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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 .

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo