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Misc 2 For each of the exercise given below , verify that the given function (𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑜𝑟 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡) is a solution of the corresponding differential equation . (iv) 𝑥^2=2𝑦^2 log⁡𝑦 : (𝑥^2+𝑦^2 ) 𝑑𝑦/𝑑𝑥−𝑥𝑦=0 𝑥^2=2𝑦^2 log⁡𝑦 Differentiating Both sides w.r.t. x (𝑥^2 )^′=(2𝑦^2 log⁡𝑦 )′ 2𝑥=(2𝑦^2 )^′ log⁡𝑦 +2𝑦^2 (log⁡𝑦 )^′ 2𝑥=2×2𝑦𝑦^′ log⁡𝑦 + 2𝑦^2× 1/𝑦 𝑦′ 2𝑥=4𝑦𝑦^′ log⁡𝑦 + 2𝑦𝑦′ 𝑥=2𝑦𝑦^′ log⁡𝑦 + 𝑦𝑦′ 𝒙=〖𝒚𝒚〗^′ (𝟐 𝒍𝒐𝒈⁡𝒚+𝟏) Now, from our equation 𝑥^2=2𝑦^2 log⁡𝑦 𝑥^2/(2𝑦^2 )=log⁡𝑦 𝒍𝒐𝒈⁡𝒚=𝒙^𝟐/(𝟐𝒚^𝟐 ) Putting value of 𝑙𝑜𝑔⁡𝑦 in (1) 𝑥=〖𝑦𝑦〗^′ (2 𝒍𝒐𝒈⁡𝒚+1) 𝑥=〖𝑦𝑦〗^′ (2×𝒙^𝟐/(𝟐𝒚^𝟐 ) " " +1) 𝑥=〖𝑦𝑦〗^′ (𝑥^2/𝑦^2 " " +1) 𝒙=〖𝒚𝒚〗^′ ((𝒙^𝟐 + 𝒚^𝟐)/𝒚^𝟐 ) 𝑥=𝑦^′ ((𝑥^2 + 𝑦^2)/𝑦) 𝑥𝑦=𝑦^′ (𝑥^2 + 𝑦^2 ) 𝒚^′ (𝒙^𝟐 + 𝒚^𝟐 )−𝒙𝒚=𝟎 Thus, Given Function is a solution of the Differential Equation

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.