# Ex 9.5, 4 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 4show that the given differential equation is homogeneous and solve each of them.(๐ฅ^2โ๐ฆ^2 )๐๐ฅ+2๐ฅ๐ฆ ๐๐ฆ=0 Step 1: Find ๐๐ฆ/๐๐ฅ (๐ฅ^2โ๐ฆ^2 )๐๐ฅ+2๐ฅ๐ฆ ๐๐ฆ=0 2xy dy = โ (๐ฅ^2โ๐ฆ^2 ) dx 2xy dy = (๐ฆ^2โ๐ฅ^2 ) dx ๐๐ฆ/๐๐ฅ = (๐ฆ^2 โ ๐ฅ^2)/2๐ฅ๐ฆ Step 2: Putting F(x, y) = ๐๐ฆ/๐๐ฅ and finding F(๐x, ๐y) F(x, y) = (๐ฆ^2 โ ๐ฅ^2)/2๐ฅ๐ฆ F(๐x, ๐y) = ((๐๐ฆ)^2โ(๐๐ฅ)^2)/(2๐๐ฅ. ๐๐ฆ)= (๐ฅ^2 ๐ฆ^2 โ๐^2 ๐ฅ^2)/(๐^2.2๐ฅ๐ฆ)= (๐^2 (๐ฆ^2 โ ๐ฅ^2))/(๐^2.2๐ฅ๐ฆ) = (๐ฆ^2 โ๐ฅ^2)/2๐ฅ๐ฆ = F(x, y) โด F(๐x, ๐y) = F(x, y) = ๐ยฐ F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, ๐๐ฆ/๐๐ฅ is a homogenous differential equation. Step 3. Solving ๐๐ฆ/๐๐ฅ by putting y = vx Put y = vx. differentiating w.r.t.x ๐๐ฆ/๐๐ฅ = x ๐๐ฃ/๐๐ฅ+๐ฃ๐๐ฅ/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐ฅ ๐๐ฃ/๐๐ฅ + v Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐๐ฆ/๐๐ฅ = (๐ฆ^2 โ ๐ฅ^2)/2๐ฅ๐ฆ (๐ฅ ๐๐ฃ)/๐๐ฅ+๐ฃ = (ใ(๐ฅ๐ฃ)ใ^2 โ ๐ฅ^2)/(2๐ฅ(๐ฅ ๐ฃ)) (๐ฅ ๐๐ฃ)/๐๐ฅ+๐ฃ = (๐ฅ^2 ๐ฃ^2โ๐ฅ^2)/(2๐ฅ^2 ๐ฃ) (๐ฅ ๐๐ฃ)/๐๐ฅ = (๐ฅ^2 ๐ฃ^2 โ ๐ฅ^2)/(2๐ฅ^2 ๐ฃ) โ v x ๐๐ฃ/๐๐ฅ = (๐ฅ^2 ๐ฃ^2 โ ๐ฅ^2 โ 2๐ฅ^2 ๐ฃ^2)/(2๐ฅ^2 ๐ฃ) x ๐๐ฃ/๐๐ฅ = (ใโ๐ฅใ^2 ๐ฃ^2 โ ๐ฅ^2)/(2๐ฅ^2 ๐ฃ) ๐๐ฃ/๐๐ฅ = โ 1/๐ฅ ((๐ฅ^2 ใ(๐ฃใ^2 + 1))/(2๐ฅ^2 ๐ฃ)) ๐๐ฃ/๐๐ฅ = โ 1/๐ฅ ((๐ฃ^2 + 1)/2๐ฃ) (2๐ฃ ๐๐ฃ)/(๐ฃ^2 + 1) = (โ๐๐ฅ)/๐ฅ Integrating both sides โซ1โ2๐ฃ/(๐ฃ^2 + 1) ๐๐ฃ = โซ1โ(โ๐๐ฅ)/๐ฅ โซ1โ2๐ฃ/(๐ฃ^2 + 1) ๐๐ฃ = โlogโก|๐ฅ|+๐ถ Putting t = v2 + 1 diff.w.r.t v. ๐/๐๐ฃ (v2 + 1) = ๐๐ก/๐๐ฃ 2v = ๐๐ก/๐๐ฃ dv = ๐๐ก/2๐ฃ Now, From (2) โซ1โ2๐ฃ/๐ก ๐๐ก/2๐ฃ = โ log|๐ฅ|+๐ ( from (2) ) โซ1โ๐๐ก/๐ก = โ log|๐ฅ|+๐ log|๐ก| = โ log|๐ฅ|+๐ Putting t = v2 + 1 log|"v2 + 1" | = โlog|"x" |+๐ log |"v2 + 1" | + log|"x" |=๐ log |"x(v2 + 1)" | = C Putting v = ๐ฆ/๐ฅ log |[(๐ฆ/๐ฅ)^2+1]๐ฅ|=๐ log|(๐ฆ^2 + ๐ฅ^2)/๐ฅ^2 ร๐ฅ|=๐ Putting ๐ = log c log |(๐ฆ^2 + ๐ฅ^2)/๐ฅ| = log c ๐ฆ^2+๐ฅ^2 = cx ๐^๐+๐^๐=๐๐ is the general solution of the given differential equation.

Chapter 9 Class 12 Differential Equations

Serial order wise

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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.