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Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

Last updated at Dec. 11, 2019 by Teachoo

Misc 10 Solve the differential equation π¦ π^(π₯/π¦) ππ₯= (π^(π₯/π¦)+π¦^2 )ππ¦ (π¦β 0) π¦ π^(π₯/π¦) ππ₯= (π^(π₯/π¦)+π¦^2 )ππ¦ ππ₯/ππ¦ = (π₯π^(π₯/π¦) + π¦^2)/(π¦^(π^(π₯/π¦) ) ) We can see that it is not homogeneous, so letβs try something else π¦π^(π₯/π¦) ππ₯/ππ¦=π₯π^(π₯/π¦)+π¦^2 π¦π^(π₯/π¦) ππ₯/ππ¦βπ₯π^(π₯/π¦)=π¦^2 π^(π₯/π¦) (π¦ ππ₯/ππ¦βπ₯)=π¦^2 π^(π₯/π¦) ((π¦ ππ₯/ππ¦ β π₯)/π¦^2 )=π¦^2 Let π^(π₯/π¦) = z Diff w.r.t. y. π^(π₯/π¦) π(π₯/π¦)/ππ¦ = ππ§/ππ¦ π^(π₯/π¦) ((π¦ ππ₯/ππ¦ β π₯ ππ¦/ππ¦)/π¦^2 )=ππ§/ππ¦ " " β¦(1) π^(π₯/π¦) ((π¦ ππ₯/ππ¦ β π₯)/π¦^2 )=ππ§/ππ¦ " " From (1) ππ§/ππ¦ = 1 dz = dy Integrating on both sides β«1βππ§ = β«1βππ¦ z = y + c Putting π^(π₯/π¦) = z π^(π/π) = y + c