Ex 9.5, 6 - Chapter 9 Class 12 Differential Equations

Transcript

Ex 9.5, 6 show that the given differential equation is homogeneous and solve each of them. 𝑥 𝑑𝑦−𝑦 𝑑𝑥= ﷮ 𝑥﷮2﷯+ 𝑦﷮2﷯﷯ 𝑑𝑥 Step 1: Find 𝑑𝑦﷮𝑑𝑥﷯ x dy − y dx = ﷮ 𝑥﷮2﷯+ 𝑦﷮2﷯﷯ dx x dy = ﷮ 𝑥﷮2﷯+ 𝑦﷮2﷯﷯ dx + ydx x dy = ﷮ 𝑥﷮2﷯+ 𝑦﷮2﷯﷯+𝑦﷯ dx 𝑑𝑦﷮𝑑𝑥﷯ = ﷮ 𝑥﷮2﷯ + 𝑦﷮2﷯﷯ + 𝑦﷮𝑥﷯ Step 2: Put 𝑑𝑦﷮𝑑𝑥﷯ = F(x, y) and find F(𝜆x, 𝜆y) ∴ F(x, y) = 𝑑𝑦﷮𝑑𝑥﷯ = ﷮ 𝑥﷮2 ﷯+ 𝑦﷮2﷯﷯+ 𝑦﷮𝑥﷯ F(𝜆 x, 𝜆y) = ﷮ (𝜆𝑥)﷮2﷯ + 𝜆﷮2﷯ 𝑦﷮2﷯﷯﷯+ 𝜆𝑦﷮𝜆𝑥﷯ = ﷮ 𝜆﷮2﷯ 𝑥﷮2﷯ + 𝜆﷮2﷯ 𝑦﷮2﷯﷯ + 𝜆𝑦﷮𝜆𝑥﷯ = ﷮ 𝜆﷮2﷯( 𝑥﷮2﷯ + 𝑦﷮2﷯)﷯ + 𝜆𝑦﷮𝜆𝑥﷯= 𝜆 ﷮ 𝑥﷮2﷯ + 𝑦﷮2﷯﷯ + 𝜆𝑦﷮𝜆𝑥﷯ = ﷮ 𝑥﷮2﷯ + 𝑦﷮2﷯﷯ + 𝜆𝑦﷮𝑥﷯ = F(x, y) Hence, 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﷯ = 1 − 2 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯ + 𝑦﷮𝑥﷯ 𝑥 𝑑𝑣﷮𝑑𝑥﷯ + v =1 −2 (𝑣𝑥)﷮2﷯﷮ 𝑥﷮2﷯﷯ + 𝑣𝑥﷮𝑥﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v =1 − 2 𝑣﷮2﷯ 𝑥﷮2﷯﷮ 𝑥﷮2﷯﷯ + 𝑣 x 𝑑𝑣﷮𝑑𝑥﷯ = 1 − 2v2 + v − v x 𝑑𝑣﷮𝑑𝑥﷯ = 1 − 2v2 Putting value of 𝑑𝑦﷮𝑑𝑥﷯ and y = vx in 𝑑𝑦﷮𝑑𝑥﷯= ﷮ 𝑥﷮2﷯ + 𝑦﷮2﷯ ﷯+𝑦﷮𝑥﷯ x 𝑑𝑣﷮𝑑𝑥﷯+𝑣= ﷮ 𝑥﷮2﷯ + 𝑣𝑥﷯﷮2﷯﷯ + (𝑣𝑥)﷮𝑥﷯ x 𝑑𝑣﷮𝑑𝑥﷯+𝑣= ﷮ 𝑥﷮2﷯ + 𝑥﷮2﷯ 𝑣﷮2﷯﷯ + 𝑣𝑥﷮𝑥﷯ x 𝑑𝑣﷮𝑑𝑥﷯+𝑣 = ﷮ 𝑥﷮2﷯(1+ 𝑣﷮2﷯)﷯ + 𝑣𝑥﷮𝑥﷯ x 𝑑𝑣﷮𝑑𝑥﷯+𝑣 = 𝑥 ﷮1+ 𝑣﷮2﷯﷯ + 𝑣𝑥﷮𝑥﷯ x 𝑑𝑣﷮𝑑𝑥﷯+𝑣 = 𝑥( ﷮1+ 𝑣﷮2﷯﷯ + 𝑣)﷮𝑥﷯ x 𝑑𝑣﷮𝑑𝑥﷯+𝑣= ﷮1+ 𝑣﷮2﷯﷯+𝑣 x 𝑑𝑣﷮𝑑𝑥﷯= ﷮1+ 𝑣﷮2﷯﷯+𝑣 − 𝑣 x 𝑑𝑣﷮𝑑𝑥﷯= ﷮1+ 𝑣﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯= ﷮1 + 𝑣﷮2﷯﷯﷮𝑥﷯ 𝑑𝑣﷮ ﷮1 + 𝑣﷮2﷯﷯﷯= 𝑑𝑥﷮𝑥﷯ Integrating both sides. ﷮﷮ 𝑑𝑣﷮ ﷮1 + 𝑣﷮2﷯﷯﷯﷯ = ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮ 𝑑𝑣﷮ ﷮1 + 𝑣﷮2﷯﷯﷯﷯ = log 𝑥﷯+𝑐 log 𝑣+ ﷮ 𝑣﷮2﷯+1﷯﷯ =𝑙𝑜𝑔 𝑥﷯+𝑐 log 𝑣+ ﷮ 𝑣﷮2﷯+1﷯﷯ =𝑙𝑜𝑔 𝑐𝑥﷯ v + ﷮ 𝑣﷮2﷯+1﷯ = cx Putting v = 𝑦﷮𝑥﷯ 𝑦﷮𝑥﷯+ ﷮ 𝑦﷮𝑥﷯﷯﷮2﷯+1﷯=𝑐𝑥 𝑦﷮𝑥﷯+ ﷮ 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+1﷯=𝑐𝑥 𝑦+ ﷮ 𝑥﷮2﷯ 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯﷯+ 𝑥﷮2﷯﷯=𝑐 𝑥﷮2﷯ 𝑦+ ﷮ 𝑥﷮2﷯ 𝑦﷮2﷯﷮ 𝑥﷮2﷯﷯+ 𝑥﷮2﷯﷯=𝑐 𝑥﷮2﷯ 𝒚+ ﷮ 𝒚﷮𝟐﷯ + 𝒙﷮𝟐﷯﷯ =𝒄 𝒙﷮𝟐﷯ is the general solution of the given differential equation