# Ex 9.5, 2 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 24, 2018 by Teachoo

Last updated at Dec. 24, 2018 by Teachoo

Transcript

Ex 9.5, 2 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. ๐ฆ^โฒ=(๐ฅ+๐ฆ)/๐ฅ Step 1: Find ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ = (๐ฅ + ๐ฆ)/๐ฅ Step 2. Putting F(x, y) = ๐๐ฆ/๐๐ฅ and find F(๐x, ๐y) So, F(x, y) = (๐ฅ + ๐ฆ)/๐ฅ F(๐x, ๐y) = (๐๐ฅ +๐๐ฆ)/๐๐ฅ = (๐(๐ฅ +๐ฆ))/๐๐ฅ = (๐ฅ + ๐ฆ)/๐ฅ = F(x, y) = ๐ยฐF(x, y) Therefore F(x, y) Is a homogenous function of degree zero. Hence 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 = (๐ฅ + ๐ฆ)/๐ฅ ๐ฅ ( ๐๐ฃ)/๐๐ฅ + v = (๐ฅ + ๐ฃ๐ฅ)/๐ฅ ๐ฅ ( ๐๐ฃ)/๐๐ฅ + v = 1+๐ฃ ๐ฅ (๐ฅ ๐๐ฃ)/๐๐ฅ = 1+๐ฃโ๐ฃ ๐ฅ ( ๐๐ฃ)/๐๐ฅ = 1 ( ๐๐ฃ)/๐๐ฅ = 1/๐ฅ Integrating both sides โซ1โใ๐๐ฃ=โซ1โใ๐๐ฅ/๐ฅ ใ ใ v = log|๐ฅ|+๐ Putting v = ๐ฆ/๐ฅ ๐ฆ/๐ฅ = log|๐ฅ| + c y = x log|๐| + cx

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.