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Solve the differential equation: ye^(x/y) dx=(xe^(x/y)+y^2 )dy,(y≠0)

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y𝑒^(𝑥/𝑦) 𝑑𝑥=(𝑥𝑒^(𝑥/𝑦)+𝑦^2 )𝑑𝑦 Step 1: Finding 𝑑𝑥/𝑑𝑦 y𝑒^(𝑥/𝑦) 𝑑𝑥=(𝑥𝑒^(𝑥/𝑦)+𝑦^2 )𝑑𝑦 𝒅𝒙/𝒅𝒚=(𝒙𝒆^(𝒙/𝒚) + 𝒚^𝟐)/(𝒚𝒆^(𝒙/𝒚) ) Step 2 : Solving 𝑑𝑥/𝑑𝑦 by Putting 𝑥=𝑣𝑦 𝑑𝑥/𝑑𝑦=(𝑥𝑒^(𝑥/𝑦) + 𝑦^2)/(𝑦𝑒^(𝑥/𝑦) ) Put 𝒙=𝒗𝒚 Diff. w.r.t. 𝑦 𝑑𝑥/𝑑𝑦=𝑑/𝑑𝑦 (𝑣𝑦) 𝑑𝑥/𝑑𝑦=𝑦 . 𝑑𝑣/𝑑𝑦+𝑣 𝑑𝑦/𝑑𝑦 𝒅𝒙/𝒅𝒚=𝒚 . 𝒅𝒗/𝒅𝒚+𝒗 Putting values of 𝑑𝑥/𝑑𝑦 and x in (1) 𝑑𝑥/𝑑𝑦=(𝑥𝑒^(𝑥/𝑦)+𝑦^2)/(𝑦𝑒^(𝑥/𝑦) ) 𝒗+𝒚 𝒅𝒗/𝒅𝒚=(𝒗𝒚𝒆^𝒗+𝑦^2)/(𝒚𝒆^𝒗 ) 𝑣+𝑦 𝑑𝑣/𝑑𝑦=(𝑣〖𝑦𝑒〗^𝑣)/(𝑦𝑒^𝑣 ) + 𝑦^2/(𝑦𝑒^𝑣 ) v+𝑦 𝑑𝑣/𝑑𝑦=𝑣+ 𝑦/𝑒^𝑣 𝑦 𝑑𝑣/𝑑𝑦=𝑦/𝑒^𝑣 𝑑𝑣/𝑑𝑦=1/〖 𝑒〗^𝑣 〖 𝒆〗^𝒗 𝒅𝒗=𝒅𝒚 Integrating Both Sides ∫1▒〖〖 𝒆〗^𝒗 𝑑𝑣〗= ∫1▒𝑑𝑦 〖 𝒆〗^𝒗=𝒚+𝒄 Putting back 𝑣=𝑥/𝑦 𝒆^(𝒙/𝒚)=𝒚+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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.