# Misc 14 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 11, 2019 by Teachoo

Last updated at Dec. 11, 2019 by Teachoo

Transcript

Misc 14 Find a particular solution of the differential equation (๐ฅ+1) ๐๐ฆ/๐๐ฅ=2 ๐^(โ๐ฆ)โ1 , given that ๐ฆ=0 when ๐ฅ=0 (๐ฅ+1) ๐๐ฆ/๐๐ฅ=2๐^(โ๐ฆ)โ1 The variables are separable ๐๐ฆ/(2๐^( โ๐ฆ ) โ 1) = ๐๐ฅ/(๐ฅ + 1) Integrating both sides โซ1โ๐๐ฆ/(2๐^(โ๐ฆ) โ 1) = โซ1โ๐๐ฅ/(๐ฅ + 1) โซ1โ๐๐ฆ/(2/๐^๐ฆ โ 1) = log (x + 1) + c โซ1โ๐๐ฆ/((2 โ ๐^๐ฆ)/๐^๐ฆ ) = log (x + 1) + C โซ1โ๐^๐ฆ/(2 โใ ๐ใ^๐ฆ ) dy = log (x + 1) + C Putting t = 2โ๐^๐ฆ dt = โ๐^๐ฆdy โdt = ๐^๐ฆdy Putting value of t & dt in equation โซ1โ(โ๐๐ก)/๐ก = log (x + 1) + c โ log t = log (x + 1) + c Putting back value of t โฆ(1) โ log (2 โ ๐^๐ฆ) = log (x + 1) + C 0 = log (x + 1) + log (2 โ ๐^๐ฆ) + C log (x + 1) + log (2 โ ๐^๐ฆ) + C = 0 Given y = 0 when x = 0 Putting x = 0 & y = 0 in (1) log (0 + 1) + log (2 โ e0) + C = 0 log 1 + log (2 โ 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 โ ๐^๐ฆ) + 0 = 0 log (x + 1) + log (2 โ ๐^๐ฆ) = 0 log (2 โ ey) = โ log (x + 1) log (2 โ ey) = log (x + 1)โ1 log (2 โ ey) = log (1/(๐ฅ + 1)) 2 โ ey = 1/(๐ฅ + 1) ey = 2โ1/(๐ฅ + 1) ey = (2๐ฅ + 2 โ 1)/(๐ฅ + 1) ey = (2๐ฅ + 1)/(๐ฅ + 1) Taking log both sides y = log |(๐๐ + ๐)/(๐ + ๐)| , x โ โ1 (๐ด๐ ๐ logโกใ๐ฅ=logโกใ๐ฅ^๐ ใ ใ )

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Deleted for CBSE Board 2021 Exams only

Misc 4 Important

Misc 5 Important Deleted for CBSE Board 2021 Exams only

Misc 6

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10 Important

Misc 11

Misc 12

Misc 13

Misc 14 Important You are here

Misc 15 Important

Misc 16

Misc 17 Important Deleted for CBSE Board 2021 Exams only

Misc 18

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.