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Misc 10 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 10 - Chapter 9 Class 12 Differential Equations - Part 3

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Transcript

Misc 10 Solve the differential equation 𝑦 𝑒^(π‘₯/𝑦) 𝑑π‘₯= (x𝑒^(π‘₯/𝑦)+𝑦^2 )𝑑𝑦 (𝑦≠0) 𝑦 𝑒^(π‘₯/𝑦) 𝑑π‘₯= (𝑒^(π‘₯/𝑦)+𝑦^2 )𝑑𝑦 𝑑π‘₯/𝑑𝑦 = (π‘₯𝑒^(π‘₯/𝑦) + 𝑦^2)/(𝑦^(𝑒^(π‘₯/𝑦) ) ) We can see that it is not homogeneous, so let’s try something else 𝑦𝑒^(π‘₯/𝑦) 𝑑π‘₯/𝑑𝑦=π‘₯𝑒^(π‘₯/𝑦)+𝑦^2 𝑦𝑒^(π‘₯/𝑦) 𝑑π‘₯/π‘‘π‘¦βˆ’π‘₯𝑒^(π‘₯/𝑦)=𝑦^2 𝑒^(π‘₯/𝑦) (𝑦 𝑑π‘₯/π‘‘π‘¦βˆ’π‘₯)=𝑦^2 𝑒^(π‘₯/𝑦) ((𝑦 𝑑π‘₯/𝑑𝑦 βˆ’ π‘₯)/𝑦^2 )=𝑦^2 Let 𝑒^(π‘₯/𝑦) = z Diff w.r.t. y. 𝑒^(π‘₯/𝑦) 𝑑(π‘₯/𝑦)/𝑑𝑦 = 𝑑𝑧/𝑑𝑦 𝑒^(π‘₯/𝑦) ((𝑦 𝑑π‘₯/𝑑𝑦 βˆ’ π‘₯ 𝑑𝑦/𝑑𝑦)/𝑦^2 )=𝑑𝑧/𝑑𝑦 " " …(1) 𝑒^(π‘₯/𝑦) ((𝑦 𝑑π‘₯/𝑑𝑦 βˆ’ π‘₯)/𝑦^2 )=𝑑𝑧/𝑑𝑦 " " From (1) 𝑑𝑧/𝑑𝑦 = 1 dz = dy Integrating on both sides ∫1▒𝑑𝑧 = ∫1▒𝑑𝑦 z = y + c Putting 𝑒^(π‘₯/𝑦) = z 𝒆^(𝒙/π’š) = 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.