# Misc 13 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Nov. 14, 2019 by

Last updated at Nov. 14, 2019 by

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

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 1 (iii) Important

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii)

Misc 2 (iv) Important

Misc 3 Deleted for CBSE Board 2022 Exams

Misc 4 Important

Misc 5 Important Deleted for CBSE Board 2022 Exams

Misc 6

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13 You are here

Misc 14 Important

Misc 15 Important

Misc 16 (MCQ)

Misc 17 (MCQ) Important Deleted for CBSE Board 2022 Exams

Misc 18 (MCQ)

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.