# Ex 9.6, 19 (MCQ) - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Aug. 20, 2021 by Teachoo

Last updated at Aug. 20, 2021 by Teachoo

Transcript

Ex 9.6, 19 The integrating Factor of the differential equation (1โ๐ฆ^2 ) ๐๐ฅ/๐๐ฆ+๐ฆ๐ฅ=๐๐ฆ (โ1<๐ฆ<1) is (A) 1/(๐ฆ^2โ1) (B) 1/โ(๐ฆ^2โ1) (C) 1/(1โ๐ฆ^2 ) (D) 1/โ(1โ๐ฆ^2 ) (1โ๐ฆ^2 ) ๐๐ฅ/๐๐ฆ+๐ฆ๐ฅ=๐๐ฆ Dividing both sides by 1 โ y2 ๐๐ฅ/๐๐ฆ + ๐ฆ๐ฅ/(1โ๐ฆ^2 ) = ๐๐ฆ/(1โ๐ฆ^2 ) Differential equation is of the form ๐ ๐/๐ ๐ + P1x = Q1 where P1 = ๐ฆ/(1 โ ๐ฆ^2 ) & Q1 = ๐๐ฆ/(1 โ ๐ฆ^2 ) IF = ๐^โซ1โ๐1๐๐ฆ Finding โซ1โใ๐ท๐ ๐ ๐ใ โซ1โใ๐1 ๐๐ฆ=ใ โซ1โใ๐ฆ/(1โ๐ฆ^2 ) ๐๐ฆ ใ Putting 1 โ y2 = t โ2y dy = dt y dy = (โ1)/2 dt โด Our equation becomes โซ1โใ๐1 ๐๐ฆ= (โ1)/2 ใ โซ1โใ๐๐ก/๐ก ใ โซ1โใ๐1 ๐๐ฆ= (โ1)/2 ใ logโก๐ก Putting back value of t โซ1โใ๐1 ๐๐ฆ= (โ1)/2 ใ logโก(1โ๐ฆ^2) โซ1โใ๐1 ๐๐ฆ=ใ ใlogโก(1โ๐ฆ^2 )ใ^((โ1)/2) โซ1โใ๐1 ๐๐ฆ= ใ log 1/โ(1 โ ๐ฆ^2 ) Thus, IF = ๐^โซ1โ๐1๐๐ฅ IF = ๐^(๐๐๐ 1/โ(1 โ ๐ฆ^2 )) IF = ๐/โ(๐ โ ๐^๐ ) So, the correct answer is (d)

Ex 9.6

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Ex 9.6, 10 Deleted for CBSE Board 2022 Exams

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Ex 9.6, 12 Important Deleted for CBSE Board 2022 Exams

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Ex 9.6, 18 (MCQ)

Ex 9.6, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams You are here

Chapter 9 Class 12 Differential Equations (Term 2)

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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.