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


Transcript

Example 22 (Introduction) Solve the differential equation ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ+๐‘ฅ/(1+๐‘ฆ^2 )=tan^(โˆ’1)โก๐‘ฆ/(1+๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ+๐‘ฅ/(1 + ๐‘ฆ^2 )=tan^(โˆ’1)โก๐‘ฆ/(1 + ๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=tan^(โˆ’1)โก๐‘ฆ/(1 + ๐‘ฆ^2 ) โ€“ ๐‘ฅ/(1 + ๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=(tan^(โˆ’1)โก๐‘ฆ โˆ’ ๐‘ฅ)/(1 + ๐‘ฆ^2 ) ๐’…๐’š/๐’…๐’™ = ((๐Ÿ + ๐’š^๐Ÿ))/ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โกใ€–๐’š โˆ’ ๐’™ใ€— The variables cannot be separated. So variable separable method is not possible Now, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((1 + ๐‘ฆ^2))/tan^(โˆ’1)โกใ€–๐‘ฆโˆ’๐‘ฅใ€— Put F(x, y) = ๐’…๐’š/๐’…๐’™ F(x, y) = (1 + ๐‘ฆ^2)/(tan^(โˆ’1)โก๐‘ฆโˆ’๐‘ฅ) F(๐œ†x, ๐œ†y) = (1 + ๐œ†^2 ๐‘ฆ^2)/(tan^(โˆ’1)โก๐œ†๐‘ฆโˆ’๐œ†๐‘ฅ) F(๐œ†x, ๐œ†y) โ‰  ๐œ†ยฐ F(x, y) Hence, the equation is not homogenous. So we use the integrating factor method ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((1 + ๐‘ฆ^2))/(tan^(โˆ’1) y โˆ’ x) This is not of the form ๐’…๐’š/๐’…๐’™+๐‘ท๐’š=๐‘ธ โˆด We need to find ๐’…๐’™/๐’…๐’š ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ = (tan^(โˆ’1)โก๐‘ฆ โˆ’ ๐‘ฅ)/(1 + ๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ = tan^(โˆ’1)โก๐‘ฆ/(1 + ๐‘ฆ^2 ) โˆ’ ๐‘ฅ/(1 + ๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ + ๐‘ฅ/(1 + ๐‘ฆ^2 ) โˆ’ (tan^(โˆ’1)โก๐‘ฆ )/(1 + ๐‘ฆ^2 ) Differential equation is of the form ๐’…๐’™/๐’…๐’š + P1 x = Q1 Thus, we solve question by integrating factor method taking ๐’…๐’™/๐’…๐’š Example 22 Solve the differential equation ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ+๐‘ฅ/(1+๐‘ฆ^2 )=tan^(โˆ’1)โก๐‘ฆ/(1+๐‘ฆ^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ+๐‘ฅ/(1 + ๐‘ฆ^2 )=tan^(โˆ’1)โก๐‘ฆ/(1 + ๐‘ฆ^2 ) Differential equation is of the form ๐’…๐’™/๐’…๐’š + P1 x = Q1 where P1 = 1/(1 + ๐‘ฆ^2 ) & Q1 = (tan^(โˆ’1)โก๐‘ฆ )/(1 + ๐‘ฆ^2 ) Now, IF = ๐‘’^โˆซ1โ–’ใ€–๐‘ƒ_1 ๐‘‘๐‘ฆใ€— IF = ๐‘’^โˆซ1โ–’ใ€–1/(1 + ๐‘ฆ^2 ) ๐‘‘๐‘ฆใ€— IF = ๐’†^ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก๐’š Solution is x (IF) = โˆซ1โ–’ใ€–(๐‘„ร—๐ผ๐น)๐‘‘๐‘ฆ+๐ถใ€— x๐’†^ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก๐’š = โˆซ1โ–’ใ€–ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก๐’š/(๐Ÿ + ๐’š^๐Ÿ )ร—๐’†^ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก๐’š ๐’…๐’šใ€— + C Let I = โˆซ1โ–’ใ€–tan^(โˆ’1)โก๐‘ฆ/(1 + ๐‘ฆ^2 )ร—๐‘’^tan^(โˆ’1)โก๐‘ฆ ๐‘‘๐‘ฆใ€— Let ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โกใ€–๐’š ใ€—= t 1/(1 + ๐‘ฆ^2 ) dy = dt Putting values of t & dt in I I = โˆซ1โ–’ใ€–๐’•๐’†^๐’• ๐’…๐’•ใ€— Integrating by parts with โˆซ1โ–’ใ€–๐‘“(๐‘ก) ๐‘”(๐‘ก) ๐‘‘๐‘ก=๐‘“(๐‘ก) โˆซ1โ–’ใ€–๐‘”(๐‘ก) ๐‘‘๐‘ก โˆ’โˆซ1โ–’ใ€–[๐‘“^โ€ฒ (๐‘ก) โˆซ1โ–’ใ€–๐‘”(๐‘ก) ๐‘‘๐‘ก] ๐‘‘๐‘กใ€—ใ€—ใ€—ใ€— Take f (t) = t & g (t) = ๐‘’^"t" I = t.โˆซ1โ–’ใ€–๐‘’^๐‘ก ๐‘‘๐‘กโˆ’โˆซ1โ–’[1โˆซ1โ–’ใ€–๐‘’^๐‘ก ๐‘‘๐‘กใ€—] ๐‘‘๐‘กใ€— I = t๐‘’^๐‘ก โˆ’ โˆซ1โ–’ใ€–๐‘’^๐‘ก ๐‘‘๐‘กใ€— I = t๐’†^๐’• โˆ’ ๐’†^๐’• I = ๐‘’^๐‘ก (t โˆ’ 1) Putting value of t = tan^(โˆ’1)โก๐‘ฆ I = ๐’†^(ใ€–๐ญ๐š๐งใ€—^(โˆ’๐Ÿ)โก๐’š ) (ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ) ๐’šโˆ’๐Ÿ) Putting value of I in (1) ใ€–๐‘ฅ๐‘’ใ€—^(tan^(โˆ’1)โก๐‘ฆ ) = ๐ผ + ๐ถ ใ€–๐‘ฅ๐‘’ใ€—^(tan^(โˆ’1) ๐‘ฆ )= ๐‘’^(tan^(โˆ’1) ๐‘ฆ ) (tan^(โˆ’1)โก๐‘ฆโˆ’1) + c Divide by ๐’†^(ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ) ๐’š ) ใ€–๐‘ฅ๐‘’ใ€—^(tan^(โˆ’1) ๐‘ฆ )/๐‘’^tan^(โˆ’1)โกใ€–๐‘ฆ ใ€— = (๐‘’^(tan^(โˆ’1) ๐‘ฆ) (tan^(โˆ’1)โก๐‘ฆ โˆ’ 1))/๐‘’^tan^(โˆ’1)โกใ€–๐‘ฆ ใ€— + ๐‘/๐‘’^tan^(โˆ’1)โกใ€–๐‘ฆ ใ€— ๐’™ = (ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก๐’šโˆ’๐Ÿ) + c๐’†^(โˆ’ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ) ๐’š ) Which is the required 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.