**NCERT Solutions for Maths Class 12 Exercise 9.2**

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Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

**NCERT Solutions for Maths Class 12 Exercise 9.2** helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

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**NCERT Solutions for Maths Class 12 Exercise 9.2** are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

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**NCERT Solutions for Maths Class 12 Exercise 9.1**

**NCERT Solutions for Maths Class 12 Exercise 9.3**

**NCERT Solutions for Maths Class 12 Exercise 9.4**

**NCERT Solutions for Maths Class 12 Exercise 9.5**

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**NCERT Solutions for Maths Class 12 Exercise 9.6**

**NCERT Solutions for Maths Class 12 Exercise 9.2**** **

**In each of the
Exercises 1 to 10, verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation.**

**Maths Class
12 Ex 9.2 Question 1.**

*y* = *e ^{x}* + 1 :

*y*’’ –

*y*’ = 0

**Solution:**

The given
function is *y* = *e ^{x}* + 1 … (i)

Differentiating equation (i) w.r.t. x, we get

*y*’ = *e ^{x }* … (ii) and

*y*’’ =

*e*… (iii)

^{x }

^{ }Subtracting equation (ii) from equation (iii), we get *y*’’ – *y*’ = 0

Hence, the function *y* = *e ^{x}* + 1 is the solution of

*y*’’ –

*y*’ = 0.

**Maths Class
12 Ex 9.2 Question 2.**

*y* = *x*^{2} + 2*x* + C : *y*’
– 2*x* – 2 = 0

**Solution:**

The given
function is *y* = *x*^{2} + 2*x* + C … (i)

Differentiating equation (i) w.r.t. x, we get

*y*’ = 2*x* + 2* ^{
}* ⇒

*y*’ – 2

*x*– 2 = 0 … (ii)

Hence, the function *y* = *x*^{2} + 2*x* + C is the solution of *y*’
– 2*x* – 2 = 0.

**Maths Class
12 Ex 9.2 Question 3.**

*y* = cos* x* + C :
*y*’ + sin *x* = 0

**Solution:**

The given
function is *y* = cos* x* + C … (i)

Differentiating equation (i) w.r.t. x, we get

*y*’ = –sin *x* ⇒ *y*’ + sin* x* = 0 … (ii)

Hence, the function *y* = cos* x* + C is the solution of *y*’ + sin *x* = 0.

**Maths Class
12 Ex 9.2 Question 4.**

**Solution:**

The
given function is *y* = √(1 + *x*^{2}) … (1)

Differentiating equation (i) w.r.t. x, we get

Hence, the function *y* = √(1 + *x*^{2}) is the
solution of *y*’ = *xy*/(1 + *x*^{2}).

**Maths Class
12 Ex 9.2 Question 5.**

*y* = A*x*
: *xy*’ = *y* (*x*
≠ 0)

**Solution:**

The given
function is *y* = A*x* … (i)

Differentiating equation (i) w.r.t. x, we get

*y*’ = A* ^{ }* … (ii)

Dividing equation (ii) by equation (i), we get

*y*’/*y *= 1/*x *⇒ *x**y*’ = *y *^{}

Hence, the function *y* = A*x* is the solution of *xy*’ = *y* (*x* ≠ 0).

**Maths Class
12 Ex 9.2 Question 6.**

**Solution:**

The given
function is *y* = *x *sin*
x* … (i)

Differentiating equation (i) w.r.t. x, we get

*y*’ = *x *cos*
x *+* *sin* x ^{ }* … (ii)

Multiplying
equation (ii) by *x*, we get

*xy*’ = *x*^{2}* *cos* x *+* x *sin* x*

**Maths Class 12 Ex 9.2 Question 7.**

**Solution:**

The given
function is *xy* = log *y* + C … (i)

Differentiating equation (i) w.r.t. x, we get

*y* – cos* y* = *x* :
(*y* sin *y *+ cos *y *+* x*) *y*’
= *y*

**Solution:**

The given
function is *y* – cos* y* = *x* … (i)

Differentiating equation (i) w.r.t. x, we get

*y*’ – (–sin *y*)*
y*’ = 1* ^{ }* ⇒

*y*’(1 + sin

*y*) = 1 … (ii)

^{}Multiplying equation (ii) by *y*,
we get

*y*’*y*(1 + sin* y*) = *y* ⇒ *y*’(*y* + *y*
sin* y*) = *y * ⇒ *y*’(*x*
+ cos* y *+ *y* sin* y*) = *y *

Hence, the function *y* – cos* y* = *x*
is the solution of *y*’(*x* + cos* y *+ *y* sin* y*) = *y*.

**Maths Class
12 Ex 9.2 Question 9.**

*x* +* y* = tan^{-1} *y*
: *y*^{2 }*y*’* *+ *y*^{2}* *+*
*1 = 0

**Solution:**

The given
function is *x* +* y* = tan^{-1} *y* … (i)

Differentiating equation (i) w.r.t. x, we get

1 + *y*’ = 1/(1 + *y*^{2}) × *y*’* ^{ }* ⇒ (1 +

*y*’) (1 +

*y*

^{2}) =

*y*’

⇒ 1 + *y*^{2} + *y*’ + *y*^{2 }*y*’ = *y*’ ⇒ *y*^{2 }*y*’* *+ *y*^{2}* *+*
*1 = 0 * ^{}* … (ii)

Hence, the function *x* +* y* = tan^{-1} *y* is the solution of *y*^{2 }*y*’* *+ *y*^{2}* *+* *1 = 0.

**Maths Class
12 Ex 9.2 Question 10.**

**Solution:**

**Maths Class 12 Ex 9.2 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:**

(D) The general
solution of a differential equation of fourth order has 4 arbitrary constants.

Because it contains the same number of arbitrary constants as the order of
differential equation.

**Maths Class
12 Ex 9.2 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:**

(D) Number of arbitrary
constants = 0

Because particular solution is free from arbitrary constants.

**NCERT Solutions for Maths Class 12 Exercise 9.1**

**NCERT Solutions for Maths Class 12 Exercise 9.3**

**NCERT Solutions for Maths Class 12 Exercise 9.4**

**NCERT Solutions for Maths Class 12 Exercise 9.5**

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**NCERT Solutions for Maths Class 12 Exercise 9.6**