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Last updated at May 29, 2018 by Teachoo

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Ex 9.5, 12 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 𝑥2𝑑𝑦+ 𝑥𝑦+ 𝑦2 𝑑𝑥=0;𝑦=1 When 𝑥=1 The differential equation can be written 𝑎s 𝑥2𝑑𝑦 = −(xy + y2) dx 𝑑𝑦𝑑𝑥 = − 𝑥𝑦 + 𝑦2 𝑥2 Let F(x, y) = 𝑑𝑦𝑑𝑥 = − 𝑥𝑦 + 𝑦2 𝑥2 Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = − 𝜆𝑥𝜆𝑦 + 𝜆2 𝑦2 𝜆2𝑥2 = − 𝜆2 𝑥𝑦 + 𝑦2 𝜆2𝑥2= − 𝑥𝑦 + 𝑦2 𝑥2 = 𝜆° F(x, y) = F(x, y) = − 𝑥𝑦 + 𝑦2 𝑥2 ∴ F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v Putting value of 𝑑𝑦𝑑𝑥 and y = vx in (1) 𝑑𝑦𝑑𝑥 = − 𝑥𝑦 + 𝑦2 𝑥2 v + 𝑑𝑣𝑑𝑥 = − 𝑥 𝑣𝑥 + 𝑣𝑥2 𝑥2 v + 𝑥 𝑑𝑣𝑑𝑥 = −( 𝑥2𝑣 + 𝑥2 𝑣2) 𝑥2 + 𝑥 𝑑𝑣𝑑𝑥 = − 𝑥2 (𝑣 + 𝑣2) 𝑥2 v + 𝑥 𝑑𝑣𝑑𝑥 = −(v2 + v) 𝑥 𝑑𝑣𝑑𝑥 = − v2 − v − v 𝑥 𝑑𝑣𝑑𝑥 = − 𝑣2+2𝑣 𝑑𝑣 𝑣2 + 2𝑣 = − 𝑑𝑥𝑥 Integrating both sides 𝑑𝑣 𝑣2 + 2𝑣 = − 𝑑𝑥𝑥 𝑑𝑣 𝑣2 + 2𝑣 = − log x + log c 𝑑𝑣 (𝑣2 + 2𝑣 + 1) − 1 = − log x + log c 𝑑𝑣 𝑣 + 12 − 12 = − log x + log c 12 log 𝑣 + 1 − 1𝑣 + 1 + 1 = − log x + log c 12 log 𝑣𝑣 + 2 = − log x + log c log 𝑣 𝑣 + 2 = − log x + log C log 𝑣 𝑣 + 2 + log x = log C log 𝑥 𝑣 𝑣 + 2 = log C 𝑥 𝑣 𝑣 + 2 = C Putting value of v i.e 𝑦𝑥 𝑥 𝑦𝑥 𝑦𝑥 + 2 = C 𝑥2 × 𝑦𝑥 𝑦𝑥 + 2 = C 𝑥𝑦 𝑦 + 2𝑥𝑥 = C 𝑥 𝑦 𝑦 + 2𝑥 = C 𝑥 𝑦 = C 𝑦+2𝑥 Squaring both sides x2y = c2(y + 2x) Put x = 1 & y = 1 12(1) = C2(1 + 2) 1 = 3C2 C2 = 13 Putting value in (2) x2 y = 13(y + 2x) 3x2y = y + 2x y + 2x = 3x2y

Chapter 9 Class 12 Differential Equations

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 9 years. He provides courses for Maths and Science at Teachoo.