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Misc 14 - Find particular solution (x + 1) dy/dx = 2e-y - 1

Misc 14 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 14 - Chapter 9 Class 12 Differential Equations - Part 3 Misc 14 - Chapter 9 Class 12 Differential Equations - Part 4

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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 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.