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

Transcript

Misc 14 Find a particular solution of the differential equation (๐‘ฅ+1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2 ๐‘’^(โˆ’๐‘ฆ)โˆ’1 , given that ๐‘ฆ=0 when ๐‘ฅ=0 (๐‘ฅ+1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2๐‘’^(โˆ’๐‘ฆ)โˆ’1 The variables are separable ๐‘‘๐‘ฆ/(2๐‘’^( โˆ’๐‘ฆ ) โˆ’ 1) = ๐‘‘๐‘ฅ/(๐‘ฅ + 1) Integrating both sides โˆซ1โ–’๐‘‘๐‘ฆ/(2๐‘’^(โˆ’๐‘ฆ) โˆ’ 1) = โˆซ1โ–’๐‘‘๐‘ฅ/(๐‘ฅ + 1) โˆซ1โ–’๐‘‘๐‘ฆ/(2/๐‘’^๐‘ฆ โˆ’ 1) = log (x + 1) + c โˆซ1โ–’๐‘‘๐‘ฆ/((2 โˆ’ ๐‘’^๐‘ฆ)/๐‘’^๐‘ฆ ) = log (x + 1) + C โˆซ1โ–’๐‘’^๐‘ฆ/(2 โˆ’ใ€– ๐‘’ใ€—^๐‘ฆ ) dy = log (x + 1) + C Putting t = 2โˆ’๐‘’^๐‘ฆ dt = โˆ’๐‘’^๐‘ฆdy โ€“dt = ๐‘’^๐‘ฆdy Putting value of t & dt in equation โˆซ1โ–’(โˆ’๐‘‘๐‘ก)/๐‘ก = log (x + 1) + c โˆ’ log t = log (x + 1) + c Putting back value of t โ€ฆ(1) โˆ’ log (2 โˆ’ ๐‘’^๐‘ฆ) = log (x + 1) + C 0 = log (x + 1) + log (2 โˆ’ ๐‘’^๐‘ฆ) + C log (x + 1) + log (2 โˆ’ ๐‘’^๐‘ฆ) + C = 0 Given y = 0 when x = 0 Putting x = 0 & y = 0 in (1) log (0 + 1) + log (2 โˆ’ e0) + C = 0 log 1 + log (2 โˆ’ 1) + C = 0 log 1 + log 1 + C = 0 0 + 0 + C = 0 C = 0 Putting value of C in (1) (As log 1 = 0) log (x + 1) + log (2 โˆ’ ๐‘’^๐‘ฆ) + 0 = 0 log (x + 1) + log (2 โˆ’ ๐‘’^๐‘ฆ) = 0 log (2 โˆ’ ey) = โ€“ log (x + 1) log (2 โˆ’ ey) = log (x + 1)โ€“1 log (2 โˆ’ ey) = log (1/(๐‘ฅ + 1)) 2 โˆ’ ey = 1/(๐‘ฅ + 1) ey = 2โˆ’1/(๐‘ฅ + 1) ey = (2๐‘ฅ + 2 โˆ’ 1)/(๐‘ฅ + 1) ey = (2๐‘ฅ + 1)/(๐‘ฅ + 1) Taking log both sides y = log |(๐Ÿ๐’™ + ๐Ÿ)/(๐’™ + ๐Ÿ)| , x โ‰  โˆ’1 (๐ด๐‘  ๐‘Ž logโกใ€–๐‘ฅ=logโกใ€–๐‘ฅ^๐‘Ž ใ€— ใ€— )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.