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Last updated at Dec. 11, 2019 by Teachoo
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Ex 9.5, 15 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 2๐ฅ๐ฆ+๐ฆ^2โ2๐ฅ^2 ๐๐ฆ/๐๐ฅ=0;๐ฆ=2 When ๐ฅ=1 Differential equation can be written ๐s 2๐ฅ๐ฆ+๐ฆ^2โ2๐ฅ^2 ๐๐ฆ/๐๐ฅ=0 2๐ฅ๐ฆ+๐ฆ^2= 2๐ฅ^2 ๐๐ฆ/๐๐ฅ 2๐ฅ^2 ๐๐ฆ/๐๐ฅ=2๐ฅ๐ฆ+๐ฆ^2 ๐๐ฆ/๐๐ฅ= (2๐ฅ๐ฆ + ๐ฆ^2)/(2๐ฅ^2 ) ๐๐ฆ/๐๐ฅ= ๐ฆ/๐ฅ + ๐ฆ^2/(2๐ฅ^2 ) โฆ(1) Let F(x, y) = ๐๐ฆ/๐๐ฅ = ๐ฆ/๐ฅ + ๐ฆ^2/(2๐ฅ^2 ) Finding F(๐x, ๐y) F(๐x, ๐y) = ๐๐ฆ/๐๐ฅ + ใ(๐๐ฆ)ใ^2/(2ใ(๐๐ฅ)ใ^(2 ) ) = ๐ฆ/๐ฅ + ๐ฆ^2/(2๐ฅ^2 ) = ๐ยฐ F(x, y) โด F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x ๐๐ฆ/๐๐ฅ = x ๐๐ฃ/๐๐ฅ + v Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐๐ฆ/๐๐ฅ= ๐ฆ/๐ฅ + ๐ฆ^2/(2๐ฅ^2 ) ๐ฃ+๐ฅ ๐๐ฃ/๐๐ฅ = ๐ฃ๐ฅ/๐ฅ + 1/2 (๐ฃ^2 ๐ฅ^2)/๐ฅ^2 ๐ฃ+๐ฅ ๐๐ฃ/๐๐ฅ = ๐ฃ+ ๐ฃ^2/2 ๐๐ฆ/๐๐ฅ= ๐ฆ/๐ฅ + ๐ฆ^2/(2๐ฅ^2 ) ๐ฃ+๐ฅ ๐๐ฃ/๐๐ฅ = ๐ฃ+ ๐ฃ^2/2 ๐ฅ๐๐ฃ/๐๐ฅ = ๐ฃ^2/2 2๐๐ฃ/๐ฃ^2 = ๐๐ฅ/๐ฅ Integrating both sides 2โซ1โ๐๐ฃ/๐ฃ^2 "=" โซ1โ๐๐ฅ/๐ฅ 2โซ1โใ๐ฃ^(โ2) ๐๐ฃ=logโก|๐ฅ|+๐ใ 2 (๐ฃ^(โ2 + 1) )/(โ2 + 1) =logโก|๐ฅ|+๐ 2 (ใ๐ฃ ใ^(โ1) )/(โ1) =logโก|๐ฅ|+๐ (โ2 )/๐ฃ =logโก|๐ฅ|+๐ 2 (ใ๐ฃ ใ^(โ1) )/(โ1) =logโก|๐ฅ|+๐ (โ2 )/๐ฃ =logโก|๐ฅ|+๐ Putting value of v = (๐ฆ )/๐ฅ (โ2๐ฅ)/๐ฆ = log |๐ฅ| + C Now, Putting x = 1 & y = 2 in (2) (โ2(1))/2 = log |1| + C โ1 = 0 + C C = โ1 โฆ(2) (As log 1 = 0) Putting value of C in (2) (โ2๐ฅ)/๐ฆ = log |๐ฅ| + C (โ2๐ฅ)/๐ฆ = log |๐ฅ| โ 1 y = (โ2๐ฅ)/(ใlog ใโก|๐ฅ| โ 1) y = ๐๐/ใ๐ โ ๐ฅ๐จ๐ ใโก|๐|" "
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