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Ex 9.4, 6 Show that the given differential equation is homogeneous and solve each of them. 𝑥 𝑑𝑦−𝑦 𝑑𝑥=√(𝑥^2+𝑦^2 ) 𝑑𝑥 Step 1: Find 𝑑𝑦/𝑑𝑥 x dy − y dx = √(𝑥^2+𝑦^2 ) dx x dy = √(𝑥^2+𝑦^2 ) dx + y dx x dy = (√(𝑥^2+𝑦^2 )+𝑦) dx 𝒅𝒚/𝒅𝒙 = (√(𝒙^𝟐 + 𝒚^𝟐 ) + 𝒚)/𝒙 Step 2: Put 𝑑𝑦/𝑑𝑥 = F(x, y) and find F(𝜆x, 𝜆y) F(x, y) = 𝑑𝑦/𝑑𝑥 = (√(𝑥^(2 )+ 𝑦^2 ) + 𝑦)/𝑥 F(𝜆 x, 𝜆y) = (√(〖(𝜆𝑥)〗^2 + (𝜆^2 𝑦^2 ) )+ 𝜆𝑦)/𝜆𝑥 = (√(𝜆^2 𝑥^2 + 𝜆^2 𝑦^2 ) + 𝜆𝑦)/𝜆𝑥 = (√(𝜆^2 (𝑥^2 + 𝑦^2)) + 𝜆𝑦)/𝜆𝑥= (𝜆√(𝑥^2 + 𝑦^2 ) + 𝜆𝑦)/𝜆𝑥 = (√(𝑥^2 + 𝑦^2 ) + 𝑦)/𝑥 = F(x, y) Hence, F(𝜆x, 𝜆y) = F(x, y) = 𝜆° F(x, y) Hence, F(x, y) is a homogenous Function of with degree 0 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+𝑣 𝑑𝑥/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + 𝒗 Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥=(√(𝑥^2 + 𝑦^2 )+ 𝑦)/𝑥 x 𝑑𝑣/𝑑𝑥+𝑣=(√(𝑥^2 + (𝑣𝑥)^2 ) + (𝑣𝑥))/𝑥 x 𝑑𝑣/𝑑𝑥+𝑣=(√(𝑥^2 + 𝑥^2 𝑣^2 ) + 𝑣𝑥)/𝑥 x 𝑑𝑣/𝑑𝑥+𝑣 =(√(𝑥^2 (1 + 𝑣^2)) + 𝑣𝑥)/𝑥 x 𝑑𝑣/𝑑𝑥+𝑣 =(𝑥√(1 + 𝑣^2 ) + 𝑣𝑥)/𝑥 x 𝑑𝑣/𝑑𝑥+𝑣 =(𝑥(√(1 + 𝑣^2 ) + 𝑣))/𝑥 x 𝒅𝒗/𝒅𝒙+𝒗= √(𝟏+𝒗^𝟐 )+𝒗 x 𝑑𝑣/𝑑𝑥= √(1+𝑣^2 )+𝑣 − 𝑣 x 𝑑𝑣/𝑑𝑥= √(1+𝑣^2 ) 𝑑𝑣/𝑑𝑥= √(1 + 𝑣^2 )/𝑥 𝒅𝒗/√(𝟏 + 𝒗^𝟐 )= 𝒅𝒙/𝒙 Integrating both sides. ∫1▒𝑑𝑣/√(1 + 𝑣^2 ) = ∫1▒𝑑𝑥/𝑥 ∫1▒𝒅𝒗/√(𝟏 + 𝒗^𝟐 ) = log |𝒙|+𝒄 We know that ∫1▒𝑑𝑣/√(𝑎^2 + 𝑥^2 ) =𝑙𝑜𝑔|𝑥+√(𝑥^2+𝑎^2 )|+𝑐 Putting a = 1, x = v log |𝑣+√(𝑣^2+1)| =𝑙𝑜𝑔|𝑥|+𝑐 log |𝑣+√(𝑣^2+1)| =𝑙𝑜𝑔|𝑐𝑥| v + √(𝒗^𝟐+𝟏) = cx Putting v = 𝑦/𝑥 𝒚/𝒙+√((𝒚/𝒙)^𝟐+𝟏)=𝒄𝒙 𝑦/𝑥+√(𝑦^2/𝑥^2 +1)=𝑐𝑥 𝑦/𝑥+√((𝑦^2 + 𝑥^2)/𝑥^2 )=𝑐𝑥 𝑦/𝑥+√(𝑦^2 + 𝑥^2 )/𝑥=𝑐𝑥 𝒚+√(𝒚^𝟐 〖+ 𝒙〗^𝟐 ) =𝒄𝒙^𝟐 ∴ General solution is 𝒚+√(𝒚^𝟐 〖+ 𝒙〗^𝟐 ) =𝒄𝒙^𝟐

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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