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Misc 4 Find the general solution of the differential equation 𝑑𝑦/𝑑π‘₯+√((1βˆ’π‘¦^2)/(1βˆ’π‘₯^2 ))=0𝑑𝑦/𝑑π‘₯+√((1 βˆ’ 𝑦^2)/(1 βˆ’ π‘₯^2 )) = 0 𝑑𝑦/𝑑π‘₯ = βˆ’ √((1 + 𝑦^2)/(1 + π‘₯^2 )) π’…π’š/√(𝟏 βˆ’ π’š^𝟐 )=𝒅𝒙/√(𝟏 βˆ’ 𝒙^𝟐 ) Integrating both sides ∫1▒𝑑𝑦/√(1 βˆ’ 𝑦^2 ) = ∫1▒𝑑π‘₯/√(1 βˆ’ π‘₯^2 ) sinβˆ’1 y = βˆ’ sinβˆ’1 x + C sinβˆ’1x + sinβˆ’1 y = C

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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.