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Ex 9.5, 7 - Show homogeneous: {x cos (y/x) + y sin (y/x)} y dx - 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, 7 show that the given differential equation is homogeneous and solve each of them. 𝑥𝑐𝑜𝑠 𝑦﷮𝑥﷯﷯+𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯﷯𝑦 𝑑𝑥= 𝑦𝑠𝑖𝑛 𝑦﷮𝑥﷯﷯−𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯﷯𝑥 𝑑𝑦 Step 1: Find 𝑑𝑦﷮𝑑𝑥﷯ 𝑥𝑐𝑜𝑠 𝑦﷮𝑥﷯﷯+𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯﷯𝑦 𝑑𝑥= 𝑦𝑠𝑖𝑛 𝑦﷮𝑥﷯﷯−𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯﷯𝑥 𝑑𝑦 𝑑𝑦﷮𝑑𝑥﷯= 𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯ + 𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯﷮𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯ − 𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯﷯﷯ 𝑦﷮𝑥﷯ Step 2. Put 𝑑𝑦﷮𝑑𝑥﷯ = F (x, y) and find F(𝜆x, 𝜆y) ⇒ F(x, y) = 𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯ + 𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯﷮𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯ − 𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯﷯﷯ 𝑦﷮𝑥﷯ F(𝜆x, 𝜆y) = 𝜆𝑥 cos﷮ 𝜆𝑦﷮𝜆𝑥﷯﷯﷯ + 𝜆𝑦 sin﷮ 𝜆𝑦﷮𝜆𝑥﷯ ﷯﷯﷮𝜆𝑥 sin﷮ 𝜆𝑦﷮𝜆𝑥﷯﷯﷯ − 𝜆𝑥 cos﷮ 𝜆𝑦﷮𝜆𝑥﷯ ﷯﷯﷯﷯ 𝜆𝑦﷮𝜆𝑥﷯ = 𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯ + 𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯﷮𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯ − 𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯﷯﷯ 𝑦﷮𝑥﷯ = F(x, y) ∴ F(𝜆x, 𝜆y) = F(x, y) = 𝜆° 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 Putting 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 cos﷮ 𝑦﷮𝑥﷯﷯﷯ + 𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯﷮𝑦 sin﷮ 𝑦﷮𝑥﷯﷯﷯ − 𝑥 cos﷮ 𝑦﷮𝑥﷯﷯﷯﷯﷯ 𝑦﷮𝑥﷯ v + x 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥 cos﷮ 𝑣𝑥﷮𝑥﷯﷯﷯ + 𝑣𝑥 sin﷮ 𝑣𝑥﷮𝑥﷯﷯﷯﷮𝑣𝑥 sin﷮ 𝑣𝑥﷮𝑥﷯﷯ − 𝑥 cos﷮ 𝑣𝑥﷮𝑥﷯﷯﷯﷯﷯ × 𝑣𝑥﷮𝑥﷯ v + x 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥𝑐𝑜𝑠 𝑣 + 𝑣𝑥 𝑠𝑖𝑛 𝑣﷯𝑣﷮𝑣𝑥 sin﷮𝑣 − 𝑥 cos﷮𝑣﷯ ﷯﷯ v + x 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥 𝑐𝑜𝑠 𝑣 + 𝑣 𝑠𝑖𝑛 𝑣﷯𝑣﷮𝑥 𝑣 𝑠𝑖𝑛 𝑣 − cos﷮𝑣﷯﷯﷯ v + x 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 𝑐𝑜𝑠 𝑣 + 𝑣 𝑠𝑖𝑛 𝑣﷯﷮𝑣 𝑠𝑖𝑛 𝑣 − cos﷮𝑣﷯﷯ x 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 cos﷮𝑣+ 𝑣﷮2﷯﷯ sin﷮𝑣﷯﷮𝑣 𝑠𝑖𝑛 𝑣 − cos﷮𝑣﷯﷯ − v x 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 cos﷮ 𝑣﷮2﷯﷯ sin﷮𝑣﷯ − 𝑣﷮2﷯ sin﷮𝑣 +𝑣 cos﷮𝑣﷯﷯﷮𝑣 𝑠𝑖𝑛 𝑣 − cos﷮𝑣﷯﷯ 𝑥𝑑𝑣﷮𝑑𝑥﷯ = 2𝑣 cos﷮𝑣﷯﷮𝑣 𝑠𝑖𝑛 𝑣 − cos﷮𝑣﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 1﷮𝑥﷯ 2𝑣 cos﷮𝑣﷯﷮𝑣 𝑠𝑖𝑛 𝑣 − cos﷮𝑣﷯﷯﷯ 𝑣 sin﷮𝑣 − cos﷮𝑣﷯﷯﷮𝑣 cos﷮𝑣﷯﷯ dv = 2 𝑑𝑥﷮𝑥﷯ Integrating both sides. ﷮﷮ 𝑣 sin﷮𝑣 − cos﷮𝑣﷯﷯﷮𝑣 cos﷮𝑣﷯﷯﷯ dv = 2 𝑑𝑥﷮𝑥﷯ ﷮﷮ 𝑣 sin﷮𝑣 ﷯﷮𝑣 cos﷮𝑣﷯﷯﷯ dv − ﷮﷮ cos﷮𝑣﷯﷮𝑣 cos﷮𝑣﷯﷯﷯ dv = 2 ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮ sin﷮𝑣 ﷯﷮ cos﷮𝑣﷯﷯﷯ dv − ﷮﷮ 𝑑𝑣﷮𝑣﷯﷯ dv = 2 ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮𝑡𝑎𝑛﷯ dv − ﷮﷮ 𝑑𝑣﷮𝑣﷯﷯ dv = 2 ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮𝑡𝑎𝑛﷯ dv − ﷮﷮ 𝑑𝑣﷮𝑣﷯﷯ dv = 2 ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ log﷮ sec﷮𝑥﷯﷯− log﷮ 𝑣﷯﷯=2 log﷮ 𝑥﷯﷯ ﷯ + C log﷮ sec﷮𝑥﷯﷮𝑣﷯﷯= log﷮ 𝑥﷮2﷯﷯+𝑐 ﷯ log﷮ sec﷮𝑥﷯﷮𝑣﷯﷯= log﷮𝑐 𝑥﷮2﷯﷯ ﷯ Put v = 𝑦﷮𝑥﷯ log sec﷮ 𝑦﷮𝑥﷯﷯﷯﷮ 𝑦﷮𝑥﷯﷯﷯ = log (cx2) sec ﷮ 𝑦﷮𝑥﷯﷯ ﷯﷮ 𝑦﷮𝑥﷯﷯ = Cx2 sec ﷮ 𝑦﷮𝑥﷯﷯ ﷯﷯ = C x2 × 𝑦﷮𝑥﷯ sec ﷮ 𝑦﷮𝑥﷯﷯ ﷯﷯= C xy 1﷮ cos﷮ 𝑦﷮𝑥﷯﷯ ﷯﷯﷯ = C xy xy 𝑐𝑜𝑠 𝑦﷮𝑥﷯﷯﷯ = c1 xy 𝒄𝒐𝒔 𝒚﷮𝒙﷯﷯ = c1

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Davneet Singh
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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