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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Ex 9.4, 5 For each of the differential equations in Exercises 1 to 10, find the general solution : (๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ) )๐‘‘๐‘ฆโˆ’(๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) )๐‘‘๐‘ฅ=0 (๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ) )๐‘‘๐‘ฆโˆ’(๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) )๐‘‘๐‘ฅ=0 (๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ) ) dy = (๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) ) dx ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/(๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ) ) dx ๐‘‘๐‘ฆ = (๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/(๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ) ) dx Integrating both sides. โˆซ1โ–’๐‘‘๐‘ฆ = โˆซ1โ–’(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/(๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ) ) dx ๐‘ฆ = โˆซ1โ–’(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/(๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ) ) dx Let t = ๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ) ๐‘‘๐‘ก/๐‘‘๐‘ฅ = (๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) ) dx dx = ๐‘‘๐‘ก/(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ) ) Substituting values in (1), we get โˆซ1โ–’๐‘‘๐‘ฆ = โˆซ1โ–’(๐‘’^(๐‘ฅ )โˆ’ใ€– ๐‘’ใ€—^(โˆ’๐‘ฅ))/๐‘ก ๐‘‘๐‘ก/(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ) ) . โˆซ1โ–’๐‘‘๐‘ฆ = โˆซ1โ–’ใ€–๐‘‘๐‘ก/๐‘ก " " ใ€— y = log |๐‘ก|+๐‘ Putting back t = ๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ) y = log |๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ) | + C y = log (๐’†^๐’™+๐’†^(โˆ’๐’™)) + C

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.