Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations

Transcript

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