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


Transcript

Ex 9.3, 9 For each of the differential equations in Exercises 1 to 10, find the general solution : ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=sin^(โˆ’1)โก๐‘ฅ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=sin^(โˆ’1)โก๐‘ฅ ๐‘‘๐‘ฆ = sin^(โˆ’1)โก๐‘ฅ dx Integrating both sides โˆซ1โ–’ใ€–๐’…๐’š ใ€—= โˆซ1โ–’ใ€–ใ€–๐ฌ๐ข๐งใ€—^(โˆ’๐Ÿ)โกใ€–๐’™.๐Ÿ ๐’…๐’™ใ€— ใ€— Integrating by parts, using formula โˆซ1โ–’ใ€–๐‘“ (๐‘ฅ)๐‘”(๐‘ฅ)๐‘‘๐‘ฅ ใ€—= ๐‘“(๐‘ฅ) โˆซ1โ–’ใ€–๐‘”(๐‘ฅ)๐‘‘๐‘ฅ โˆ’โˆซ1โ–’ใ€–[๐‘“โ€ฒ(๐‘ฅ)โˆซ1โ–’๐‘”(๐‘ฅ)๐‘‘๐‘ฅ] ๐‘‘๐‘ฅ ใ€— ใ€— Take f(x) = sinโˆ’1 x and g(x) = 1 y = x ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) ๐’™ โˆ’ โˆซ1โ–’๐’™/โˆš(๐Ÿ โˆ’ ๐’™^๐Ÿ ) dx Let t = 1 โˆ’ x2 dt = โˆ’2xdx x dx = (โˆ’๐‘‘๐‘ก)/2 Hence, our equation becomes y = x sinโˆ’1 x โˆ’ โˆซ1โ–’(โˆ’๐‘‘๐‘ก)/(2โˆš๐‘ก) y = x sinโˆ’1 x + โˆซ1โ–’๐‘‘๐‘ก/(2โˆš๐‘ก) y = x sinโˆ’1 x + ๐Ÿ/๐Ÿ โˆซ1โ–’ใ€–๐’•^((โˆ’๐Ÿ)/๐Ÿ) ๐’…๐’•ใ€— y = x sinโˆ’1 x + ๐Ÿ/๐Ÿ ๐’•^((โˆ’๐Ÿ)/๐Ÿ + ๐Ÿ)/((โˆ’๐Ÿ)/๐Ÿ + ๐Ÿ) + C y = x sinโˆ’1 x + 1/2 (๐‘ก^(1/2) )/((1/2) )+๐ถ y = x sinโˆ’1 x + โˆš๐‘ก + C Putting back value of t y = x sinโˆ’1 x + โˆš(๐Ÿโˆ’๐’™^๐Ÿ ) + C y = sinโˆ’1 x โˆซ1โ–’ใ€–๐Ÿ ๐’…๐’™ โˆ’โˆซ1โ–’[๐Ÿ/โˆš(๐Ÿ โˆ’ ๐’™^๐Ÿ ) โˆซ1โ–’ใ€–๐Ÿ.๐’…๐’™ ใ€—] ใ€— dx

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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.