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Ex 9.5, 9 - Show homogeneous: y dx + x log (y/x) dy - 2x dy = 0 - Solving homogeneous differential equation

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.5, 9 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. 𝑦 𝑑𝑥+𝑥𝑙𝑜𝑔 𝑦﷮𝑥﷯﷯𝑑𝑦−2𝑥 𝑑𝑦=0 Step 1 : Find 𝑑𝑦﷮𝑑𝑥﷯ 𝑦 𝑑𝑥+𝑥𝑙𝑜𝑔 𝑦﷮𝑥﷯﷯𝑑𝑦−2𝑥 𝑑𝑦=0 dy 𝑥 log﷮ 𝑦﷮𝑥﷯﷯−2𝑥 ﷯﷯ = − y dx 𝑑𝑦﷮𝑑𝑥﷯ = −𝑦﷮𝑥 log﷮ 𝑦﷮𝑥﷯﷯ − 2𝑥 ﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = −𝑦﷮−𝑥 2 − log﷮ 𝑦﷮𝑥﷯﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦﷮𝑥﷯﷮2 − log﷮ 𝑦﷮𝑥﷯﷯﷯﷯ Step 2. Putting F(x , y) = 𝑑𝑦﷮𝑑𝑥﷯ and finding F(𝜆x, 𝜆y) F(x, y) = 𝑦﷮𝑥﷯﷮2 − log﷮ 𝑦﷮𝑥﷯﷯﷯﷯ 𝐹(𝜆𝑥,𝜆𝑦) = 𝜆𝑦﷮𝜆𝑥﷯﷮2 − log﷮ 𝜆𝑦﷮𝜆𝑥﷯﷯﷯﷯ = 𝑦﷮𝑥﷯﷮2 − log﷮ 𝑦﷮𝑥﷯﷯﷯﷯ = 𝜆° 𝐹(𝑥, 𝑦)﷯ Thus, F(x, y) is a homogenous equation function of order zero Therefore 𝑑𝑦﷮𝑑𝑥﷯ is a homogenous differential equation Step 3 : Solving 𝑑𝑦﷮𝑑𝑥﷯ by putting y = vx Putting y = vx Diff w.r.t.x 𝑑𝑦﷮𝑑𝑥﷯ = x 𝑑𝑣﷮𝑑𝑥﷯ + v 𝑑𝑥﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = x 𝑑𝑣﷮𝑑𝑥﷯ + v Putting value of 𝑑𝑦﷮𝑑𝑥﷯ and y = vx in (1) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦﷮𝑥﷯﷮2 − log﷮ 𝑦﷮𝑥﷯﷯﷯﷯ v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣𝑥﷮𝑥﷯﷮2 − 𝑙𝑜𝑔 𝑣𝑥﷮𝑥﷯﷯﷯ v + 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣﷮2 − log﷮𝑣﷯﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣﷮2 − log﷮𝑣﷯﷯ − v 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 − 2𝑣 + 𝑣 log﷮𝑣﷯﷮2 − log﷮𝑣﷯﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 log﷮𝑣﷯ − 𝑣﷮2 − log﷮𝑣﷯﷯ 2 − log﷮𝑣﷯﷮𝑣 log﷮𝑣﷯﷯ dv = 𝑑𝑥﷮𝑥﷯ Integrating both sides ﷮﷮ 2 − log﷮𝑣﷯﷮𝑣 log﷮𝑣﷯ − 𝑣﷯﷯ dv = ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮ 2 − log﷮𝑣﷯﷮−𝑣 (1 − log﷮𝑣﷯)﷯﷯ dv = log x + log c ﷮﷮ 1 + 1 − log﷮𝑣﷯﷮−𝑣 (1 − log﷮𝑣﷯)﷯﷯ 𝑑𝑣= log x + log c ﷮﷮ 1﷮(−𝑣)(1 − log﷮𝑣)﷯﷯﷯𝑑𝑣 − ﷮﷮ 1﷮𝑣﷯﷯𝑑𝑣 = log x + log c ﷮﷮ 1﷮𝑣( log﷮𝑣﷯ − 1)﷯﷯𝑑𝑣 − ﷮﷮ 1﷮𝑣﷯﷯𝑑𝑣 = log x + log c ﷮﷮ 𝑑𝑣﷮𝑣( log﷮𝑣 − 1)﷯﷯﷯ – log v = log x + log c Put t = log v − 1 dt = 1﷮𝑣﷯ dv So, our equation becomes ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯ − log v = log x + log c log t − log v = log x + log c Putting value of t log (log v − 1) −log v + = log x + log c log (log v − 1) = log x + log c + log v log (log v − 1) = log C xv Putting value of v = 𝑦﷮𝑥﷯ log log 𝑦﷮𝑥﷯−1﷯= log﷮𝑥 𝑐 𝑦﷮𝑥﷯﷯ log log 𝑦﷮𝑥﷯−1﷯= log﷮ 𝑐𝑦﷯ log 𝑦﷮𝑥﷯ − 1 = cy cy = log 𝒚﷮𝒙﷯﷯ − 1

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