Ex 9.5, 5 - Show homogeneous: x2 dy/dx = x2 - 2y2 + xy - Solving homogeneous differential equation

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.5, 5 show that the given differential equation is homogeneous and solve each of them. ๐‘ฅ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘ฅ^2โˆ’2๐‘ฆ^2+๐‘ฅ๐‘ฆ Step 1 Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฅ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘ฅ^2โˆ’2๐‘ฆ^2+๐‘ฅ๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= (๐‘ฅ^2 โˆ’ 2๐‘ฆ^2 + ๐‘ฅ๐‘ฆ)/๐‘ฅ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= 1โˆ’(2๐‘ฆ^2)/๐‘ฅ^2 + ๐‘ฅ๐‘ฆ/๐‘ฅ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= 1โˆ’(2๐‘ฆ^2)/๐‘ฅ + ๐‘ฆ/๐‘ฅ Step 2. Put ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = F (x, y) and find F(๐œ†x, ๐œ†y) ๐น(๐‘ฅ, ๐‘ฆ) = 1 โˆ’ (2๐‘ฆ^2)/๐‘ฅ^2 + ๐‘ฆ/๐‘ฅ Finding F(๐œ†x, ๐œ†y) F(๐œ†x, ๐œ†y) = 1 โˆ’ (2ใ€–(๐œ†๐‘ฆ)ใ€—^2)/(๐œ†๐‘ฅ)^2 + ๐œ†๐‘ฆ/๐œ†๐‘ฅ = 1 โˆ’ (2๐œ†^2 ๐‘ฆ^2)/(๐œ†^2 ๐‘ฅ^2 ) + ๐‘ฆ/๐‘ฅ = 1 โˆ’ (2๐‘ฆ^2)/๐‘ฅ^2 + ๐‘ฆ/๐‘ฅ = F(x, y) โˆด F(๐œ†x, ๐œ†y) = F(x, y) = ๐œ†ยฐ F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ is a homogenous differential equation. Step 3. Solving ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ by putting y = vx Putting y = vx. Differentiating w.r.t.x ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + v Putting value of ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ and y = vx in (1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1 โˆ’ (2๐‘ฆ^2)/๐‘ฅ^2 + ๐‘ฆ/๐‘ฅ ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + v =1 โˆ’2 ใ€–(๐‘ฃ๐‘ฅ)ใ€—^2/๐‘ฅ^2 + ๐‘ฃ๐‘ฅ/๐‘ฅ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + v =1 โˆ’ (2๐‘ฃ^2 ๐‘ฅ^2)/๐‘ฅ^2 + ๐‘ฃ x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = 1 โˆ’ 2v2 + v โˆ’ v x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = 1 โˆ’ 2v2 ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (1 โˆ’ 2๐‘ฃ^2)/๐‘ฅ ๐‘‘๐‘ฃ/(1 โˆ’ 2๐‘ฃ^2 ) = ๐‘‘๐‘ฅ/๐‘ฅ Integrating both sides. โˆซ1โ–’๐‘‘๐‘ฃ/(1โˆ’2๐‘ฃ^2 ) = โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ โˆซ1โ–’((๐‘‘๐‘ฃ))/((1โˆ’2๐‘ฃ^2))ร— (1/2)/(1/2) = โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ 1/2 โˆซ1โ–’๐‘‘๐‘ฃ/(1/2 โˆ’ ๐‘ฃ^2 ) = โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ 1/2 ร—1/2( 1/โˆš2 ) log |(1/โˆš2 + ๐‘ฃ)/(1/โˆš2 โˆ’ ๐‘ฃ)|= log |๐‘ฅ| + c โˆš2/4 log |(1 + โˆš2 ๐‘ฃ)/(1 โˆ’ โˆš2 ๐‘ฃ)| = log |๐‘ฅ| + c Putting v = ๐‘ฆ/๐‘ฅ 1/(2โˆš2) log |(1 + โˆš2 ๐‘ฆ/๐‘ฅ)/(1 โˆ’ โˆš2 ๐‘ฆ/๐‘ฅ)| = log |๐‘ฅ| + c 1/(2โˆš2) log |((๐‘ฅ + โˆš2 ๐‘ฆ)/๐‘ฅ)/((๐‘ฅ โˆ’ โˆš2 ๐‘ฆ)/๐‘ฅ)| = log |๐‘ฅ| + c ๐Ÿ/(๐Ÿโˆš๐Ÿ) log |(๐’™+โˆš๐Ÿ ๐’š)/(๐’™โˆ’โˆš๐Ÿ ๐’š)| = log |๐’™| + c .

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