# 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

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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.