1. Chapter 9 Class 12 Differential Equations
2. Serial order wise
3. Ex 9.5

Transcript

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^2โ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^2โ2๐ฆ^2+๐ฅ๐ฆ ๐๐ฆ/๐๐ฅ= (๐ฅ^2 โ 2๐ฆ^2 + ๐ฅ๐ฆ)/๐ฅ^2 ๐๐ฆ/๐๐ฅ= 1โ(2๐ฆ^2)/๐ฅ^2 + ๐ฅ๐ฆ/๐ฅ^2 ๐๐ฆ/๐๐ฅ= 1โ(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x, ๐y) ๐น(๐ฅ, ๐ฆ) = 1 โ (2๐ฆ^2)/๐ฅ^2 + ๐ฆ/๐ฅ Finding F(๐x, ๐y) F(๐x, ๐y) = 1 โ (2ใ(๐๐ฆ)ใ^2)/(๐๐ฅ)^2 + ๐๐ฆ/๐๐ฅ = 1 โ (2๐^2 ๐ฆ^2)/(๐^2 ๐ฅ^2 ) + ๐ฆ/๐ฅ = 1 โ (2๐ฆ^2)/๐ฅ^2 + ๐ฆ/๐ฅ = 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 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 โ (2๐ฆ^2)/๐ฅ^2 + ๐ฆ/๐ฅ ๐ฅ ๐๐ฃ/๐๐ฅ + v =1 โ2 ใ(๐ฃ๐ฅ)ใ^2/๐ฅ^2 + ๐ฃ๐ฅ/๐ฅ x ๐๐ฃ/๐๐ฅ + v =1 โ (2๐ฃ^2 ๐ฅ^2)/๐ฅ^2 + ๐ฃ x ๐๐ฃ/๐๐ฅ = 1 โ 2v2 + v โ v x ๐๐ฃ/๐๐ฅ = 1 โ 2v2 ๐๐ฃ/๐๐ฅ = (1 โ 2๐ฃ^2)/๐ฅ ๐๐ฃ/(1 โ 2๐ฃ^2 ) = ๐๐ฅ/๐ฅ Integrating both sides. โซ1โ๐๐ฃ/(1โ2๐ฃ^2 ) = โซ1โ๐๐ฅ/๐ฅ โซ1โ((๐๐ฃ))/((1โ2๐ฃ^2))ร (1/2)/(1/2) = โซ1โ๐๐ฅ/๐ฅ 1/2 โซ1โ๐๐ฃ/(1/2 โ ๐ฃ^2 ) = โซ1โ๐๐ฅ/๐ฅ 1/2 ร1/2( 1/โ2 ) log |(1/โ2 + ๐ฃ)/(1/โ2 โ ๐ฃ)|= log |๐ฅ| + c โ2/4 log |(1 + โ2 ๐ฃ)/(1 โ โ2 ๐ฃ)| = log |๐ฅ| + c Putting v = ๐ฆ/๐ฅ 1/(2โ2) log |(1 + โ2 ๐ฆ/๐ฅ)/(1 โ โ2 ๐ฆ/๐ฅ)| = log |๐ฅ| + c 1/(2โ2) log |((๐ฅ + โ2 ๐ฆ)/๐ฅ)/((๐ฅ โ โ2 ๐ฆ)/๐ฅ)| = log |๐ฅ| + c ๐/(๐โ๐) log |(๐+โ๐ ๐)/(๐โโ๐ ๐)| = log |๐| + c .

Ex 9.5