Check sibling questions

Ex 9.5, 5 - Show homogeneous: x2 dy/dx = x2 - 2y2 + xy - Solving homogeneous differential equation

Ex 9.5, 5 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 5 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.5, 5 - Chapter 9 Class 12 Differential Equations - Part 4 Ex 9.5, 5 - Chapter 9 Class 12 Differential Equations - Part 5


Transcript

Ex 9.5, 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 x 𝑑𝑣/𝑑𝑥 = 1 − 2v2 + v − v x 𝑑𝑣/𝑑𝑥 = 1 − 2v2 𝑑𝑣/𝑑𝑥 = (1 − 2𝑣^2)/𝑥 𝑑𝑣/(1 − 2𝑣^2 ) = 𝑑𝑥/𝑥 Integrating both sides. ∫1▒𝑑𝑣/(1−2𝑣^2 ) = ∫1▒𝑑𝑥/𝑥 ∫1▒((𝑑𝑣))/((1−2𝑣^2))× (1/2)/(1/2) = ∫1▒𝑑𝑥/𝑥 1/2 ∫1▒𝑑𝑣/(1/2 − 𝑣^2 ) = ∫1▒𝑑𝑥/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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.