# Misc 14 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 14 Find a particular solution of the differential equation 𝑥+1 𝑑𝑦𝑑𝑥=2 𝑒−𝑦−1 , given that 𝑦=0 when 𝑥=0 𝑥+1 𝑑𝑦𝑑𝑥=2 𝑒−𝑦−1 The variables are separable 𝑑𝑦2 𝑒 −𝑦 − 1 = 𝑑𝑥𝑥 + 1 Integrating both sides 𝑑𝑦2 𝑒−𝑦 − 1 = 𝑑𝑥𝑥 + 1 𝑑𝑦 2 𝑒𝑦 − 1 = log (x + 1) + c 𝑑𝑦 2 − 𝑒𝑦 𝑒𝑦 = log (x + 1) + c 𝑒𝑦2 − 𝑒𝑦 dy = log (x + 1) + C Put 2− 𝑒𝑦 = t 𝑒𝑦dy = –dt Putting value of t & dt in equation −𝑑𝑡𝑡 = log (x + 1) + c − log t = log (x + 1) + c Putting back value of t − log (2 − 𝑒𝑦) = log (x + 1) + c Given y = 0 when x = 0 Putting x = 0 & y = 0 in (1) log (0 + 1) + log (2 − e0) = − C log 1 + log (2 − 1) = −C 0 + log 1 = −C 0 = −C C = 0 Putting value of C in (1) − log (2 − ey) = log (x + 1) + c −log (2 − ey) = log (x + 1) + 0 log (2 − ey) = −log (x + 1) log (2 − ey) = log 1𝑥 + 1 2 − ey = 1𝑥 + 1 ey = 2− 1𝑥 + 1 ey = 2𝑥 + 2 − 1𝑥 + 1 ey = 2𝑥 + 1𝑥 + 1 Taking log both sides y = log 𝟐𝒙 + 𝟏𝒙+𝟏 , x ≠ − 1

Chapter 9 Class 12 Differential Equations

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.