# Ex 9.4, 5 - 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, 5 show that the given differential equation is homogeneous and solve each of them. 𝑥^2 𝑑𝑦/𝑑𝑥=𝑥^2−2𝑦^2+𝑥𝑦 Step 1: Find 𝑑𝑦/𝑑𝑥 𝑥^2 𝑑𝑦/𝑑𝑥=𝑥^2−2𝑦^2+𝑥𝑦 𝑑𝑦/𝑑𝑥= (𝑥^2 − 2𝑦^2 + 𝑥𝑦)/𝑥^2 𝑑𝑦/𝑑𝑥= 1−(2𝑦^2)/𝑥^2 + 𝑥𝑦/𝑥^2 𝒅𝒚/𝒅𝒙= 𝟏−(𝟐𝒚^𝟐)/𝒙 + 𝒚/𝒙 Step 2: Put 𝑑𝑦/𝑑𝑥 = F (x, y) and find F(𝜆x, 𝜆y) 𝐹(𝑥, 𝑦) = 1 − (2𝑦^2)/𝑥^2 + 𝑦/𝑥 Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = 1 − (2〖(𝜆𝑦)〗^2)/(𝜆𝑥)^2 + 𝜆𝑦/𝜆𝑥 = 1 − (2𝜆^2 𝑦^2)/(𝜆^2 𝑥^2 ) + 𝑦/𝑥 = 1 − (2𝑦^2)/𝑥^2 + 𝑦/𝑥 = F(x, y) ∴ F(𝜆x, 𝜆y) = F(x, y) = 𝜆° F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, 𝑑𝑦/𝑑𝑥 is a homogenous differential equation. Step 3: Solving 𝑑𝑦/𝑑𝑥 by putting y = vx Putting y = vx. Differentiating w.r.t.x 𝑑𝑦/𝑑𝑥 = x 𝑑𝑣/𝑑𝑥+𝑣𝑑𝑥/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥 = 1 − (2𝑦^2)/𝑥^2 + 𝑦/𝑥 𝑥 𝒅𝒗/𝒅𝒙 + v = 1 − 2 〖(𝒗𝒙)〗^𝟐/𝒙^𝟐 + 𝒗𝒙/𝒙 x 𝑑𝑣/𝑑𝑥 + v = 1 − (2𝑣^2 𝑥^2)/𝑥^2 + 𝑣 Putting y = vx. Differentiating w.r.t.x 𝑑𝑦/𝑑𝑥 = x 𝑑𝑣/𝑑𝑥+𝑣𝑑𝑥/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥 = 1 − (2𝑦^2)/𝑥^2 + 𝑦/𝑥 𝑥 𝒅𝒗/𝒅𝒙 + v = 1 − 2 〖(𝒗𝒙)〗^𝟐/𝒙^𝟐 + 𝒗𝒙/𝒙 x 𝑑𝑣/𝑑𝑥 + v = 1 − (2𝑣^2 𝑥^2)/𝑥^2 + 𝑣 Using ∫1▒1/(𝑎^2 − 𝑥^2 ) dx = 1/2𝑎 log |(𝑎 + 𝑥)/(𝑎 − 𝑥)| 𝟏/𝟐 ×𝟏/𝟐( 𝟏/√𝟐 ) log |(𝟏/√𝟐 + 𝒗)/(𝟏/√𝟐 − 𝒗)|= log |𝑥| + c √2/4 log |(1 + √2 𝑣)/(1 − √2 𝑣)| = log |𝑥| + c Putting v = 𝑦/𝑥 𝟏/(𝟐√𝟐) log |(𝟏 + √𝟐 𝒚/𝒙)/(𝟏 − √𝟐 𝒚/𝒙)| = log |𝒙| + c 1/(2√2) log |((𝑥 + √2 𝑦)/𝑥)/((𝑥 − √2 𝑦)/𝑥)| = log |𝑥| + c 𝟏/(𝟐√𝟐) log |(𝒙+√𝟐 𝒚)/(𝒙−√𝟐 𝒚)| = log |𝒙| + c .