# Ex 9.4, 15 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 9.4, 15 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 2𝑥𝑦+𝑦^2−2𝑥^2 𝑑𝑦/𝑑𝑥=0;𝑦=2 When 𝑥=1 Differential equation can be written 𝑎s 2𝑥𝑦+𝑦^2−2𝑥^2 𝑑𝑦/𝑑𝑥=0 2𝑥𝑦+𝑦^2= 2𝑥^2 𝑑𝑦/𝑑𝑥 2𝑥^2 𝑑𝑦/𝑑𝑥=2𝑥𝑦+𝑦^2 𝑑𝑦/𝑑𝑥= (2𝑥𝑦 + 𝑦^2)/(2𝑥^2 ) 𝒅𝒚/𝒅𝒙= 𝒚/𝒙 + 𝒚^𝟐/(𝟐𝒙^𝟐 ) Let F(x, y) = 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) Finding F(𝝀x, 𝝀y) F(𝜆x, 𝜆y) = 𝜆𝑦/𝜆𝑥 + 〖(𝜆𝑦)〗^2/(2〖(𝜆𝑥)〗^(2 ) ) = 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) = 𝜆° F(x, y) ∴ F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x 𝒅𝒚/𝒅𝒙 = x 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥= 𝑦/𝑥 + 𝑦^2/(2𝑥^2 ) 𝑣+𝑥 𝑑𝑣/𝑑𝑥 = 𝑣+ 𝑣^2/2 𝑥𝑑𝑣/𝑑𝑥 = 𝑣^2/2 𝟐𝒅𝒗/𝒗^𝟐 = 𝒅𝒙/𝒙 Integrating both sides 2∫1▒𝑑𝑣/𝑣^2 "=" ∫1▒𝑑𝑥/𝑥 2∫1▒〖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 (−𝟐𝒙)/𝒚 = log |𝒙| + C Putting x = 1 & y = 2 in (2) (−2(1))/2 = log |𝟏| + C −1 = 0 + C C = −1 Putting value in (2) (−2𝑥)/𝑦 = log |𝑥| + C (−2𝑥)/𝑦 = log |𝑥| − 1 y = (−2𝑥)/〖log 〗|𝑥|" − 1 " y = 𝟐𝒙/〖𝟏 − 𝐥𝐨𝐠 〗|𝒙|" "