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

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. ๐ฆ๏ทฎโฒ๏ทฏ= ๐ฅ+๐ฆ๏ทฎ๐ฅ๏ทฏ Step 1: Find ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = ๐ฅ + ๐ฆ๏ทฎ๐ฅ๏ทฏ Step 2. Putting F(x, y) = ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ and find F(๐x, ๐y) So, F(x, y) = ๐ฅ + ๐ฆ๏ทฎ๐ฅ๏ทฏ F(๐x, ๐y) = ๐๐ฅ +๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = ๐(๐ฅ +๐ฆ)๏ทฎ๐๐ฅ๏ทฏ = ๐ฅ + ๐ฆ๏ทฎ๐ฅ๏ทฏ = F(x, y) = ๐ยฐF(x, y) Therefore F(x, y) Is a homogenous function of degree zero. Hence ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ is a homogenous differential equation Step 3: Solving ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ by putting y = vx Put y = vx. differentiating w.r.t.x ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = x ๐๐ฃ๏ทฎ๐๐ฅ๏ทฏ+ ๐ฃ๐๐ฅ๏ทฎ๐๐ฅ๏ทฏ ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = ๐ฅ ๐๐ฃ๏ทฎ๐๐ฅ๏ทฏ + v Putting value of ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ and y = vx in (1) ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = ๐ฅ + ๐ฆ๏ทฎ๐ฅ๏ทฏ ๐ฅ ๐๐ฃ๏ทฎ๐๐ฅ๏ทฏ + v = ๐ฅ + ๐ฃ๐ฅ๏ทฎ๐ฅ๏ทฏ ๐ฅ ๐๐ฃ๏ทฎ๐๐ฅ๏ทฏ + v = 1+๐ฃ ๐ฅ ๐ฅ ๐๐ฃ๏ทฎ๐๐ฅ๏ทฏ = 1+๐ฃโ๐ฃ ๐ฅ ๐๐ฃ๏ทฎ๐๐ฅ๏ทฏ = 1 ๐๐ฃ๏ทฎ๐๐ฅ๏ทฏ = 1๏ทฎ๐ฅ๏ทฏ Integrating both sides ๏ทฎ๏ทฎ๐๐ฃ= ๏ทฎ๏ทฎ ๐๐ฅ๏ทฎ๐ฅ๏ทฏ ๏ทฏ ๏ทฏ v = log ๐ฅ๏ทฏ+๐ Putting v = ๐ฆ๏ทฎ๐ฅ๏ทฏ ๐๏ทฎ๐๏ทฏ = x log ๐๏ทฏ + cx is the general solution of the given differential equation