Ex 9.3, 2 For each of the differential equations in Exercises 1 to 10, find the general solution : ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=โˆš(4โˆ’๐‘ฆ^2 ) (โˆ’2<๐‘ฆ<2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆš(4โˆ’๐‘ฆ2) ๐‘‘๐‘ฆ/โˆš(4 โˆ’ ๐‘ฆ^2 ) = dx. ๐’…๐’š/โˆš(๐Ÿ^๐Ÿ โˆ’ ๐’š^๐Ÿ ) = dx Integrating both sides โˆซ1โ–’ใ€–๐‘‘๐‘ฆ/โˆš(2^2 โˆ’ ๐‘ฆ^2 )=โˆซ1โ–’๐‘‘๐‘ฅใ€— We know that โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )=sin^(โˆ’1)โกใ€–๐‘ฅ/๐‘Žใ€— ใ€— Putting ๐‘Ž = 2, x = y ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โกใ€–๐’š/๐Ÿใ€— = x + c ๐‘ฆ/2 = sin(x + c) y = 2 sin(x + c) โˆด y = 2 sin (x + c) is the general solution

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.