1. Chapter 9 Class 12 Differential Equations
2. Serial order wise

Transcript

Ex 9.5, 4 show that the given differential equation is homogeneous and solve each of them. (๐ฅ^2โ๐ฆ^2 )๐๐ฅ+2๐ฅ๐ฆ ๐๐ฆ=0 Step 1: Find ๐๐ฆ/๐๐ฅ (๐ฅ^2โ๐ฆ^2 )๐๐ฅ+2๐ฅ๐ฆ ๐๐ฆ=0 2xy dy = โ (๐ฅ^2โ๐ฆ^2 ) dx 2xy dy = (๐ฆ^2โ๐ฅ^2 ) dx ๐๐ฆ/๐๐ฅ = (๐ฆ^2 โ ๐ฅ^2)/2๐ฅ๐ฆ Step 2: Putting F(x, y) = ๐๐ฆ/๐๐ฅ and finding F(๐x, ๐y) F(x, y) = (๐ฆ^2 โ ๐ฅ^2)/2๐ฅ๐ฆ F(๐x, ๐y) = ((๐๐ฆ)^2โ(๐๐ฅ)^2)/(2๐๐ฅ. ๐๐ฆ)= (๐ฅ^2 ๐ฆ^2 โ๐^2 ๐ฅ^2)/(๐^2.2๐ฅ๐ฆ)= (๐^2 (๐ฆ^2 โ ๐ฅ^2))/(๐^2.2๐ฅ๐ฆ) = (๐ฆ^2 โ๐ฅ^2)/2๐ฅ๐ฆ = F(x, y) โด F(๐x, ๐y) = F(x, y) = ๐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 โ ๐ฅ^2)/2๐ฅ๐ฆ (๐ฅ ๐๐ฃ)/๐๐ฅ+๐ฃ = (ใ(๐ฅ๐ฃ)ใ^2 โ ๐ฅ^2)/(2๐ฅ(๐ฅ ๐ฃ)) (๐ฅ ๐๐ฃ)/๐๐ฅ+๐ฃ = (๐ฅ^2 ๐ฃ^2โ๐ฅ^2)/(2๐ฅ^2 ๐ฃ) (๐ฅ ๐๐ฃ)/๐๐ฅ = (๐ฅ^2 ๐ฃ^2 โ ๐ฅ^2)/(2๐ฅ^2 ๐ฃ) โ v x ๐๐ฃ/๐๐ฅ = (๐ฅ^2 ๐ฃ^2 โ ๐ฅ^2 โ 2๐ฅ^2 ๐ฃ^2)/(2๐ฅ^2 ๐ฃ) x ๐๐ฃ/๐๐ฅ = (ใโ๐ฅใ^2 ๐ฃ^2 โ ๐ฅ^2)/(2๐ฅ^2 ๐ฃ) ๐๐ฃ/๐๐ฅ = โ 1/๐ฅ ((๐ฅ^2 ใ(๐ฃใ^2 + 1))/(2๐ฅ^2 ๐ฃ)) ๐๐ฃ/๐๐ฅ = โ 1/๐ฅ ((๐ฃ^2 + 1)/2๐ฃ) (2๐ฃ ๐๐ฃ)/(๐ฃ^2 + 1) = (โ๐๐ฅ)/๐ฅ Integrating both sides โซ1โ2๐ฃ/(๐ฃ^2 + 1) ๐๐ฃ = โซ1โ(โ๐๐ฅ)/๐ฅ โซ1โ2๐ฃ/(๐ฃ^2 + 1) ๐๐ฃ = โ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) โซ1โ2๐ฃ/๐ก ๐๐ก/2๐ฃ = โ log|๐ฅ|+๐ ( from (2) ) โซ1โ๐๐ก/๐ก = โ log|๐ฅ|+๐ log|๐ก| = โ 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+๐^๐=๐๐ is the general solution of the given differential equation.