneous equation. To simplify it, we need to make the substitution as:
Substituting the values of y and in equation (1), we get:

104. Class XII Chapter 9 – Differential Equations Maths
Page 104 of 120
Integrating both sides, we get:
Substituting the values of I1 and I2 in equation (3), we get:
Therefore, equation (2) becomes:

105. Class XII Chapter 9 – Differential Equations Maths
Page 105 of 120
Hence, the given result is proved.
Question 5:
Form the differential equation of the family of circles in the first quadrant which touch
the coordinate axes.
Answer
The equation of a circle in the first quadrant with centre (a, a) and radius (a) which
touches the coordinate axes is:

106. Class XII Chapter 9 – Differential Equations Maths
Page 106 of 120
Differentiating equation (1) with respect to x, we get:
Substituting the value of a in equation (1), we get:
Hence, the required differential equation of the family of circles is

107. Class XII Chapter 9 – Differential Equations Maths
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Question 6:
Find the general solution of the differential equation
Answer
Integrating both sides, we get:
Question 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
Answer
Integrating both sides, we get:

108. Class XII Chapter 9 – Differential Equations Maths
Page 108 of 120
Hence, the given result is proved.

109. Class XII Chapter 9 – Differential Equations Maths
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Question 8:
Find the equation of the curve passing through the point whose differential
equation is,
Answer
The differential equation of the given curve is:
Integrating both sides, we get:
The curve passes through point
On substituting in equation (1), we get:
Hence, the required equation of the curve is

110. Class XII Chapter 9 – Differential Equations Maths
Page 110 of 120
Question 9:
Find the particular solution of the differential equation
, given that y = 1 when x = 0
Answer
Integrating both sides, we get:
Substituting these values in equation (1), we get:
Now, y = 1 at x = 0.
Therefore, equation (2) becomes:
Substituting in equation (2), we get:

111. Class XII Chapter 9 – Differential Equations Maths
Page 111 of 120
This is the required particular solution of the given differential equation.
Question 10:
Solve the differential equation
Answer
Differentiating it with respect to y, we get:
From equation (1) and equation (2), we get:
Integrating both sides, we get:

112. Class XII Chapter 9 – Differential Equations Maths
Page 112 of 120
Question 11:
Find a particular solution of the differential equation , given that
y = – 1, when x = 0 (Hint: put x – y = t)
Answer
Substituting the values of x – y and in equation (1), we get:

113. Class XII Chapter 9 – Differential Equations Maths
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Integrating both sides, we get:
Now, y = –1 at x = 0.
Therefore, equation (3) becomes:
log 1 = 0 – 1 + C
⇒ C = 1
Substituting C = 1 in equation (3) we get:
This is the required particular solution of the given differential equation.
Question 12:
Solve the differential equation
Answer

114. Class XII Chapter 9 – Differential Equations Maths
Page 114 of 120
This equation is a linear differential equation of the form
The general solution of the given differential equation is given by,
Question 13:
Find a particular solution of the differential equation ,
given that y = 0 when
Answer
The given differential equation is:
This equation is a linear differential equation of the form

115. Class XII Chapter 9 – Differential Equations Maths
Page 115 of 120
The general solution of the given differential equation is given by,
Now,
Therefore, equation (1) becomes:
Substituting in equation (1), we get:
This is the required particular solution of the given differential equation.
Question 14:
Find a particular solution of the differential equation , given that y = 0
when x = 0
Answer

116. Class XII Chapter 9 – Differential Equations Maths
Page 116 of 120
Integrating both sides, we get:
Substituting this value in equation (1), we get:
Now, at x = 0 and y = 0, equation (2) becomes:
Substituting C = 1 in equation (2), we get:

117. Class XII Chapter 9 – Differential Equations Maths
Page 117 of 120
This is the required particular solution of the given differential equation.
Question 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 20000 in 1999
and 25000 in the year 2004, what will be the population of the village in 2009?
Answer
Let the population at any instant (t) be y.
It is given that the rate of increase of population is proportional to the number of
inhabitants at any instant.
Integrating both sides, we get:
log y = kt + C … (1)
In the year 1999, t = 0 and y = 20000.
Therefore, we get:
log 20000 = C … (2)
In the year 2004, t = 5 and y = 25000.
Therefore, we get:

118. Class XII Chapter 9 – Differential Equations Maths
Page 118 of 120
In the year 2009, t = 10 years.
Now, on substituting the values of t, k, and C in equation (1), we get:
Hence, the population of the village in 2009 will be 31250.
Question 16:
The general solution of the differential equation is
A. xy = C
B. x = Cy2
C. y = Cx
D. y = Cx2
Answer
The given differential equation is:

119. Class XII Chapter 9 – Differential Equations Maths
Page 119 of 120
Integrating both sides, we get:
Hence, the correct answer is C.
Question 17:
The general solution of a differential equation of the type is
A.
B.
C.
D.
Answer
The integrating factor of the given differential equation
The general solution of the differential equation is given by,
Hence, the correct answer is C.

120. Class XII Chapter 9 – Differential Equations Maths
Page 120 of 120
Question 18:
The general solution of the differential equation is
A. xey
+ x2
= C
B. xey
+ y2
= C
C. yex
+ x2
= C
D. yey
+ x2
= C
Answer
The given differential equation is:
This is a linear differential equation of the form
The general solution of the given differential equation is given by,
Hence, the correct answer is C.