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Last updated at Dec. 11, 2019 by Teachoo
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 )