Misc 8 - Solve y ex/y dx = (ex/y + y2)dy - Chapter 9 Class 12 - Miscellaneous

part 2 - Misc 8 - Miscellaneous - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Misc 8 - Miscellaneous - Serial order wise - Chapter 9 Class 12 Differential Equations

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Misc 8 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 𝒅𝒙/π’…π’š = (𝒙𝒆^(𝒙/π’š) + π’š^𝟐)/(π’š^(𝒆^(𝒙/π’š) ) ) 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^(𝒙/π’š) ((π’š 𝒅𝒙/π’…π’š βˆ’ 𝒙)/π’š^𝟐 )=π’š^𝟐 Let 𝒆^(𝒙/π’š) = 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 " " 𝒆^(𝒙/π’š) ((π’š 𝒅𝒙/π’…π’š βˆ’ 𝒙)/π’š^𝟐 )=𝒅𝒛/π’…π’š " " From (1) 𝒅𝒛/π’…π’š = 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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