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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

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Misc 5 Show that the general solution of the differential equation 𝑑𝑦/𝑑𝑥+(𝑦^2+𝑦+1)/(𝑥^2+𝑥+1)=0 is given by (𝑥+𝑦+1)=A(1−𝑥−𝑦−2𝑥𝑦), where A is parameter. 𝑑𝑦/𝑑𝑥+(𝑦^2 + 𝑦 + 1)/(𝑥^2 + 𝑥 + 1) = 0 𝑑𝑦/𝑑𝑥=(−(𝑦^2 + 𝑦 + 1))/(𝑥^2 + 𝑥 + 1) 𝑑𝑦/(𝑦^2 + 𝑦 + 1)=(−𝑑𝑥)/(𝑥^2 + 𝑥 + 1) 𝑑𝑦/(𝑦^2 +2(1/2)𝑦 + (1/2)^2− (1/2)^2+ 1)=(−𝑑𝑥)/(𝑥^2 + 2(1/2)𝑥 + (1/2)^2− (1/2)^2+ 1) 𝑑𝑦/((𝑦 + 1/2)^2+ 3/4)=(−𝑑𝑥)/((𝑥 + 1/2)^2+ 3/4) 𝑑𝑦/((𝑦 + 1/2)^2+ (√3/2)^2 )=(−𝑑𝑥)/((𝑥 + 1/2)^2+ (√3/2)^2 ) Integrating both sides ∫1▒𝑑𝑦/((𝑦 + 1/2)^2 +(√3/2)^2 ) = − ∫1▒𝑑𝑥/((𝑥 + 1/2)^2 +(√3/2)^2 ) 2/√3 tan−1 ((𝑦 + 1/2)/(√3/2)) = (−2)/√3 tan−1 ((𝑥 + 1/2)/(√3/2)) + C 2/√3 ["tan−1 " ((2𝑦 + 1)/√3)" + tan−1 " ((2𝑥 + 1)/√3)] = C (∫1▒𝑑𝑥/(𝑥^2 + 𝑎^2 )=1/𝑎 "tan−1" 𝑥/𝑎) (Using tan−1 A + tan−1 B = tan−1 ((𝐴 + 𝐵)/(1 − 𝐴𝐵)) ) 2/√3 "tan−1" ⌈((2𝑦 + 1)/√3 + (2𝑥 + 1)/√3)/(1 − (2𝑦 − 1)/√3 ×(2𝑥 + 1)/√3 )⌉=𝐶 "tan−1" [((2𝑦 + 1 + 2𝑥 + 1)/√3)/(1 − ((2𝑦 + 1)(2𝑥 + 1))/√3)] = √3/2 𝐶 ((2𝑦 + 1 + 2𝑥 + 1)/√3)/(1 − ((2𝑦 + 1)(2𝑥 + 1))/3) = tan (√3/2 𝐶) ((2𝑦 + 2𝑥 +2)/√3)/((3 − (2𝑦 + 1)(2𝑥 + 1))/3) = C1 (Where C1 = tan (√3/2 𝐶)) (√3(2𝑦 + 2𝑥 + 2))/(3 − (4𝑥𝑦 + 2𝑦 + 2𝑥 + 1) ) = C 2√3 (x + y + 1) = C1 |3−4𝑥𝑦−2𝑥−2𝑦−1| 2√3 (x + y + 1) = C1 |2−4𝑥𝑦−2𝑥−2𝑦| 2√3 (x + y + 1) = C1 × 2 |1−𝑥−𝑦−𝑥𝑦| √𝟑 (x + y + 1) = C1 ( 1 − x − y − 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.