# Misc 14 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by

Last updated at Dec. 11, 2019 by

Transcript

Misc 14 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 …(1) − 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) (As log 1 = 0) 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 (𝐴𝑠 𝑎 log〖𝑥=log〖𝑥^𝑎 〗 〗 )

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 1 (iii) Important

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii)

Misc 2 (iv) Important

Misc 3 Deleted for CBSE Board 2022 Exams

Misc 4 Important

Misc 5 Important Deleted for CBSE Board 2022 Exams

Misc 6

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13

Misc 14 Important You are here

Misc 15 Important

Misc 16 (MCQ)

Misc 17 (MCQ) Important Deleted for CBSE Board 2022 Exams

Misc 18 (MCQ)

Chapter 9 Class 12 Differential Equations (Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.