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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Misc 10 Solve the differential equation 𝑦 𝑒^(π‘₯/𝑦) 𝑑π‘₯= (𝑒^(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

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.