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# NCERT Solutions for Class 12 Maths Differential Equations

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Differential Equations Exercise 9.2*.

As we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the **CBSE NCERT Syllabus**. Thus, * NCERT Solutions* helps the students to solve the exercise questions as given in

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**NCERT Solutions for Class 12 Maths Differential Equations Exercise 9.2**The PDF books of * NCERT Solutions for Class 12* are the first step towards the learning and understanding the each sections of Maths, Physics, Chemistry, Biology as it all help in engineering medical entrance exams. To solve it, you just need to click on download links of

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**CBSE Class 12**^{th}Maths## NCERT Solutions for Class 12 Maths Differential Equations

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

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 1.**

Check whether the function y = e^{x} + 1 is the solution of y″ – y′ = 0.

**NCERT Solutions:**

Given function is y = e^{x} + 1 …(1)

Differentiating the equation (1) with respect to x, we get

Thus, the given function y = e^{x} + 1 is the solution of y″ – y′ = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 2.**

Check whether the function y = x^{2} + 2x + C is the solution of y′ – 2x – 2 = 0.

**NCERT Solutions:**

Given function is y = x^{2} + 2x + C …(1)

Differentiating the equation (1) with respect to x, we get

Thus, the given function y = x^{2} + 2x + C is the solution of y′ – 2x – 2 = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 3.**

Check whether the function y = cos x + C is the solution of y′ + sin x = 0.

**NCERT Solutions:**

Given function is y = cos x + C …(1)

Differentiating the equation (1) with respect to x, we get

Thus, the given function y = cos x + C is the solution of y′ + sin x = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 4.**

Check whether the function y = √(1 + x^{2}) is the solution of y′ = xy/(1 + x^{2}).

**NCERT Solutions:**

Given function is y = √(1 + x^{2}) …(1)

Differentiating the equation (1) with respect to x, we get

…(2)

Dividing (2) by (1), we get

Thus, the given function y = √(1 + x^{2}) is the solution of y′ = xy/(1 + x^{2}).

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 5.**

Check whether the function y = Ax is the solution of xy′ = y (x ≠ 0).

**NCERT Solutions:**

Given function is y = Ax …(1)

Differentiating the equation (1) with respect to x, we get

y′ = A …(2)

Dividing (2) by (1), we get

Thus, the given function y = Ax is the solution of xy′ = y (x ≠ 0).

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 6.**

Check whether the function y = x sin x is the solution of xy′ = y + x√(x^{2} – y^{2}), (x ≠ 0 and x > y or x < – y).

**NCERT Solutions:**

Given function is y = x sin x …(1)

Differentiating the equation (1) with respect to x, we get

y′ = x cos x + sin x …(2)

Multiply x in (2), we get

xy′ = x^{2} cos x + x sin x

Thus, the given function y = x sin x is the solution of xy′ = y + x√(x^{2} – y^{2}), (x ≠ 0 and x > y or x < – y).

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 7.**

Check whether the function xy = log y + C is the solution of , xy ≠ 1.

**NCERT Solutions:**

Given function is xy = log y + C …(1)

Differentiating the equation (1) with respect to x, we get

…(2)

Thus, the given function xy = log y + C is the solution of , xy ≠ 1.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 8.**

Check whether the function y – cos y = x is the solution of (y sin y + cos y + x) y′ = y.

**NCERT Solutions:**

Given function is y – cos y = x …(1)

Differentiating the equation (1) with respect to x, we get

y’ – (–sin y)y’ = 1 ⇒ y'(1 + sin y) = 1 …(2)

Multiply y in (2), we get

Thus, the given function y – cos y = x is the solution of (y sin y + cos y + x) y′ = y.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 9.**

Check whether the function x + y = tan^{–1}y is the solution of y^{2}y′ + y^{2} + 1 = 0.

**NCERT Solutions:**

Given function is x + y = tan^{–1}y …(1)

Differentiating the equation (1) with respect to x, we get

Multiply y in (2), we get

Thus, the given function x + y = tan^{–1}y is the solution of y^{2}y′ + y^{2} + 1 = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 10.**

Check whether the function is the solution of .

**NCERT Solutions:**

Given function is …(1)

Differentiating the equation (1) with respect to x, we get

Thus, the given function is the solution of .

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 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

**NCERT Solutions:**

(D)

Given function The number of arbitrary constant in the general solution of differential equation of nth order is n. So, the differential equation of fourth order have 4 constant.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.2: Ques No 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

**NCERT Solutions:**

(D)

In a particular solution of a differential equation of third order, there is no arbitrary constant.