# Ex 9.5, 12 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 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.