# NCERT Solutions for Exercise 9.6 Class 12 Maths Chapter 9- Differential Equations

NCERT solutions for exercise 9.6 Class 12 Maths chapter 9 looks into the questions related to first-order linear differential equations. NCERT solutions for Class 12 Maths chapter 9 exercise 9.6 gives an insight into the concepts of linear differential equations and steps involved in solving linear differential equations. There are a few solved examples before the exercise 9.6 Class 12 Maths. The given sample questions give an idea about the steps involved in solving. Once these example questions are understood, students can solve Class 12 Maths chapter 9 exercise 9.6. The solutions to the Class 12th Maths chapter 6 exercise 9.6 are designed by expert Mathematics faculties and the same can be used to prepare for board exams that follow NCERT and also for competitive exams like JEE Main. Other practice exercises of the chapter given in the NCERT Book are listed below.

**Also check - **

Differential Equations exercise 9.1

Differential Equations exercise 9.2

Differential Equations exercise 9.3

Differential Equations exercise 9.4

Differential Equations exercise 9.5

Differential Equations miscellaneous exercise

## ** ****Differential Equations**** Class 12 Chapter 9 Exercise**: 9.6

**Differential Equations**

**Class 12 Chapter 9 Exercise**: 9.6

** Question:1 ** Find the general solution:

** Answer: **

Given equation is

This is type where p = 2 and Q = sin x

Now,

Now, the solution of given differential equation is given by relation

Let

Put the value of I in our equation

Now, our equation become

Therefore, the general solution is

** Question:2 ** Solve for general solution:

** Answer: **

Given equation is

This is type where p = 3 and

Now,

Now, the solution of given differential equation is given by the relation

Therefore, the general solution is

** Question:3 ** Find the general solution

** Answer: **

Given equation is

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Therefore, the general solution is

** Question:4 ** Solve for General Solution.

** Answer: **

Given equation is

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Therefore, the general solution is

** Question:5 ** Find the general solution.

** Answer: **

Given equation is

we can rewrite it as

This is where and

Now,

Now, the solution of given differential equation is given by relation

take

Now put again

Put this value in our equation

Therefore, the general solution is

** Question:6 ** Solve for General Solution.

** Answer: **

Given equation is

Wr can rewrite it as

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Let

Put this value in our equation

Therefore, the general solution is

** Question:7 ** Solve for general solutions.

** Answer: **

Given equation is

we can rewrite it as

This is type where and

Now,

Now, the solution of given differential equation is given by relation

take

Put this value in our equation

Therefore, the general solution is

** Question:8 ** Find the general solution.

** Answer: **

Given equation is

we can rewrite it as

This is type where and

Now,

Now, the solution of the given differential equation is given by the relation

Therefore, the general solution is

** Question:9 ** Solve for general solution.

** Answer: **

Given equation is

we can rewrite it as

This is type where and

Now,

Now, the solution of the given differential equation is given by the relation

Lets take

Put this value in our equation

Therefore, the general solution is

** Question:10 ** Find the general solution.

** Answer: **

Given equation is

we can rewrite it as

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Lets take

Put this value in our equation

Therefore, the general solution is

** Question:11 ** Solve for general solution.

** Answer: **

Given equation is

we can rewrite it as

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Therefore, the general solution is

** Question:12 ** Find the general solution.

** Answer: **

Given equation is

we can rewrite it as

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Therefore, the general solution is

** Question:13 ** Solve for particular solution.

** Answer: **

Given equation is

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Now, by using boundary conditions we will find the value of C

It is given that y = 0 when

at

Now,

Therefore, the particular solution is

** Question:14 ** Solve for particular solution.

** Answer: **

Given equation is

we can rewrite it as

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Now, by using boundary conditions we will find the value of C

It is given that y = 0 when x = 1

at x = 1

Now,

Therefore, the particular solution is

** Question:15 ** Find the particular solution.

** Answer: **

Given equation is

This is type where and

Now,

Now, the solution of given differential equation is given by relation

Now, by using boundary conditions we will find the value of C

It is given that y = 2 when

at

Now,

Therefore, the particular solution is

** 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) is the curve passing through origin

Then, the slope of tangent to the curve at point (x , y) is given by

Now, it is given that

It is type of equation where

Now,

Now,

Now, Let

Put this value in our equation

Now, by using boundary conditions we will find the value of C

It is given that curve passing through origin i.e. (x , y) = (0 , 0)

Our final equation becomes

Therefore, the required equation of the curve 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) is the curve passing through point (0 , 2)

Then, the slope of tangent to the curve at point (x , y) is given by

Now, it is given that

It is type of equation where

Now,

Now,

Now, Let

Put this value in our equation

Now, by using boundary conditions we will find the value of C

It is given that curve passing through point (0 , 2)

Our final equation becomes

Therefore, the required equation of curve is

** Question:18 ** The Integrating Factor of the differential equation is

(A)

(B)

(C)

(D)

** Answer: **

Given equation is

we can rewrite it as

Now,

It is type of equation where

Now,

Therefore, the correct answer is (C)

** Question:19 ** The Integrating Factor of the differential equation is

(A)

(B)

(C)

(D)

** Answer: **

Given equation is

we can rewrite it as

It is type of equation where

Now,

Therefore, the correct answer is (D)

**More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6**

A few examples are solved in the book just before the exercise 9.6 Class 12 Maths to get an idea of how these types of differential equations are solved. After the examples Class, 12 Maths chapter 9 exercise 9.6 are given and the NCERT solutions for Class 12 Maths chapter 9 exercise 9.6 are given here. There are 19 questions in the NCERT syllabus Class 12th Maths chapter 6 exercise 9.6. A few of them is to find the particular solutions of linear differential equations and the rest to find the general solutions. Two of the given questions are of multiple-choice type.

**Also Read| **Differential Equations Class 12th Notes

**Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6**

These solutions of the exercise 9.6 Class 12 Maths are useful to get an idea of topic 9.5.3

Students can expect one question of the types explained in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.6 for the CBSE board exam.

**Also see-**

NCERT Exemplar Solutions Class 12 Maths Chapter 9

NCERT Solutions for Class 12 Maths Chapter 9

**NCERT Solutions Subject Wise**

NCERT Solutions Class 12 Chemistry

NCERT Solutions for Class 12 Physics

NCERT Solutions for Class 12 Biology

NCERT Solutions for Class 12 Mathematics

**Subject Wise NCERT Exemplar Solutions**

- NCERT Exemplar Class 12 Maths
- NCERT Exemplar Class 12 Physics
NCERT Exemplar Class 12 Chemistry

NCERT Exemplar Class 12 Biology