Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Ex 9.3, 13 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition : ๐‘๐‘œ๐‘ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)=๐‘Ž(๐‘Žโˆˆ๐‘);๐‘ฆ=2 When ๐‘ฅ=0 cos (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = ๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = cosโˆ’1 ๐‘Ž ๐’…y = cosโˆ’1 ๐‘Ž dx Integrating both sides โˆซ1โ–’ใ€–๐‘‘๐‘ฆ=ใ€— โˆซ1โ–’ใ€–cos^(โˆ’1)โกใ€–๐‘Ž ๐‘‘๐‘ฅใ€— ใ€— y = x ใ€–๐’„๐’๐’”ใ€—^(โˆ’๐Ÿ) ๐‘Ž + c Putting x = 0 and y = 2 2 = 0 cos^(โˆ’1) ๐‘Ž + c 2 = c c = 2 Putting value of c in (1) y = x cos^(โˆ’1) ๐‘Ž + 2 y โˆ’ 2 = x ใ€–๐’„๐’๐’”ใ€—^(โˆ’๐Ÿ) ๐‘Ž (๐‘ฆ โˆ’ 2)/๐‘ฅ = cos^(โˆ’1) ๐‘Ž cos ((๐’š โˆ’ ๐Ÿ)/๐’™)= ๐‘Ž

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.