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


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Ex 9.4, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : [๐‘ฅ sin^2โกใ€–(๐‘ฆ/๐‘ฅ)โˆ’๐‘ฆใ€— ]๐‘‘๐‘ฅ+๐‘ฅ๐‘‘๐‘ฆ=0;๐‘ฆ=๐œ‹/4 When ๐‘ฅ=1 Ex 9.4, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : [๐‘ฅ sin^2โกใ€–(๐‘ฆ/๐‘ฅ)โˆ’๐‘ฆใ€— ]๐‘‘๐‘ฅ+๐‘ฅ๐‘‘๐‘ฆ=0;๐‘ฆ=๐œ‹/4 When ๐‘ฅ=1 Differential equation can be written as [๐‘ฅ sin^2โกใ€–(๐‘ฆ/๐‘ฅ)โˆ’๐‘ฆใ€— ]๐‘‘๐‘ฅ+๐‘ฅ๐‘‘๐‘ฆ=0 x dy = โˆ’ [๐‘ฅ sin^2โกใ€–(๐‘ฆ/๐‘ฅ)โˆ’๐‘ฆใ€— ]dx ๐’…๐’š/๐’…๐’™ = โˆ’[๐’™/๐’™ ใ€–๐ฌ๐ข๐งใ€—^๐Ÿโกใ€–๐’š/๐’™ โˆ’๐’š/๐’™ใ€— ] Let F(x, y) = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ =โˆ’[๐‘ฅ/๐‘ฅ sin^2โกใ€–๐‘ฆ/๐‘ฅ โˆ’๐‘ฆ/๐‘ฅใ€— ] Finding F(๐€x, ๐€y) F(๐œ†x, ๐œ†y) = โˆ’[sin^2โกใ€–๐œ†๐‘ฆ/๐œ†๐‘ฅโˆ’๐œ†๐‘ฆ/๐œ†๐‘ฅใ€— ] = โˆ’[sin^2โกใ€–๐‘ฆ/๐‘ฅโˆ’๐‘ฆ/๐‘ฅใ€— ] = ๐œ†ยฐ F (x, y) โˆด F(x, y) is ๐‘Ž homogenous function of degree zero Putting y = vx Diff w.r.t. x ๐’…๐’š/๐’…๐’™ = x ๐’…๐’—/๐’…๐’™ + v Putting value of ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ and y = vx in (1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’[๐‘ฅ/๐‘ฅ sin^2โกใ€–๐‘ฆ/๐‘ฅ โˆ’๐‘ฆ/๐‘ฅใ€— ] v + (๐’™ ๐’…๐’—)/๐’…๐’™ = โˆ’[ใ€–๐ฌ๐ข๐งใ€—^๐Ÿ ๐’—๐’™/๐’™โˆ’๐’—๐’™/๐’™] v + (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = โˆ’[sin^2 ๐‘ฃโˆ’๐‘ฃ] (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = v โˆ’ sin2 v โˆ’ v (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = โˆ’ sin2 v ๐’…๐’—/ใ€–๐’”๐’Š๐’ใ€—^๐Ÿโก๐’— = โˆ’๐’…๐’™/๐’™ Integrating both sides โˆซ1โ–’๐‘‘๐‘ฃ/(sin^2 ๐‘ฃ) = โˆ’โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ โˆซ1โ–’ใ€–๐‘๐‘œ๐‘ ๐‘’๐‘^2 ๐‘ฃ ๐‘‘๐‘ฃใ€— = โˆ’โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ โˆ’ cot v = โˆ’ log |๐‘ฅ|+๐‘ log |๐’™| โˆ’ cot v = C Putting value of v = ๐‘ฆ/๐‘ฅ log |๐‘ฅ| โˆ’ cot (๐‘ฆ/๐‘ฅ) = C Putting x = 1 & y = ๐…/๐Ÿ’ log 1 โˆ’ cot (๐œ‹/4) = C 0 โˆ’ 1 = C C = โˆ’1 Putting value of C in (2) log |๐‘ฅ| โˆ’ cot ๐‘ฆ/๐‘ฅ = โˆ’1 cot ๐‘ฆ/๐‘ฅ โˆ’ log |๐‘ฅ| = 1 cot ๐‘ฆ/๐‘ฅ โˆ’ log |๐‘ฅ| = log e cot ๐‘ฆ/๐‘ฅ = log |๐‘ฅ| + log e cot ๐’š/๐’™ = log |๐’†๐’™|

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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.