# Misc 13

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 13 Find a particular solution of the differential equation 𝑑𝑦𝑑𝑥+𝑦 cot𝑥=4𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 𝑥≠0 , given that 𝑦=0 when 𝑥= 𝜋2 Given 𝑑𝑦𝑑𝑥+𝑦 cot𝑥=4𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 This of the form 𝑑𝑦𝑑𝑥+𝑃𝑦=𝑄 where P = cot x & Q = 4x cosec x IF = 𝑒 𝑃𝑑𝑥 IF = 𝑒 cot𝑥 𝑑𝑥 IF = 𝑒log( sin𝑥) IF = sin x Solution is y (IF) = 𝑄×𝐼.𝐹𝑑𝑥+𝑐 y sin x = 4𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 sin𝑥 𝑑𝑥+𝑐 y sin x = 4𝑥 1 sin𝑥 sin𝑥𝑑𝑥+𝑐 y sin x = 4𝑥 𝑑𝑥+𝑐 y sin x = 4 𝑥22+𝑐 y sin x = 2x2 + C Given that 𝑦=0 when 𝑥= 𝜋2 Put x = 𝜋2 & y = 0 in (1) 0 × sin 𝜋2 = 2 𝜋22+𝐶 0 = 2 𝜋42 + C 0 = 𝜋22 + C C = −𝜋22 Putting value of C in (1) y sin x = 2x2 + c y sin x = 2 𝒙𝟐 − 𝝅𝟐𝟐

Chapter 9 Class 12 Differential Equations

Example 1
Important

Ex 9.1, 11 Important

Ex 9.1, 12 Important

Example 7 Important

Ex 9.3, 7 Important

Ex 9.3, 10 Important

Example 13 Important

Ex 9.4, 14 Important

Example 17 Important

Example 18 Important

Ex 9.5, 8 Important

Ex 9.5, 15 Important

Example 22 Important

Ex 9.6, 7 Important

Ex 9.6, 13 Important

Ex 9.6, 14 Important

Example 25 Important

Example 27 Important

Example 28 Important

Misc 6 Important

Misc 11 Important

Misc 12 Important

Misc 13 Important You are here

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.