 # NCERT Solutions For Class 12 Maths Chapter 9- Differential Equations Exercise 9.2

## NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.2 – CBSE Term II Free PDF Download

Exercise 9.2 of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations is based on the topic “General and Particular Solutions of a Differential Equation”. The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation. 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. Solve the problems present in this exercise to understand these topics better.

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations Exercise 9.2        ### Access Answers to NCERT Class 12 Maths Chapter 9- Differential Equations Exercise 9.2 Page Number 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, ⇒ 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, 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, 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:-  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, 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:-   Solution:-  8. y – cos y = x : (y sin y + cos y + x) y′ = y

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

Solution:-  Therefore, the given function is the solution of the corresponding differential equation. Solution:-  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.

#### Access Other Exercise Solutions of Class 12 Maths Chapter 9

Exercise 9.1 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