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Last updated at Dec. 11, 2019 by Teachoo
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Ex 9.5, 6 Show that the given differential equation is homogeneous and solve each of them. ๐ฅ ๐๐ฆโ๐ฆ ๐๐ฅ=โ(๐ฅ^2+๐ฆ^2 ) ๐๐ฅ Step 1: Find ๐๐ฆ/๐๐ฅ x dy โ y dx = โ(๐ฅ^2+๐ฆ^2 ) dx x dy = โ(๐ฅ^2+๐ฆ^2 ) dx + y dx x dy = (โ(๐ฅ^2+๐ฆ^2 )+๐ฆ) dx ๐๐ฆ/๐๐ฅ = (โ(๐ฅ^2 + ๐ฆ^2 ) + ๐ฆ)/๐ฅ Step 2: Put ๐๐ฆ/๐๐ฅ = F(x, y) and find F(๐x, ๐y) F(x, y) = ๐๐ฆ/๐๐ฅ = (โ(๐ฅ^(2 )+ ๐ฆ^2 ) + ๐ฆ)/๐ฅ F(๐ x, ๐y) = (โ(ใ(๐๐ฅ)ใ^2 + (๐^2 ๐ฆ^2 ) )+ ๐๐ฆ)/๐๐ฅ = (โ(๐^2 ๐ฅ^2 + ๐^2 ๐ฆ^2 ) + ๐๐ฆ)/๐๐ฅ = (โ(๐^2 (๐ฅ^2 + ๐ฆ^2)) + ๐๐ฆ)/๐๐ฅ= (๐โ(๐ฅ^2 + ๐ฆ^2 ) + ๐๐ฆ)/๐๐ฅ = (โ(๐ฅ^2 + ๐ฆ^2 ) + ๐ฆ)/๐ฅ = F(x, y) Hence, F(๐x, ๐y) = F(x, y) = ๐ยฐ F(x, y) Hence, F(x, y) is a homogenous Function of with degree 0 So, ๐๐ฆ/๐๐ฅ is a homogenous differential equation. Step 3 - Solving ๐๐ฆ/๐๐ฅ by putting y = vx Putting y = vx. Differentiating w.r.t.x ๐๐ฆ/๐๐ฅ = ๐ฅ ๐๐ฃ/๐๐ฅ+๐ฃ ๐๐ฅ/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐ฅ ๐๐ฃ/๐๐ฅ + ๐ฃ Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐๐ฆ/๐๐ฅ=(โ(๐ฅ^2 + ๐ฆ^2 )+ ๐ฆ)/๐ฅ x ๐๐ฃ/๐๐ฅ+๐ฃ=(โ(๐ฅ^2 + (๐ฃ๐ฅ)^2 ) + (๐ฃ๐ฅ))/๐ฅ x ๐๐ฃ/๐๐ฅ+๐ฃ=(โ(๐ฅ^2 + ๐ฅ^2 ๐ฃ^2 ) + ๐ฃ๐ฅ)/๐ฅ x ๐๐ฃ/๐๐ฅ+๐ฃ =(โ(๐ฅ^2 (1 + ๐ฃ^2)) + ๐ฃ๐ฅ)/๐ฅ x ๐๐ฃ/๐๐ฅ+๐ฃ =(๐ฅโ(1 + ๐ฃ^2 ) + ๐ฃ๐ฅ)/๐ฅ x ๐๐ฃ/๐๐ฅ+๐ฃ =(๐ฅ(โ(1 + ๐ฃ^2 ) + ๐ฃ))/๐ฅ x ๐๐ฃ/๐๐ฅ+๐ฃ= โ(1+๐ฃ^2 )+๐ฃ x ๐๐ฃ/๐๐ฅ= โ(1+๐ฃ^2 )+๐ฃ โ ๐ฃ x ๐๐ฃ/๐๐ฅ= โ(1+๐ฃ^2 ) ๐๐ฃ/๐๐ฅ= โ(1 + ๐ฃ^2 )/๐ฅ ๐๐ฃ/โ(1 + ๐ฃ^2 )= ๐๐ฅ/๐ฅ Integrating both sides. โซ1โ๐๐ฃ/โ(1 + ๐ฃ^2 ) = โซ1โ๐๐ฅ/๐ฅ โซ1โ๐๐ฃ/โ(1 + ๐ฃ^2 ) = log |๐ฅ|+๐ log |๐ฃ+โ(๐ฃ^2+1)| =๐๐๐|๐ฅ|+๐ log |๐ฃ+โ(๐ฃ^2+1)| =๐๐๐|๐๐ฅ| v + โ(๐ฃ^2+1) = cx We know that โซ1โ๐๐ฃ/โ(๐^2 + ๐ฅ^2 ) =๐๐๐|๐ฅ+โ(๐ฅ^2+๐^2 )|+๐ Putting a = 1, x = v (As log ๐ + log b = log ab) Putting v = ๐ฆ/๐ฅ ๐ฆ/๐ฅ+โ((๐ฆ/๐ฅ)^2+1)=๐๐ฅ ๐ฆ/๐ฅ+โ(๐ฆ^2/๐ฅ^2 +1)=๐๐ฅ ๐ฆ/๐ฅ+โ((๐ฆ^2 + ๐ฅ^2)/๐ฅ^2 )=๐๐ฅ ๐ฆ/๐ฅ+โ(๐ฆ^2 + ๐ฅ^2 )/๐ฅ=๐๐ฅ ๐+โ(๐^๐ ใ+ ๐ใ^๐ ) =๐๐^๐ โด General solution is ๐+โ(๐^๐ ใ+ ๐ใ^๐ ) =๐๐^๐
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