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Transcript

Misc 9 Find a particular solution of the differential equation (๐‘ฅโˆ’๐‘ฆ)(๐‘‘๐‘ฅ+๐‘‘๐‘ฆ)=๐‘‘๐‘ฅโˆ’๐‘‘๐‘ฆ , given that ๐‘ฆ=โˆ’1 , when ๐‘ฅ=0 (๐ป๐‘–๐‘›๐‘ก:๐‘๐‘ข๐‘ก ๐‘ฅโˆ’๐‘ฆ=๐‘ก) (๐‘ฅโˆ’๐‘ฆ)(๐‘‘๐‘ฅ+๐‘‘๐‘ฆ)=๐‘‘๐‘ฅโˆ’๐‘‘๐‘ฆ ๐‘ฅ๐‘‘๐‘ฅ + ๐‘ฅ๐‘‘๐‘ฆ โˆ’ y dx โˆ’ y dy = dx โˆ’ dy x dx โˆ’ y dx โˆ’ dx = โˆ’ xdy + y dy โˆ’ dy (x โˆ’ y โˆ’ 1) dx = (โˆ’x + y โˆ’ 1)dy ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ฅ โˆ’ ๐‘ฆ โˆ’ 1)/(โˆ’๐‘ฅ + ๐‘ฆ โˆ’ 1) ๐’…๐’š/๐’…๐’™ = ((๐’™ โˆ’ ๐’š โˆ’ ๐Ÿ))/(โˆ’(๐’™ โˆ’๐’š + ๐Ÿ)) Let x โˆ’ y = t Diff w.r.t.x 1 โˆ’ ๐’…๐’š/๐’…๐’™ = ๐’…๐’•/๐’…๐’™ Putting value of t & dt in (1) 1 โˆ’ ๐‘‘๐‘ก/๐‘‘๐‘ฅ = โˆ’ ((๐‘ก โˆ’ 1))/(๐‘ก + 1) ๐‘‘๐‘ก/๐‘‘๐‘ฅ = 1 + ((๐‘ก โˆ’ 1))/(๐‘ก + 1) ๐‘‘๐‘ก/๐‘‘๐‘ฅ = (๐‘ก + 1 + ๐‘ก โˆ’ 1)/(๐‘ก + 1) ๐‘‘๐‘ก/๐‘‘๐‘ฅ = 2๐‘ก/(๐‘ก โˆ’ 1) (๐’• โˆ’ ๐Ÿ)/๐Ÿ๐’• dt = ๐’…๐’™ Integrating both sides โˆซ1โ–’ใ€–(๐‘ก + 1 )/2๐‘ก ๐‘‘๐‘กใ€— = โˆซ1โ–’๐‘‘๐‘ฅ โˆซ1โ–’ใ€–(๐‘ก/2๐‘ก+1/2๐‘ก) ๐‘‘๐‘กใ€— = x + c โˆซ1โ–’(1/2+1/2๐‘ก)๐‘‘๐‘ก = x + c ๐’•/๐Ÿ + ๐Ÿ/๐Ÿ log |๐’•| = x + c Putting value of t = x โ€“ y (๐‘ฅ โˆ’ ๐‘ฆ)/2 + 1/2 log |๐‘ฅโˆ’๐‘ฆ| = x + C Given y = โ€“1 when x = 0 Put x = 0 & y = โˆ’1 in (2) (0 + 1)/2 + 1/2 log 1 = 0 + C 1/2 = C C = 1/2 Putting value in (2) (๐‘ฅ โˆ’ ๐‘ฆ)/2+1/2 logโกใ€–|๐‘ฅ โˆ’ ๐‘ฆ|=๐‘ฅ+1/2ใ€— (๐‘ฅ โˆ’ ๐‘ฆ)/2+1/2 logโกใ€–|๐‘ฅ โˆ’ ๐‘ฆ|=(2๐‘ฅ + 1)/2ใ€— x โˆ’ y + log |๐‘ฅโˆ’๐‘ฆ|=2๐‘ฅ+1 log |๐‘ฅโˆ’๐‘ฆ| = 2x + 1 โˆ’ x + y log |๐’™โˆ’๐’š| = x + y + 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.