# Misc 5 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1 (i)

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 You are here

Misc 6

Misc 7 Important

Misc 8 Important

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 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 5 Show that the general solution of the differential equation ππ¦/ππ₯+(π¦^2+π¦+1)/(π₯^2+π₯+1)=0 is given by (π₯+π¦+1)=A(1βπ₯βπ¦β2π₯π¦), where A is parameter. ππ¦/ππ₯+(π¦^2 + π¦ + 1)/(π₯^2 + π₯ + 1) = 0 ππ¦/ππ₯=(β(π¦^2 + π¦ + 1))/(π₯^2 + π₯ + 1) π π/(π^π + π + π)=(βπ π)/(π^π + π + π) ππ¦/(π¦^2 +2(1/2)π¦ + (1/2)^2β (1/2)^2+ 1)=(βππ₯)/(π₯^2 + 2(1/2)π₯ + (1/2)^2β (1/2)^2+ 1) ππ¦/((π¦ + 1/2)^2+ 3/4)=(βππ₯)/((π₯ + 1/2)^2+ 3/4) π π/((π + π/π)^π+ (βπ/π)^π )=(βπ π)/((π + π/π)^π+ (βπ/π)^π ) Integrating both sides β«1βππ¦/((π¦ + 1/2)^2 +(β3/2)^2 ) = β β«1βππ₯/((π₯ + 1/2)^2 +(β3/2)^2 ) π/βπ tanβ1 ((π + π/π)/(βπ/π)) = (βπ)/βπ tanβ1 ((π + π/π)/(βπ/π)) + C 2/β3 ["tanβ1 " ((2π¦ + 1)/β3)" + tanβ1 " ((2π₯ + 1)/β3)] = C (Using tanβ1 A + tanβ1 B = tanβ1 ((π΄ + π΅)/(1 β π΄π΅)) ) 2/β3 "tanβ1" β((2π¦ + 1)/β3 + (2π₯ + 1)/β3)/(1 β (2π¦ β 1)/β3 Γ(2π₯ + 1)/β3 )β=πΆ "tanβ1" [((2π¦ + 1 + 2π₯ + 1)/β3)/(1 β ((2π¦ + 1)(2π₯ + 1))/β3)] = β3/2 πΆ ((ππ + π + ππ + π)/βπ)/(π β ((ππ + π)(ππ + π))/π) = tan (βπ/π πͺ) ((2π¦ + 2π₯ +2)/β3)/((3 β (2π¦ + 1)(2π₯ + 1))/3) = C1 (β3(2π¦ + 2π₯ + 2))/(3 β (4π₯π¦ + 2π¦ + 2π₯ + 1) ) = C 2βπ (x + y + 1) = C1 (πβπππβππβππβπ) 2β3 (x + y + 1) = C1 (2β4π₯π¦β2π₯β2π¦) 2β3 (x + y + 1) = C1 Γ 2 (1βπ₯βπ¦βπ₯π¦) βπ (x + y + 1) = C1 (1 β x β y β 2xy) is the required general solution