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NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations Exercise 9.4 Question 1 to 23 Answer: Downlaod pdf

For each of the differential equations in Exercises 1 to 10, find the general solution: 1. 1 cos 1 cos dy x dx x 2. 2 4 ( 2 2) dy y y d 3. 1 ( 1) dy y y dx 4. sec2 x tan y dx + sec2 y tan x dy = 0 5. (e x + e x ) dy (e x e x ) dx = 0 6. 2 2 (1 ) (1 ) dy x y dx 7. y log y dx x dy = 0 8. 5 5 dy x y dx 9. 1 sin dy x dx 10. e x tan y dx + (1 e x ) sec2 y dy = 0 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition: 11. 3 2 ( 1) dy xxx dx = 2x 2 + x; y = 1 when x = 0 12. 2 ( 1) 1 dy x x dx ; y = 0 when x = 2 13. cos dy a dx (a R); y = 1 when x = 0 14. tan dy y x

ncert solutions for class 12 Math Chapter 9 ex.9.4

Other EXERCISE for Class 12 Chapter 9 Differential Equations NCERT Solutions




For each of the differential equations in Exercises 1 to 10, find the general solution:

Question 1:1 cos 1 cos dy x dx x

Question 2:2 4 ( 2 2) dy y y d

ncert class 12 maths chapter 9 Differential Equations

Question 3:1 ( 1) dy y y dx

exercise 9.4 maths class 12 Chapter 9 Differential Equations

Question 4:sec2 x tan y dx + sec2 y tan x dy = 0

Question 5:(e x + e –x ) dy – (e x – e –x ) dx = 0

cbse class 12 maths ncert solutions Chapter 9

Question 6:2 2 (1 ) (1 ) dy x y dx

Question 7:y log y dx – x dy = 0

Class 12 maths solutions Differential Equations

Question 8:5 5 dy x y dx

ncert class 12 chemistry chapter 9 exercise solutions

Question 9:1 sin dy x dx

Question 10:e x tan y dx + (1 – e x ) sec2 y dy = 0 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

Differential Equations Class 12 ncert solutions

Question 11:3 2 ( 1) dy xxx dx = 2x 2 + x; y = 1 when x = 0

Question 12:2 ( 1) 1 dy x x dx ; y = 0 when x = 2

Class 12 Maths NCERT Solutions Chapter 9 Differential Equations Exercise 9.4

Question 13:cos dy a dx (a R); y = 1 when x = 0

Exercise 9.4 class 12 Math ncert solutions Chapter 9

Question 14:tan dy y x dx ; y = 1 when x = 0

Question 15:Find the equation of a curve passing through the point (0, 0) and whose differential equation is y = e x sin x.

For each of the differential equations in Exerc

Question 16:For the differential equation ( 2) ( 2) dy xy x y dx , find the solution curve passing through the point (1, –1).

class 12 maths ncert solution pdf download Chapter 9

Question 17:Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Question 18:At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Differential Equations Math ncert solution class 12

Question 19:The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds

Question 20:In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge 2 = 0.6931).

Differential Equations ncert solutions class 12

Question 21:In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e 0 5 = 1.648).

CBSE NCERT Solutions For Class 12 Differential Equations

Question 22:In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Question 23:The general solution of the differential equation dy x y e dx✁ is (A) e x + e –y = C (B) e x + e y = C (C) e –x + e y = C (D) e –x + e –y =

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