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Transcript

Misc 5 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) π’…π’š/(π’š^𝟐 + π’š + 𝟏)=(βˆ’π’…π’™)/(𝒙^𝟐 + 𝒙 + 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) π’…π’š/((π’š + 𝟏/𝟐)^𝟐+ (βˆšπŸ‘/𝟐)^𝟐 )=(βˆ’π’…π’™)/((𝒙 + 𝟏/𝟐)^𝟐+ (βˆšπŸ‘/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/βˆšπŸ‘ tanβˆ’1 ((π’š + 𝟏/𝟐)/(βˆšπŸ‘/𝟐)) = (βˆ’πŸ)/βˆšπŸ‘ tanβˆ’1 ((𝒙 + 𝟏/𝟐)/(βˆšπŸ‘/𝟐)) + C 2/√3 ["tanβˆ’1 " ((2𝑦 + 1)/√3)" + tanβˆ’1 " ((2π‘₯ + 1)/√3)] = C (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 ((πŸπ’š + 𝟏 + πŸπ’™ + 𝟏)/βˆšπŸ‘)/(𝟏 βˆ’ ((πŸπ’š + 𝟏)(πŸπ’™ + 𝟏))/πŸ‘) = tan (βˆšπŸ‘/𝟐 π‘ͺ) ((2𝑦 + 2π‘₯ +2)/√3)/((3 βˆ’ (2𝑦 + 1)(2π‘₯ + 1))/3) = C1 (√3(2𝑦 + 2π‘₯ + 2))/(3 βˆ’ (4π‘₯𝑦 + 2𝑦 + 2π‘₯ + 1) ) = C 2βˆšπŸ‘ (x + y + 1) = C1 (πŸ‘βˆ’πŸ’π’™π’šβˆ’πŸπ’™βˆ’πŸπ’šβˆ’πŸ) 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.