# Misc 13

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 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

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .