Check sibling questions


Transcript

Ex 9.3, 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) )๐‘‘๐‘ฆ = (๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) )๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/(๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ) ) dx ๐’…๐’š = (๐’†^๐’™ โˆ’ ๐’†^(โˆ’๐’™))/(๐’†^๐’™ + ๐’†^(โˆ’๐’™) ) 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’š = โˆซ1โ–’(๐’†^๐’™ โˆ’ ๐’†^(โˆ’๐’™))/(๐’†^๐’™ + ๐’†^(โˆ’๐’™) ) dx Let t = ๐’†^๐’™+๐’†^(โˆ’๐’™) ๐‘‘๐‘ก/๐‘‘๐‘ฅ = (๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) ) dx = ๐’…๐’•/(๐’†^๐’™ โˆ’ ๐’†^(โˆ’๐’™) ) Putting value of t and dt in (1) 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’•|+๐’„ Putting back t = ๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) y = log |๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ) | + C As ๐’†^๐’™โˆ’๐’†^(โˆ’๐’™) > 0 So, its always positive Removing the modulus y = log (๐’†^๐’™โˆ’๐’†^(โˆ’๐’™)) + C

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

About the Author

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 and Computer Science at Teachoo