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


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Misc 12 Find a particular solution of the differential equation (𝑥+1) 𝑑𝑦/𝑑𝑥=2 𝑒^(−𝑦)−1 , given that 𝑦=0 when 𝑥=0 (𝑥+1) 𝑑𝑦/𝑑𝑥=2𝑒^(−𝑦)−1 The variables are separable 𝒅𝒚/(𝟐𝒆^( −𝒚 ) − 𝟏) = 𝒅𝒙/(𝒙 + 𝟏) Integrating both sides ∫1▒𝑑𝑦/(2𝑒^(−𝑦) − 1) = ∫1▒𝑑𝑥/(𝑥 + 1) ∫1▒𝑑𝑦/(2/𝑒^𝑦 − 1) = log (x + 1) + c ∫1▒𝑑𝑦/((2 − 𝑒^𝑦)/𝑒^𝑦 ) = log (x + 1) + C ∫1▒𝒆^𝒚/(𝟐 −〖 𝒆〗^𝒚 ) dy = log (x + 1) + C Putting t = 𝟐−𝒆^𝒚 dt = −𝑒^𝑦dy –dt = 𝑒^𝑦dy Putting value of t & dt in equation ∫1▒(−𝑑𝑡)/𝑡 = log (x + 1) + c − log t = log (x + 1) + c Putting back value of t − log (2 − 𝑒^𝑦) = log (x + 1) + C 0 = log (x + 1) + log (2 − 𝑒^𝑦) + C log (x + 1) + log (2 − 𝒆^𝒚) + C = 0 Given y = 0 when x = 0 Putting x = 0 & y = 0 in (1) log (0 + 1) + log (2 − e0) + C = 0 log 1 + log (2 − 1) + C = 0 log 1 + log 1 + C = 0 0 + 0 + C = 0 C = 0 Putting value of C in (1) log (x + 1) + log (2 − 𝑒^𝑦) + 0 = 0 log (x + 1) + log (2 − 𝑒^𝑦) = 0 log (2 − ey) = – log (x + 1) log (2 − ey) = log (x + 1)–1 log (2 − ey) = log (𝟏/(𝒙 + 𝟏)) 2 − ey = 𝟏/(𝒙 + 𝟏) ey = 2−1/(𝑥 + 1) ey = (2𝑥 + 2 − 1)/(𝑥 + 1) ey = (2𝑥 + 1)/(𝑥 + 1) Taking log both sides y = log |(𝟐𝒙 + 𝟏)/(𝒙 + 𝟏)| , x ≠ −1

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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.