Ex 9.4, 2 - Show homogeneous: y' = x + y / x - Chapter 9 Class 12 - Ex 9.4

part 2 - Ex 9.4, 2 - Ex 9.4 - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Ex 9.4, 2 - Ex 9.4 - Serial order wise - Chapter 9 Class 12 Differential Equations part 4 - Ex 9.4, 2 - Ex 9.4 - Serial order wise - Chapter 9 Class 12 Differential Equations

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Transcript

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

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