# Misc 7 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

Last updated at Dec. 11, 2019 by Teachoo

Transcript

Misc 7 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) ππ¦/(π¦^2 + π¦ + 1)=(βππ₯)/(π₯^2 + π₯ + 1) 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) ππ¦/((π¦ + 1/2)^2+ (β3/2)^2 )=(βππ₯)/((π₯ + 1/2)^2+ (β3/2)^2 ) Integrating both sides β«1βππ¦/((π¦ + 1/2)^2 +(β3/2)^2 ) = β β«1βππ₯/((π₯ + 1/2)^2 +(β3/2)^2 ) 2/β3 tanβ1 ((π¦ + 1/2)/(β3/2)) = (β2)/β3 tanβ1 ((π₯ + 1/2)/(β3/2)) + C 2/β3 ["tanβ1 " ((2π¦ + 1)/β3)" + tanβ1 " ((2π₯ + 1)/β3)] = C (β«1βππ₯/(π₯^2 + π^2 )=1/π "tanβ1" π₯/π) (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 ((2π¦ + 1 + 2π₯ + 1)/β3)/(1 β ((2π¦ + 1)(2π₯ + 1))/3) = tan (β3/2 πΆ) ((2π¦ + 2π₯ +2)/β3)/((3 β (2π¦ + 1)(2π₯ + 1))/3) = C1 (Where C1 = tan (β3/2 πΆ)) (β3(2π¦ + 2π₯ + 2))/(3 β (4π₯π¦ + 2π¦ + 2π₯ + 1) ) = C 2β3 (x + y + 1) = C1 |3β4π₯π¦β2π₯β2π¦β1| 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

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 Deleted for CBSE Board 2022 Exams

Misc 4 Important

Misc 5 Important Deleted for CBSE Board 2022 Exams

Misc 6

Misc 7 Important You are here

Misc 8

Misc 9 Important

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13

Misc 14 Important

Misc 15 Important

Misc 16 (MCQ)

Misc 17 (MCQ) Important Deleted for CBSE Board 2022 Exams

Misc 18 (MCQ)

Chapter 9 Class 12 Differential Equations (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.