 # NCERT Solution Class 12 Chapter 9- Differential Equations Exercise 9.6

The Exercise 9.6 of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations is based on solving the linear differential equations. A linear differential equation of the first order can be defined as a differential equation that involves the function y and its first derivative alone. Such equations are physically suitable for describing various linear phenomena in biology, economics, population dynamics, and physics. Hence, it is necessary to understand the concept given in this exercise.

## Download PDF of NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations Exercise 9.6                  ### Access Answers of Maths NCERT Class 12 Maths Chapter 9- Differential Equations Exercise 9.6 Page Number 413

For each of the differential equations given in question, find the general solution: Solution:   Solution:  Solution:  Solution:   Solution:   Solution:   Solution:  8. (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)

Solution:   Solution:   Solution:  11. y dx + (x – y2)dy = 0

Solution:   Solution: ⇒ x = 3y2 + Cy

Therefore, the required general solution of the given differential equation is x = 3y2 + Cy.

For each of the differential equations given in Exercises 13 to 15, find a particular

solution satisfying the given condition: Solution:  Solution:  Solution: 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.

Solution: Now, it is given that curve passes through origin.

Thus, equation 2 becomes

1 = C

⇒ C = 1

Substituting C = 1 in equation 2, we get,

x + y – 1 = ex

Therefore, the required general solution of the given differential equation is

x + y -1 = ex

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.

Solution: Thus, equation (2) becomes:

0 + 2 – 4 = C e0

⇒ – 2 = C

⇒ C = -2

Substituting C = -2 in equation (2), we get,

x + y – 4 =-2ex

⇒ y = 4 – x – 2ex

Therefore, the required general solution of the given differential equation is

y = 4 – x – 2ex 18. The Integrating Factor of the differential equation is

A. e–x B. e–y C. 1/x D. x

Solution:

C. 1/x

Explanation: 19. The Integrating Factor of the differential equation Solution: Explanation: #### Access other exercise solutions of Class 12 Maths Chapter 9

Exercise 9.1 Solutions 12 Questions

Exercise 9.2 Solutions 12 Questions

Exercise 9.3 Solutions 12 Questions

Exercise 9.4 Solutions 23 Questions

Exercise 9.5 Solutions 17 Questions

Miscellaneous Exercise On Chapter 9 Solutions 18 Questions