# Misc 11

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc 11Find a particular solution of the differential equation (๐ฅโ๐ฆ)(๐๐ฅ+๐๐ฆ)=๐๐ฅโ๐๐ฆ , given that ๐ฆ=โ1 , when ๐ฅ=0 (๐ป๐๐๐ก:๐๐ข๐ก ๐ฅโ๐ฆ=๐ก) (๐ฅโ๐ฆ)(๐๐ฅ+๐๐ฆ)=๐๐ฅโ๐๐ฆ ๐ฅ๐๐ฅ + ๐ฅ๐๐ฆ โ y dx โ y dy = dx โ dy x dx โ y dx โ dx = โ xdy + y dy โ dy (x โ y โ 1) dx = (โx + y โ 1)dy ๐๐ฆ/๐๐ฅ = (๐ฅ โ ๐ฆ โ 1)/(โ๐ฅ + ๐ฆ โ 1) ๐๐ฆ/๐๐ฅ = ((๐ฅ โ ๐ฆ โ 1))/(โ(๐ฅ โ๐ฆ + 1)) Put x โ y = t Diff w.r.t.x 1 โ ๐๐ฆ/๐๐ฅ = ๐๐ก/๐๐ฅ Putting value of t & dt in (1) 1 โ ๐๐ก/๐๐ฅ = โ ((๐ก โ 1))/(๐ก + 1) ๐๐ก/๐๐ฅ = 1 + ((๐ก โ 1))/(๐ก + 1) ๐๐ก/๐๐ฅ = (๐ก + 1 + ๐ก โ 1)/(๐ก + 1) ๐๐ก/๐๐ฅ = 2๐ก/(๐ก โ 1) (๐ก โ 1)/2๐ก dt = ๐๐ฅ Integrating both sides โซ1โใ(๐ก + 1 )/2๐ก ๐๐กใ = โซ1โ๐๐ฅ โซ1โใ(๐ก/2๐ก+1/2๐ก) ๐๐กใ = x + c โซ1โ(1/2+1/2๐ก)๐๐ก = x + c ๐ก/2 + 1/2 log |๐ก| = x + c Putting value of t = x โ y (๐ฅ โ ๐ฆ)/2 + 1/2 log |๐ฅโ๐ฆ| = x + C Given y = โ1 when x = 0 Put x = 0 & y = โ1 in (2) (0 + 1)/2 + 1/2 log 1 = 0 + C 1/2 = C C = 1/2 Putting value in (2) (๐ฅ โ ๐ฆ)/2+1/2 logโกใ|๐ฅ โ ๐ฆ|=๐ฅ+1/2ใ (๐ฅ โ ๐ฆ)/2+1/2 logโกใ|๐ฅ โ ๐ฆ|=(2๐ฅ + 1)/2ใ x โ y + log |๐ฅโ๐ฆ|=2๐ฅ+1 log |๐ฅโ๐ฆ| = 2x + 1 โ x + y log |๐โ๐| = x + y + 1

Chapter 9 Class 12 Differential Equations

Serial order wise

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