

NCERT Solutions for Class 12 Maths Chapter 9 Miscellaneous Exercise
Solve The Following Questions of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2:
In each of the Questions, 1 to 6 verify that the given functions (explicit) is a solution of the corresponding differential equation: Question 1. y = e x + 1 : y″ – y′ = 0 Solution : Given: y = e x + 1
Thus, the given function is the solution of the corresponding differential equation.
NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3
Question 2. y = x 2 + 2x + C : y′ – 2x – 2 = 0 Solution : Given: y = x 2 + 2x + C Differentiating both sides of this equation with respect to x , we get:
Hence, the given function is the solution of the corresponding differential equation.
NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4
Question 3. y = cos x + C : y′ + sin x = 0 Solution : Given: y = cos x + C Differentiating both sides of this equation with respect to x , we get:
Hence, the given function is the solution of the corresponding differential equation.
NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5
Question 4.
Solution :
Given: y = √1 + x
2
Hence, the given function is the solution of the corresponding differential equation.
NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6
Question 5. y = Ax : xy′ = y (x ≠ 0) Solution : Given: y = Ax Differentiating both sides with respect to x , we get:
Hence, the given function is the solution of the corresponding differential equation.
Question
6.
Solution :
Given: y = x sin x
Differentiating both sides of this equation with respect to
x
, we get:
Hence, the given function is the solution of the corresponding differential equation.
In each of the questions 7 to 10, verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
Question 7. xy = log y + C :
Solution :
Given: xy = log y + C
Differentiating both sides of this equation with respect to
x
, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question
8
. y – cos y = x : (y sin y + cos y + x) y′ = y
Solution :
Given: y – cos y = x
Differentiating both sides of the equation with respect to
x
, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question
9.
x + y = tan
-1
y : y
2
y' + y
2
+ 1 = 0
Solution :
Given: x + y = tan
-1
y
Differentiating both sides of this equation with respect to
x
, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question
10.
Solution :
Given: y = √a
2
- x
2
Differentiating both sides of this equation with respect to
x
, we get:
Hence, the given function is the solution of the corresponding differential equation.
Choose the correct answer:
Question 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 : Option (D) is correct. 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. Question 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 : The number of arbitrary constants in a particular solution of a differential equation of any order is zero (0) as a particular solution is a solution which contains no arbitrary constant. Therefore, option (D) is correct.