Ex 9.5, 1 - Class 12

Transcript

Ex 9.5, 1 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve 𝑥﷮2﷯+𝑥𝑦﷯𝑑𝑦= 𝑥﷮2﷯+ 𝑦﷮2﷯﷯𝑑𝑥 𝑥﷮2﷯+𝑥𝑦﷯𝑑𝑦 = 𝑥﷮2﷯+ 𝑦﷮2﷯﷯𝑑𝑥 Step 1: Find 𝑑𝑦﷮𝑑𝑥﷯ (x2 + xy)dy = 𝑥﷮2﷯+ 𝑦﷮2﷯﷯𝑑𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 𝑥﷮2﷯ + 𝑦﷮2﷯﷮ 𝑥﷮2﷯ + 𝑥𝑦﷯ Step 2. Putting F(x , y) = 𝑑𝑦﷮𝑑𝑥﷯ and finding F(𝜆x, 𝜆y) F(x, y) = 𝑥﷮2﷯ + 𝑦﷮2﷯﷮ 𝑥﷮2﷯ + 𝑥𝑦﷯ Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = (𝜆 𝑥)﷮2﷯+(𝜆 𝑦)﷮2﷯﷮(𝜆 𝑥)﷮2﷯+𝜆𝑥 + 𝜆𝑦﷯ = 𝜆﷮2﷯ 𝑥﷮2﷯ + 𝜆﷮2﷯ 𝑦﷮2﷯﷮ 𝜆﷮2﷯ 𝑥﷮2﷯ + 𝜆﷮2﷯𝑥𝑦﷯ = 𝜆﷮2﷯( 𝑥﷮2﷯ 𝑦﷮2﷯)﷮ 𝜆﷮2﷯( 𝑥﷮2﷯+ 𝑥𝑦)﷯ = 𝑥﷮2﷯ 𝑦﷮2﷯﷮ 𝑥﷮2﷯ + 𝑥𝑦﷯ = F(x, y) So, F(𝜆x, 𝜆y) = F(x, y) = 𝜆0 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﷯ = 𝑥﷮2﷯ + 𝑦﷮2﷯﷮ 𝑥﷮2﷯ + 𝑥𝑦﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v = 𝑥2 + 𝑣𝑥﷯﷮2﷯﷮𝑥2 + 𝑥(𝑣𝑥)﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v = 𝑥2(1 + 𝑣﷮2﷯)﷮𝑥2 + 𝑥2𝑣﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v = 𝑥2(1 + 𝑣﷮2﷯)﷮𝑥2(1 + 𝑣)﷯ x 𝑑𝑣﷮𝑑𝑥﷯ + v = 1 + 𝑣﷮2﷯﷮1 + 𝑣﷯ x 𝑑𝑣﷮𝑑𝑥﷯ = 1 + 𝑣﷮2﷯﷮1 + 𝑣﷯−𝑣 x 𝑑𝑣﷮𝑑𝑥﷯ = 1 + 𝑣﷮2﷯ − 𝑣 − 𝑣﷮2﷯﷮1+𝑣﷯ x 𝑑𝑣﷮𝑑𝑥﷯ = 1 − 𝑣﷮1 + 𝑣﷯ (1 + 𝑣)﷮(1 − 𝑣)﷯ dv = 𝑑𝑥﷮𝑥﷯ − 𝑣 + 1﷮𝑣 − 1﷯﷯ dv = 𝑑𝑥﷮𝑥﷯ 𝑣 + 1﷮𝑣 − 1﷯﷯ dv = −𝑑𝑥﷮𝑥﷯ Integrating both sides ﷮﷮ 𝑣 + 1﷮𝑣 − 1﷯﷯﷯𝑑𝑣=− ﷮﷮ 𝑑𝑥﷮𝑥﷯﷯ ﷮﷮ 𝒗 + 𝟏﷮𝒗 − 𝟏﷯﷯﷯𝒅𝒗 = −log 𝒙﷯+𝒄 Let I = ﷮﷮ 𝑣 + 1﷮𝑣 − 1﷯﷯﷯𝑑𝑣 Solving I I = ﷮﷮ 𝑣 + 1 − 1 + 1﷮𝑣 − 1﷯﷯﷯𝑑𝑣 I = ﷮﷮ 𝑣 − 1 + 2﷮𝑣 − 1﷯﷯﷯𝑑𝑣 I = ﷮﷮ 𝑣 − 1 ﷮𝑣 − 1﷯+ 2﷮𝑣 − 1﷯﷯﷯𝑑𝑣 I = ﷮﷮ 1+ 2﷮𝑣 − 1﷯﷯﷯𝑑𝑣 I = ﷮﷮𝑑𝑣﷯+ ﷮﷮ 2﷮𝑣 − 1﷯﷯𝑑𝑣 I = 𝑣+2 log﷮|𝑣−1|﷯ Putting v = y/x I = 𝑦﷮𝑥﷯+2 log﷮ 𝑦﷮𝑥﷯−1﷯﷯ I = 𝑦﷮𝑥﷯+2 log﷮ 𝑦 − 𝑥﷮𝑥﷯﷯﷯ Putting value of I in (2) 𝑦﷮𝑥﷯+2 log﷮ 𝑦 − 𝑥﷮𝑥﷯﷯﷯=− log﷮ 𝑥﷯﷯+𝐶 𝑦﷮𝑥﷯+2 log﷮ 𝑦 − 𝑥﷮𝑥﷯﷯﷯+ log﷮ 𝑥﷯﷯=𝐶 𝑦﷮𝑥﷯+ log﷮ 𝑦 − 𝑥﷯﷮2﷯﷮ 𝑥﷮2﷯﷯﷯﷯+ log﷮ 𝑥﷯﷯=𝐶 𝑦﷮𝑥﷯+ log﷮ 𝑦 − 𝑥﷯﷮2﷯﷮ 𝑥﷮2﷯﷯ ×𝑥﷯﷯=𝐶 𝑦﷮𝑥﷯+ log﷮ 𝑦 − 𝑥﷯﷮2﷯﷮𝑥﷯﷯﷯=𝐶 log﷮ 𝑦 − 𝑥﷯﷮2﷯﷮𝑥﷯﷯﷯=𝐶− 𝑦﷮𝑥﷯ 𝑦 − 𝑥﷯﷮2﷯﷮𝑥﷯= 𝑒﷮𝐶 − 𝑦﷮𝑥﷯﷯ 𝑦 − 𝑥﷯﷮2﷯﷮𝑥﷯ = 𝑒﷮𝑐﷯× 𝑒﷮− 𝑦﷮𝑥﷯﷯ 𝑦 − 𝑥﷯﷮2﷯﷮𝑥﷯ = 𝑐 𝑒﷮− 𝑦﷮𝑥﷯﷯ 𝒙−𝒚﷯﷮𝟐﷯ = 𝒄𝒙 𝒆﷮− 𝒚﷮𝒙﷯﷯