Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Question 1 Deleted for CBSE Board 2024 Exams

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Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at Aug. 14, 2023 by Teachoo

Misc 8 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¦ π π/π π = (ππ^(π/π) + π^π)/(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 π π/π π β π)/π^π )=π^π 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¦ " " π^(π/π) ((π π π/π π β π)/π^π )=π π/π π " " From (1) π π/π π = 1 dz = dy Integrating on both sides β«1βππ§ = β«1βππ¦ z = y + c Putting π^(π₯/π¦) = z π^(π/π) = y + c