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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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 (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 (√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.