# Ex 9.4, 12 - 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, 12 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 𝑥^2 𝑑𝑦+(𝑥𝑦+𝑦^2 ) 𝑑𝑥=0;𝑦=1 When 𝑥=1The differential equation can be written 𝑎s 𝑥^2 𝑑𝑦 = −(xy + y2) dx 𝑑𝑦/𝑑𝑥 = (−(𝑥𝑦 + 𝑦^2 ))/𝑥^2 "Let F(x, y) = " 𝑑𝑦/𝑑𝑥 " =" (−(𝑥𝑦 +𝑦^2 ))/𝑥^2 Finding F(𝝀x, 𝝀y) F(𝜆x, 𝜆y) = (−(𝜆𝑥𝜆𝑦 + 𝜆^2 𝑦^2 ))/〖𝜆^2 𝑥〗^2 = (−𝜆^2 (𝑥𝑦 + 𝑦^2 ))/〖𝜆^2 𝑥〗^2 = (−(𝑥𝑦 + 𝑦^2 ))/𝑥^2 = 𝜆° F(x, y) = F(x, y) = (−(𝑥𝑦 +𝑦^2 ))/𝑥^2 ∴ 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 = (−(𝑥𝑦 + 𝑦^2 ))/𝑥^2 v + (𝒙 𝒅𝒗)/𝒅𝒙 = (−(𝒙(𝒗𝒙) + (𝒗𝒙)^𝟐 ))/𝒙^𝟐 v + (𝑥 𝑑𝑣)/𝑑𝑥 = (−(𝑥^2 𝑣 + 𝑥^2 𝑣^2))/𝑥^2 v + (𝑥 𝑑𝑣)/𝑑𝑥 = −𝑥^2 ((𝑣 + 𝑣^2))/𝑥^2 v + (𝑥 𝑑𝑣)/𝑑𝑥 = −(v2 + v) (𝑥 𝑑𝑣)/𝑑𝑥 = − v2 − v − v (𝑥 𝑑𝑣)/𝑑𝑥 = −(𝑣^2+2𝑣) 𝒅𝒗/(𝒗^𝟐 + 𝟐𝒗) = − 𝒅𝒙/𝒙 Integrating both sides ∫1▒𝑑𝑣/(𝑣^2 + 2𝑣) = −∫1▒𝑑𝑥/𝑥 ∫1▒𝑑𝑣/(𝑣^2 + 2𝑣) = − log x + log c ∫1▒𝑑𝑣/(〖(𝑣〗^2 + 2𝑣 + 1) − 1) = − log x + log c ∫1▒𝒅𝒗/((𝒗 + 𝟏)^𝟐 − 𝟏^𝟐 ) = − log x + log c 𝟏/𝟐 log (𝒗 + 𝟏 − 𝟏)/(𝒗 + 𝟏 + 𝟏) = − log x + log c 1/2 log 𝑣/(𝑣 + 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 Using ∫1▒𝑑𝑥/(𝑥^2 − 𝑎^2 ) = 1/2𝑎 log |(𝑥 − 𝑎)/(𝑥 + 𝑎)|+𝐶 (𝑥√𝑣)/√(𝑣 + 2) = C Putting value of v i.e 𝑦/𝑥 (𝒙√(𝒚/𝒙))/√(𝒚/𝒙 + 𝟐) = C √(𝑥^2 × 𝑦/𝑥)/√(𝑦/𝑥 + 2) = C √𝑥𝑦/√((𝑦 + 2𝑥)/𝑥) = C (𝑥√𝑦)/√(𝑦 + 2𝑥) = C 𝑥√𝑦 = C√(𝑦+2𝑥) Squaring both sides x2y = c2(y + 2x) Putting x = 1 & y = 1 in (2) 12(1) = C2(1 + 2) 1 = 3C2 C2 = 𝟏/𝟑 Putting value in (2) x2 y = 1/3(y + 2x) 3x2y = y + 2x y + 2x = 3x2y