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Ex 9.5, 10 - Show homogeneous: (1 + ex/y) dx + e x/y (1 - x/y) - 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, 10 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. 1+ 𝑒﷮ 𝑥﷮𝑦﷯﷯﷯𝑑𝑥+ 𝑒﷮ 𝑥﷮𝑦﷯﷯ 1− 𝑥﷮𝑦﷯﷯𝑑𝑦=0 Step 1 : Find 𝑑𝑥﷮𝑑𝑦﷯ 1+ 𝑒﷮ 𝑥﷮𝑦﷯﷯﷯𝑑𝑥+ 𝑒﷮ 𝑥﷮𝑦﷯﷯ 1− 𝑥﷮𝑦﷯﷯𝑑𝑦 = 0 1+ 𝑒﷮ 𝑥﷮𝑦﷯﷯﷯ dx = − 𝑒﷮ 𝑥﷮𝑦﷯﷯ 1− 𝑥﷮𝑦﷯﷯𝑑𝑦 𝑑𝑥﷮𝑑𝑦﷯ = − 𝑒﷮ 𝑥﷮𝑦﷯﷯ 1 − 𝑥﷮𝑦﷯﷯ ﷮1 + 𝑒﷮ 𝑥﷮𝑦﷯﷯ ﷯ Step 2 : Put 𝑑𝑥﷮𝑑𝑦﷯ = F(x, y) and find F(𝜆 x, 𝜆 y) F(x, y) = − 𝑒﷮ 𝑥﷮𝑦﷯﷯ 1 − 𝑥﷮𝑦﷯﷯ ﷮1 + 𝑒﷮ 𝑥﷮𝑦﷯﷯﷯ F(𝜆x, 𝜆y) = − 𝑒﷮ 𝜆𝑥﷮𝜆𝑦﷯﷯ 1 − 𝜆𝑥﷮𝜆𝑦﷯﷯ ﷮1 + 𝑒﷮ 𝜆𝑥﷮𝜆𝑦﷯﷯ ﷯ = − 𝑒﷮ 𝑥﷮𝑦﷯﷯ 1 − 𝑥﷮𝑦﷯﷯ ﷮1 + 𝑒﷮ 𝑥﷮𝑦﷯﷯﷯= 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﷯= − 𝑒﷮ 𝑥﷮𝑦﷯﷯ 1 − 𝑥﷮𝑦﷯﷯ ﷮1 + 𝑒﷮ 𝑥﷮𝑦﷯﷯﷯ v + y 𝑑𝑣﷮𝑑𝑦﷯ = 𝑒﷮ 𝑣𝑦﷮𝑦﷯﷯ 1 − 𝑣𝑦﷮𝑦﷯﷯﷮1 + 𝑒﷮ 𝑣𝑦﷮𝑦﷯﷯﷯ v + y 𝑑𝑣﷮𝑑𝑦﷯ = 𝑒﷮𝑣﷯ 1 − 𝑣﷯﷮1 + 𝑒﷮𝑣﷯﷯ y 𝑑𝑣﷮𝑑𝑦﷯ = 𝑒﷮𝑣﷯ + 𝑣 𝑒﷮𝑣﷯﷮1 + 𝑒﷮𝑣﷯﷯ − v y 𝑑𝑣﷮𝑑𝑦﷯ = − 𝑒﷮𝑣﷯ + 𝑣 𝑒﷮𝑣﷯ − 𝑣 − 𝑣 𝑒﷮𝑣﷯﷮1 + 𝑒﷮𝑣﷯﷯ y 𝑑𝑣﷮𝑑𝑦﷯ = − 𝑣 + 𝑒﷮𝑣﷯﷯﷮1 + 𝑒﷮𝑣﷯﷯ 1 + 𝑒﷮𝑣﷯﷮𝑣 + 𝑒﷮𝑣﷯﷯ dv = −𝑑𝑦﷮𝑦﷯ Integrating both sides ﷮﷮ 1+𝑒﷮𝑣﷯﷮𝑣+ 𝑒﷮𝑣﷯﷯ dv ﷯ = ﷮﷮ −𝑑𝑦﷮𝑦﷯﷯ ﷮﷮ 1 + 𝑒﷮𝑣﷯﷮𝑣 + 𝑒﷮𝑣﷯﷯ dv ﷯=− log﷮𝑦+ log﷮𝑐﷯﷯ Put v + ev = t (1 + ev) dv = dt Thus, our equation becomes ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯=− log﷮𝑦+ log﷮𝑐﷯﷯ log﷮𝑡= log﷮𝑦+ log﷮𝑐﷯﷯ ﷯ Put value of t log (v + ev) = − log y + log c log (v + ev) + log y = log c log y (v + ev) + log c Put value of v = 𝑥﷮𝑦﷯ log y 𝑥﷮𝑦﷯﷯e 𝑥﷮𝑦﷯ log y 𝑥﷮𝑦﷯+ e﷮ 𝑥﷮𝑦﷯﷯﷯ = log c y 𝑥﷮𝑦﷯+ e﷮ 𝑥﷮𝑦﷯﷯﷯ = c x + y 𝒆﷮ 𝒙﷮𝒚﷯﷯ = C

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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