Ex 9.5, 4 - Show homogeneous: (x2 - y2) dx + 2xy dy = 0 - Ex 9.5

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.5, 4 show that the given differential equation is homogeneous and solve each of them. (๐‘ฅ^2โˆ’๐‘ฆ^2 )๐‘‘๐‘ฅ+2๐‘ฅ๐‘ฆ ๐‘‘๐‘ฆ=0 Step 1: Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (๐‘ฅ^2โˆ’๐‘ฆ^2 )๐‘‘๐‘ฅ+2๐‘ฅ๐‘ฆ ๐‘‘๐‘ฆ=0 2xy dy = โˆ’ (๐‘ฅ^2โˆ’๐‘ฆ^2 ) dx 2xy dy = (๐‘ฆ^2โˆ’๐‘ฅ^2 ) dx ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ฆ^2 โˆ’ ๐‘ฅ^2)/2๐‘ฅ๐‘ฆ Step 2: Putting F(x, y) = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ and finding F(๐œ†x, ๐œ†y) F(x, y) = (๐‘ฆ^2 โˆ’ ๐‘ฅ^2)/2๐‘ฅ๐‘ฆ F(๐œ†x, ๐œ†y) = ((๐œ†๐‘ฆ)^2โˆ’(๐œ†๐‘ฅ)^2)/(2๐œ†๐‘ฅ. ๐œ†๐‘ฆ)= (๐‘ฅ^2 ๐‘ฆ^2 โˆ’๐œ†^2 ๐‘ฅ^2)/(๐œ†^2.2๐‘ฅ๐‘ฆ)= (๐œ†^2 (๐‘ฆ^2 โˆ’ ๐‘ฅ^2))/(๐œ†^2.2๐‘ฅ๐‘ฆ) = (๐‘ฆ^2 โˆ’๐‘ฅ^2)/2๐‘ฅ๐‘ฆ = 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 Put 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 = (๐‘ฆ^2 โˆ’ ๐‘ฅ^2)/2๐‘ฅ๐‘ฆ (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ+๐‘ฃ = (ใ€–(๐‘ฅ๐‘ฃ)ใ€—^2 โˆ’ ๐‘ฅ^2)/(2๐‘ฅ(๐‘ฅ ๐‘ฃ)) (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ+๐‘ฃ = (๐‘ฅ^2 ๐‘ฃ^2โˆ’๐‘ฅ^2)/(2๐‘ฅ^2 ๐‘ฃ) (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = (๐‘ฅ^2 ๐‘ฃ^2 โˆ’ ๐‘ฅ^2)/(2๐‘ฅ^2 ๐‘ฃ) โˆ’ v x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (๐‘ฅ^2 ๐‘ฃ^2 โˆ’ ๐‘ฅ^2 โˆ’ 2๐‘ฅ^2 ๐‘ฃ^2)/(2๐‘ฅ^2 ๐‘ฃ) x ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (ใ€–โˆ’๐‘ฅใ€—^2 ๐‘ฃ^2 โˆ’ ๐‘ฅ^2)/(2๐‘ฅ^2 ๐‘ฃ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = โˆ’ 1/๐‘ฅ ((๐‘ฅ^2 ใ€–(๐‘ฃใ€—^2 + 1))/(2๐‘ฅ^2 ๐‘ฃ)) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = โˆ’ 1/๐‘ฅ ((๐‘ฃ^2 + 1)/2๐‘ฃ) (2๐‘ฃ ๐‘‘๐‘ฃ)/(๐‘ฃ^2 + 1) = (โˆ’๐‘‘๐‘ฅ)/๐‘ฅ Integrating both sides โˆซ1โ–’2๐‘ฃ/(๐‘ฃ^2 + 1) ๐‘‘๐‘ฃ = โˆซ1โ–’(โˆ’๐‘‘๐‘ฅ)/๐‘ฅ โˆซ1โ–’2๐‘ฃ/(๐‘ฃ^2 + 1) ๐‘‘๐‘ฃ = โˆ’logโก|๐‘ฅ|+๐ถ Putting t = v2 + 1 diff.w.r.t v. ๐‘‘/๐‘‘๐‘ฃ (v2 + 1) = ๐‘‘๐‘ก/๐‘‘๐‘ฃ 2v = ๐‘‘๐‘ก/๐‘‘๐‘ฃ dv = ๐‘‘๐‘ก/2๐‘ฃ Now, From (2) โˆซ1โ–’2๐‘ฃ/๐‘ก ๐‘‘๐‘ก/2๐‘ฃ = โˆ’ log|๐‘ฅ|+๐‘ ( from (2) ) โˆซ1โ–’๐‘‘๐‘ก/๐‘ก = โˆ’ log|๐‘ฅ|+๐‘ log|๐‘ก| = โˆ’ log|๐‘ฅ|+๐‘ Putting t = v2 + 1 log|"v2 + 1" | = โ€“log|"x" |+๐‘ log |"v2 + 1" | + log|"x" |=๐‘ log |"x(v2 + 1)" | = C Putting v = ๐‘ฆ/๐‘ฅ log |[(๐‘ฆ/๐‘ฅ)^2+1]๐‘ฅ|=๐‘ log|(๐‘ฆ^2 + ๐‘ฅ^2)/๐‘ฅ^2 ร—๐‘ฅ|=๐‘ Putting ๐‘ = log c log |(๐‘ฆ^2 + ๐‘ฅ^2)/๐‘ฅ| = log c ๐‘ฆ^2+๐‘ฅ^2 = cx ๐’™^๐Ÿ+๐’š^๐Ÿ=๐’„๐’™ is the general solution of the given differential equation.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.