Ex 9.6, 19 - Class 12 - Part 1 - Find Integrating Factor.jpeg

Ex 9.6, 19 - Class 12 - Part 2 - Integrating w.r.t y.jpeg
Ex 9.6, 19 - Class 12 - Part 3 - The correct answer is.jpeg

  1. Chapter 9 Class 12 Differential Equations (Term 2)
  2. Serial order wise

Transcript

Ex 9.6, 19 The integrating Factor of the differential equation (1โˆ’๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ+๐‘ฆ๐‘ฅ=๐‘Ž๐‘ฆ (โˆ’1<๐‘ฆ<1) is (A) 1/(๐‘ฆ^2โˆ’1) (B) 1/โˆš(๐‘ฆ^2โˆ’1) (C) 1/(1โˆ’๐‘ฆ^2 ) (D) 1/โˆš(1โˆ’๐‘ฆ^2 ) (1โˆ’๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ+๐‘ฆ๐‘ฅ=๐‘Ž๐‘ฆ Dividing both sides by 1 โˆ’ y2 ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ + ๐‘ฆ๐‘ฅ/(1โˆ’๐‘ฆ^2 ) = ๐‘Ž๐‘ฆ/(1โˆ’๐‘ฆ^2 ) Differential equation is of the form ๐’…๐’™/๐’…๐’š + P1x = Q1 where P1 = ๐‘ฆ/(1 โˆ’ ๐‘ฆ^2 ) & Q1 = ๐‘Ž๐‘ฆ/(1 โˆ’ ๐‘ฆ^2 ) IF = ๐‘’^โˆซ1โ–’๐‘1๐‘‘๐‘ฆ Finding โˆซ1โ–’ใ€–๐‘ท๐Ÿ ๐’…๐’šใ€— โˆซ1โ–’ใ€–๐‘ƒ1 ๐‘‘๐‘ฆ=ใ€— โˆซ1โ–’ใ€–๐‘ฆ/(1โˆ’๐‘ฆ^2 ) ๐‘‘๐‘ฆ ใ€— Putting 1 โˆ’ y2 = t โˆ’2y dy = dt y dy = (โˆ’1)/2 dt โˆด Our equation becomes โˆซ1โ–’ใ€–๐‘ƒ1 ๐‘‘๐‘ฆ= (โˆ’1)/2 ใ€— โˆซ1โ–’ใ€–๐‘‘๐‘ก/๐‘ก ใ€— โˆซ1โ–’ใ€–๐‘ƒ1 ๐‘‘๐‘ฆ= (โˆ’1)/2 ใ€— logโก๐‘ก Putting back value of t โˆซ1โ–’ใ€–๐‘ƒ1 ๐‘‘๐‘ฆ= (โˆ’1)/2 ใ€— logโก(1โˆ’๐‘ฆ^2) โˆซ1โ–’ใ€–๐‘ƒ1 ๐‘‘๐‘ฆ=ใ€— ใ€–logโก(1โˆ’๐‘ฆ^2 )ใ€—^((โˆ’1)/2) โˆซ1โ–’ใ€–๐‘ƒ1 ๐‘‘๐‘ฆ= ใ€— log 1/โˆš(1 โˆ’ ๐‘ฆ^2 ) Thus, IF = ๐‘’^โˆซ1โ–’๐‘1๐‘‘๐‘ฅ IF = ๐‘’^(๐‘™๐‘œ๐‘” 1/โˆš(1 โˆ’ ๐‘ฆ^2 )) IF = ๐Ÿ/โˆš(๐Ÿ โˆ’ ๐’š^๐Ÿ ) So, the correct answer is (d)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.