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


Transcript

Ex 9.3, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑑𝑦/𝑑𝑥=(1 − cos⁡𝑥)/(1 + cos⁡𝑥 ) 𝑑𝑦/𝑑𝑥 = (1 − cos⁡𝑥)/(1 + cos⁡𝑥 ) We know that cos 2x = 2cos2 x − 1 Putting x = 𝑥/2 cos 2𝑥/2 = 2 cos2 𝑥/2 − 1 cos x = 2 cos2 𝑥/2 − 1 1 + cos x = 2cos2 𝑥/2 We know cos 2x = 1 − 2sin2 x Putting x = 𝑥/2 cos2 2𝑥/2 = 1 − 2 sin2 𝑥/2 cos x = 1 − 2sin2 𝑥/2 1 − cos x = 2sin2 𝑥/2 Putting values in equation 𝒅𝒚/𝒅𝒙 = (𝟐 〖〖𝐬𝐢𝐧〗^𝟐 〗⁡〖𝒙/𝟐〗 )/(𝟐 〖〖𝐜𝐨𝐬〗^𝟐 〗⁡〖𝒙/𝟐〗 ) 𝑑𝑦 = (〖sin^2 〗⁡〖𝑥/2〗 )/cos^(2 )⁡〖𝑥/2〗 𝑑𝑥 𝑑𝑦 = tan2 𝑥/2 𝑑𝑥 Putting tan2 𝑥/2 = sec2 𝑥/2 − 1 𝒅𝒚 = ("sec2 " 𝒙/𝟐 " − 1" )𝒅𝒙 Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x − 1 Putting x = 𝑥/2 tan2 𝒙/𝟐 = sec2 𝒙/𝟐 − 1 ∫1▒𝒅𝒚 = ∫1▒(𝒔𝒆𝒄𝟐 𝒙/𝟐−𝟏) 𝒅𝒙 y = ∫1▒〖𝑠𝑒𝑐2 𝑥/2 𝑑𝑥− ∫1▒𝑑𝑥〗 y = 𝟏/(𝟏/𝟐) tan 𝒙/𝟐 − x + C y = 2 tan 𝑥/2 − x + c y = 2 tan 𝑥/2 − x + C y = 𝟐 tan 𝒙/𝟐 − x + C is the 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.