

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Miscellaneous
Misc 1 (ii)
Misc 1 (iii) Important
Misc 2 (i)
Misc 2 (ii) Important
Misc 2 (iii)
Misc 2 (iv) Important
Misc 3 Important
Misc 4
Misc 5 Important
Misc 6
Misc 7 Important
Misc 8 Important
Misc 9
Misc 10 Important
Misc 11 You are here
Misc 12 Important
Misc 13 (MCQ)
Misc 14 (MCQ) Important
Misc 15 (MCQ)
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Misc 11 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