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Ex 9.4, 2 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. ๐‘ฆ^โ€ฒ=(๐‘ฅ+๐‘ฆ)/๐‘ฅ Step 1: Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ฅ + ๐‘ฆ)/๐‘ฅ Step 2: Putting F(x, y) = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ and find F(๐œ†x, ๐œ†y) So, F(x, y) = (๐’™ + ๐’š)/๐’™ F(๐œ†x, ๐œ†y) = (๐œ†๐‘ฅ +๐œ†๐‘ฆ)/๐œ†๐‘ฅ = (๐œ†(๐‘ฅ +๐‘ฆ))/๐œ†๐‘ฅ = (๐‘ฅ + ๐‘ฆ)/๐‘ฅ = F(x, y) = ๐œ†ยฐF(x, y) Therefore F(x, y) is a homogenous function of degree zero. Hence 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’š/๐’…๐’™ = ๐’™ ๐’…๐’—/๐’…๐’™ + v Putting value of ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ and y = vx in (1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ฅ + ๐‘ฆ)/๐‘ฅ ๐’™ ( ๐’…๐’—)/๐’…๐’™ + v = (๐’™ + ๐’—๐’™)/๐’™ ๐‘ฅ ( ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ + v = 1+๐‘ฃ ๐‘ฅ (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = 1+๐‘ฃโˆ’๐‘ฃ ๐‘ฅ ( ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = 1 ( ๐’…๐’—)/๐’…๐’™ = ๐Ÿ/๐’™ Integrating both sides โˆซ1โ–’ใ€–๐‘‘๐‘ฃ=โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/๐‘ฅ ใ€— ใ€— v = log|๐’™|+๐’„ Putting v = ๐‘ฆ/๐‘ฅ ๐‘ฆ/๐‘ฅ = log|๐‘ฅ| + c y = x log|๐’™| + cx

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo