# Ex 9.5, 15

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 15 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 2𝑥𝑦+ 𝑦2−2 𝑥2 𝑑𝑦𝑑𝑥=0;𝑦=2 When 𝑥=1 Differential equation can be written 𝑎s 2𝑥𝑦+ 𝑦2−2 𝑥2 𝑑𝑦𝑑𝑥=0 2 𝑥2 𝑑𝑦𝑑𝑥=2𝑥𝑦+ 𝑦2 𝑑𝑦𝑑𝑥= 2𝑥𝑦 + 𝑦22 𝑥2 𝑑𝑦𝑑𝑥= 𝑦𝑥 + 𝑦22 𝑥2 Let F(x, y) = 𝑑𝑦𝑑𝑥 = 𝑦𝑥 + 𝑦22 𝑥2 Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = 𝜆𝑦𝜆𝑥 + (𝜆𝑦)22 (𝜆𝑥)2 = 𝑦𝑥 + 𝑦22 𝑥2 = 𝜆° F(x, y) ∴ 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= 𝑦𝑥 + 𝑦22 𝑥2 𝑣+𝑥 𝑑𝑣𝑑𝑥 = 𝑣𝑥𝑥 + 12 𝑣2 𝑥2 𝑥2 𝑣+𝑥 𝑑𝑣𝑑𝑥 = 𝑣+ 𝑣22 𝑥𝑑𝑣𝑑𝑥 = 𝑣 + 𝑣22 − v 𝑥𝑑𝑣𝑑𝑥 = 𝑣22 2𝑑𝑣 𝑣2 = 𝑑𝑥𝑥 Integrating both sides 2 𝑑𝑣 𝑣2 = 𝑓 𝑑𝑥𝑥 2 𝑣−2 𝑑𝑣= log 𝑥+𝑐 2 𝑣−2 + 1 −2 + 1 = log 𝑥+𝑐 2 𝑉 − 1 −1 = log 𝑥+𝑐 −2 𝑣 = log 𝑥+𝑐 Putting value of v = 𝑦 𝑥 −2𝑥𝑦 = log 𝑥 + C Put x = 1 & y = 2 −2(1)2 = log 1 + C − 1 = 0 + C C = −1 Putting value in (2) −2𝑥𝑦 = log 𝑥 − 1 y = −2𝑥 log 𝑥 − 1 y = 𝟐𝒙 𝟏 − 𝐥𝐨𝐠 𝒙

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.