

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Miscellaneous
Misc 1 (ii)
Misc 1 (iii) Important
Misc 2 (i)
Misc 2 (ii) Important
Misc 2 (iii)
Misc 2 (iv) Important
Misc 3 Important
Misc 4
Misc 5 Important
Misc 6
Misc 7 Important
Misc 8 Important You are here
Misc 9
Misc 10 Important
Misc 11
Misc 12 Important
Misc 13 (MCQ)
Misc 14 (MCQ) Important
Misc 15 (MCQ)
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Misc 8 Solve the differential equation π¦ π^(π₯/π¦) ππ₯= (xπ^(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