Ex 9.5, 3 - Show homogeneous: (x - y) dx - (x + y) dx = 0 - Ex 9.5

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.5, 3 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. 𝑥−𝑦﷯𝑑𝑥− 𝑥+𝑦﷯𝑑𝑥=0 Step 1 Find 𝑑𝑦﷮𝑑𝑥﷯ (x − y) dy − (x + y) dx = 0 (x − y) dy = (x + y) dx Step 2: Put 𝑑𝑦﷮𝑑𝑥﷯ = F(x, y) and find out F(𝜆x, 𝜆y) F(x, y) = 𝑥 + 𝑦﷮𝑥 − 𝑦﷯ F(𝜆x, 𝜆y) = 𝜆𝑥 + 𝜆𝑦﷮𝜆𝑥−𝜆𝑦﷯ = 𝜆(𝑥 + 𝑦)﷮𝜆 (𝑥 − 𝑦)﷯ = 𝑥 + 𝑦﷮𝑥 − 𝑦﷯ = F(x, y) ∴ F(𝜆x, 𝜆y) = 𝜆0 F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, 𝑑𝑦﷮𝑑𝑥﷯ is a homogenous differential equation. Step 3. Solving 𝑑𝑦﷮𝑑𝑥﷯ by putting y = vx Put y = vx. differentiating w.r.t.x 𝑑𝑦﷮𝑑𝑥﷯ = x 𝑑𝑣﷮𝑑𝑥﷯+ 𝑣𝑑𝑥﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑥 𝑑𝑣﷮𝑑𝑥﷯ + v Putting value of 𝑑𝑦﷮𝑑𝑥﷯ and y = vx in (1) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑥 + 𝑦﷮𝑥−𝑦﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v = 𝑥 + 𝑥𝑣﷮𝑥 − 𝑥𝑣﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v = 𝑥 (1 + 𝑣)﷮𝑥(1 − 𝑣)﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v = (1 + 𝑣)﷮(1 − 𝑣)﷯ x 𝑑𝑣﷮𝑑𝑥﷯ = (1 + 𝑣)﷮(1 − 𝑣)﷯ − v x 𝑑𝑣﷮𝑑𝑥﷯ = 1 + 𝑣 − 𝑣(1 − 𝑣)﷮1 − 𝑣﷯ x 𝑑𝑣﷮𝑑𝑥﷯ = 1 + 𝑣 − 𝑣 + 𝑣﷮2﷯﷮1 − 𝑣﷯ x 𝑑𝑣﷮𝑑𝑥﷯ = 1 + 𝑣﷮2﷯﷮1 − 𝑣﷯ 1 − 𝑣﷯𝑑𝑣﷮1 + 𝑣﷮2﷯﷯ = 𝑑𝑥﷮𝑥﷯ Integrating both sides ﷮﷮ 1 − 𝑣﷮1 + 𝑣﷮2﷯﷯﷯﷯𝑑𝑣= ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ tan−1 v − ﷮﷮ 𝑣﷮1 + 𝑣﷮2﷯﷯﷯ = log 𝑥﷯+𝑐 Let I = ﷮﷮ 𝑣﷮1 + 𝑣﷮2﷯﷯﷯ dv Put t = 1 + 𝑣﷮2﷯ Diff w.r.t. v 𝑑﷮𝑑𝑥﷯(1 + v2) = 𝑑𝑡﷮𝑑𝑣﷯ 2v = 𝑑𝑡﷮𝑑𝑣﷯ dv = 𝑑𝑡﷮2𝑣﷯ Therefore I = ﷮﷮ 𝑣﷮1 + 𝑣﷮2﷯﷯﷯ dv = ﷮﷮ 𝑑𝑡﷮2𝑡﷯﷯ = 1﷮2﷯𝑙𝑜𝑔 𝑡﷯+𝑐 Putting t = 1 + v2 = 1﷮2﷯𝑙𝑜𝑔 1+ 𝑣﷮2﷯﷯ + c Putting value of I in (2) tan−1 v − 1﷮2﷯ log 1+𝑣2﷯ = log 𝑥﷯ + c tan−1 v = 1﷮2﷯ log 1+𝑣2﷯ + log 𝑥﷯ + c tan−1 v = 1﷮2﷯ log 1+𝑣2﷯ + 2﷮2﷯ log 𝑥﷯ + c tan−1 v = 1﷮2﷯ log 1+𝑣2﷯ + 2 log 𝑥﷯﷯ + c tan−1 v = 1﷮2﷯ log 1+𝑣2﷯. 𝑥﷯﷮2﷯﷯ + c Putting v = 𝑦﷮𝑥﷯ tan−1 𝑦﷮𝑥﷯= 1﷮2﷯ log 1+ 𝑦﷮𝑥﷯﷯﷮2﷯﷯× 𝑥﷮2﷯﷯+c tan−1 𝑦﷮𝑥﷯= 1﷮2﷯ log 𝑥﷮2﷯ + 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯× 𝑥﷮2﷯﷯+c tan−1 𝑦﷮𝑥﷯= 1﷮2﷯ log 𝑥﷮2﷯+ 𝑦﷮2﷯﷯+c tan−1 𝒚﷮𝒙﷯= 𝟏﷮𝟐﷯ log 𝒙﷮𝟐﷯+ 𝒚﷮𝟐﷯﷯+𝐜 is the required solution

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