# Ex 9.5, 1 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Aug. 20, 2021 by Teachoo

Last updated at Aug. 20, 2021 by Teachoo

Transcript

Ex 9.5, 1 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve 𝑥2+𝑥𝑦𝑑𝑦= 𝑥2+ 𝑦2𝑑𝑥 𝑥2+𝑥𝑦𝑑𝑦 = 𝑥2+ 𝑦2𝑑𝑥 Step 1: Find 𝑑𝑦𝑑𝑥 (x2 + xy)dy = 𝑥2+ 𝑦2𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑥2 + 𝑦2 𝑥2 + 𝑥𝑦 Step 2. Putting F(x , y) = 𝑑𝑦𝑑𝑥 and finding F(𝜆x, 𝜆y) F(x, y) = 𝑥2 + 𝑦2 𝑥2 + 𝑥𝑦 Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = (𝜆 𝑥)2+(𝜆 𝑦)2(𝜆 𝑥)2+𝜆𝑥 + 𝜆𝑦 = 𝜆2 𝑥2 + 𝜆2 𝑦2 𝜆2 𝑥2 + 𝜆2𝑥𝑦 = 𝜆2( 𝑥2 𝑦2) 𝜆2( 𝑥2+ 𝑥𝑦) = 𝑥2 𝑦2 𝑥2 + 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 𝑑𝑦𝑑𝑥 is a homogenous differential equation Step 3 : Solving 𝑑𝑦𝑑𝑥 by putting y = vx Putting y = vx Diff w.r.t.x 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v 𝑑𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v Putting value of 𝑑𝑦𝑑𝑥 and y = vx in (1) 𝑑𝑦𝑑𝑥 = 𝑥2 + 𝑦2 𝑥2 + 𝑥𝑦 x 𝑑𝑣𝑑𝑥 + v = 𝑥2 + 𝑣𝑥2𝑥2 + 𝑥(𝑣𝑥) x 𝑑𝑣𝑑𝑥 + v = 𝑥2(1 + 𝑣2)𝑥2 + 𝑥2𝑣 x 𝑑𝑣𝑑𝑥 + v = 𝑥2(1 + 𝑣2)𝑥2(1 + 𝑣) x 𝑑𝑣𝑑𝑥 + v = 1 + 𝑣21 + 𝑣 x 𝑑𝑣𝑑𝑥 = 1 + 𝑣21 + 𝑣−𝑣 x 𝑑𝑣𝑑𝑥 = 1 + 𝑣2 − 𝑣 − 𝑣21+𝑣 x 𝑑𝑣𝑑𝑥 = 1 − 𝑣1 + 𝑣 (1 + 𝑣)(1 − 𝑣) dv = 𝑑𝑥𝑥 − 𝑣 + 1𝑣 − 1 dv = 𝑑𝑥𝑥 𝑣 + 1𝑣 − 1 dv = −𝑑𝑥𝑥 Integrating both sides 𝑣 + 1𝑣 − 1𝑑𝑣=− 𝑑𝑥𝑥 𝒗 + 𝟏𝒗 − 𝟏𝒅𝒗 = −log 𝒙+𝒄 Let I = 𝑣 + 1𝑣 − 1𝑑𝑣 Solving I I = 𝑣 + 1 − 1 + 1𝑣 − 1𝑑𝑣 I = 𝑣 − 1 + 2𝑣 − 1𝑑𝑣 I = 𝑣 − 1 𝑣 − 1+ 2𝑣 − 1𝑑𝑣 I = 1+ 2𝑣 − 1𝑑𝑣 I = 𝑑𝑣+ 2𝑣 − 1𝑑𝑣 I = 𝑣+2 log|𝑣−1| Putting v = y/x I = 𝑦𝑥+2 log 𝑦𝑥−1 I = 𝑦𝑥+2 log 𝑦 − 𝑥𝑥 Putting value of I in (2) 𝑦𝑥+2 log 𝑦 − 𝑥𝑥=− log 𝑥+𝐶 𝑦𝑥+2 log 𝑦 − 𝑥𝑥+ log 𝑥=𝐶 𝑦𝑥+ log 𝑦 − 𝑥2 𝑥2+ log 𝑥=𝐶 𝑦𝑥+ log 𝑦 − 𝑥2 𝑥2 ×𝑥=𝐶 𝑦𝑥+ log 𝑦 − 𝑥2𝑥=𝐶 log 𝑦 − 𝑥2𝑥=𝐶− 𝑦𝑥 𝑦 − 𝑥2𝑥= 𝑒𝐶 − 𝑦𝑥 𝑦 − 𝑥2𝑥 = 𝑒𝑐× 𝑒− 𝑦𝑥 𝑦 − 𝑥2𝑥 = 𝑐 𝑒− 𝑦𝑥 𝒙−𝒚𝟐 = 𝒄𝒙 𝒆− 𝒚𝒙

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.