Misc 11 - Find particular solution: (x - y) (dx + dy) = dx - dy - Solving homogeneous differential equation

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  1. Chapter 9 Class 12 Differential Equations
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Misc 11 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‘ฆ/๐‘‘๐‘ฅ = ((๐‘ฅ โˆ’ ๐‘ฆ โˆ’ 1))/(โˆ’(๐‘ฅ โˆ’๐‘ฆ + 1)) Put x โˆ’ y = t Diff w.r.t.x 1 โˆ’ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ก/๐‘‘๐‘ฅ 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‘ก โˆ’ 1)/2๐‘ก dt = ๐‘‘๐‘ฅ 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‘ก/2 + 1/2 log |๐‘ก| = 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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