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Misc 7 - Show that general solution is (x+y+1) = A(1-x-y-2xy)

Misc 7 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 7 - Chapter 9 Class 12 Differential Equations - Part 3 Misc 7 - Chapter 9 Class 12 Differential Equations - Part 4


Transcript

Misc 7 Show that the general solution of the differential equation 𝑑𝑦/𝑑π‘₯+(𝑦^2+𝑦+1)/(π‘₯^2+π‘₯+1)=0 is given by (π‘₯+𝑦+1)=A(1βˆ’π‘₯βˆ’π‘¦βˆ’2π‘₯𝑦), where A is parameter. 𝑑𝑦/𝑑π‘₯+(𝑦^2 + 𝑦 + 1)/(π‘₯^2 + π‘₯ + 1) = 0 𝑑𝑦/𝑑π‘₯=(βˆ’(𝑦^2 + 𝑦 + 1))/(π‘₯^2 + π‘₯ + 1) 𝑑𝑦/(𝑦^2 + 𝑦 + 1)=(βˆ’π‘‘π‘₯)/(π‘₯^2 + π‘₯ + 1) 𝑑𝑦/(𝑦^2 +2(1/2)𝑦 + (1/2)^2βˆ’ (1/2)^2+ 1)=(βˆ’π‘‘π‘₯)/(π‘₯^2 + 2(1/2)π‘₯ + (1/2)^2βˆ’ (1/2)^2+ 1) 𝑑𝑦/((𝑦 + 1/2)^2+ 3/4)=(βˆ’π‘‘π‘₯)/((π‘₯ + 1/2)^2+ 3/4) 𝑑𝑦/((𝑦 + 1/2)^2+ (√3/2)^2 )=(βˆ’π‘‘π‘₯)/((π‘₯ + 1/2)^2+ (√3/2)^2 ) Integrating both sides ∫1▒𝑑𝑦/((𝑦 + 1/2)^2 +(√3/2)^2 ) = βˆ’ ∫1▒𝑑π‘₯/((π‘₯ + 1/2)^2 +(√3/2)^2 ) 2/√3 tanβˆ’1 ((𝑦 + 1/2)/(√3/2)) = (βˆ’2)/√3 tanβˆ’1 ((π‘₯ + 1/2)/(√3/2)) + C 2/√3 ["tanβˆ’1 " ((2𝑦 + 1)/√3)" + tanβˆ’1 " ((2π‘₯ + 1)/√3)] = C (∫1▒𝑑π‘₯/(π‘₯^2 + π‘Ž^2 )=1/π‘Ž "tanβˆ’1" π‘₯/π‘Ž) (Using tanβˆ’1 A + tanβˆ’1 B = tanβˆ’1 ((𝐴 + 𝐡)/(1 βˆ’ 𝐴𝐡)) ) 2/√3 "tanβˆ’1" ⌈((2𝑦 + 1)/√3 + (2π‘₯ + 1)/√3)/(1 βˆ’ (2𝑦 βˆ’ 1)/√3 Γ—(2π‘₯ + 1)/√3 )βŒ‰=𝐢 "tanβˆ’1" [((2𝑦 + 1 + 2π‘₯ + 1)/√3)/(1 βˆ’ ((2𝑦 + 1)(2π‘₯ + 1))/√3)] = √3/2 𝐢 ((2𝑦 + 1 + 2π‘₯ + 1)/√3)/(1 βˆ’ ((2𝑦 + 1)(2π‘₯ + 1))/3) = tan (√3/2 𝐢) ((2𝑦 + 2π‘₯ +2)/√3)/((3 βˆ’ (2𝑦 + 1)(2π‘₯ + 1))/3) = C1 (Where C1 = tan (√3/2 𝐢)) (√3(2𝑦 + 2π‘₯ + 2))/(3 βˆ’ (4π‘₯𝑦 + 2𝑦 + 2π‘₯ + 1) ) = C 2√3 (x + y + 1) = C1 |3βˆ’4π‘₯π‘¦βˆ’2π‘₯βˆ’2π‘¦βˆ’1| 2√3 (x + y + 1) = C1 |2βˆ’4π‘₯π‘¦βˆ’2π‘₯βˆ’2𝑦| 2√3 (x + y + 1) = C1 Γ— 2 |1βˆ’π‘₯βˆ’π‘¦βˆ’π‘₯𝑦| βˆšπŸ‘ (x + y + 1) = C1 ( 1 βˆ’ x βˆ’ y βˆ’ 2xy) is the required general solution

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.