Ex 9.3, 5 - Find general solution: (ex + e-x) dy - (ex-e-x)dx - Ex 9.3

part 2 - Ex 9.3, 5 - Ex 9.3 - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Ex 9.3, 5 - Ex 9.3 - Serial order wise - Chapter 9 Class 12 Differential Equations

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Transcript

Ex 9.3, 5 For each of the differential equations in Exercises 1 to 10, find the general solution : (𝑒^𝑥+𝑒^(−𝑥) )𝑑𝑦−(𝑒^𝑥−𝑒^(−𝑥) )𝑑𝑥=0 (𝑒^𝑥+𝑒^(−𝑥) )𝑑𝑦−(𝑒^𝑥−𝑒^(−𝑥) )𝑑𝑥=0 (𝑒^𝑥+𝑒^(−𝑥) )𝑑𝑦 = (𝑒^𝑥−𝑒^(−𝑥) )𝑑𝑥 𝑑𝑦/𝑑𝑥 = (𝑒^𝑥 − 𝑒^(−𝑥))/(𝑒^𝑥 + 𝑒^(−𝑥) ) dx 𝒅𝒚 = (𝒆^𝒙 − 𝒆^(−𝒙))/(𝒆^𝒙 + 𝒆^(−𝒙) ) dx Integrating both sides. ∫1▒𝑑𝑦 = ∫1▒(𝑒^𝑥 − 𝑒^(−𝑥))/(𝑒^𝑥 + 𝑒^(−𝑥) ) dx 𝒚 = ∫1▒(𝒆^𝒙 − 𝒆^(−𝒙))/(𝒆^𝒙 + 𝒆^(−𝒙) ) dx Let t = 𝒆^𝒙+𝒆^(−𝒙) 𝑑𝑡/𝑑𝑥 = (𝑒^𝑥−𝑒^(−𝑥) ) dx = 𝒅𝒕/(𝒆^𝒙 − 𝒆^(−𝒙) ) Putting value of t and dt in (1) ∫1▒𝑑𝑦 = ∫1▒(𝑒^(𝑥 )−〖 𝑒〗^(−𝑥))/𝑡 𝑑𝑡/(𝑒^𝑥 − 𝑒^(−𝑥) ) . ∫1▒𝑑𝑦 = ∫1▒〖𝑑𝑡/𝑡 " " 〗 y = log |𝒕|+𝒄 Putting back t = 𝑒^𝑥−𝑒^(−𝑥) y = log |𝑒^𝑥−𝑒^(−𝑥) | + C As 𝒆^𝒙−𝒆^(−𝒙) > 0 So, its always positive Removing the modulus y = log (𝒆^𝒙−𝒆^(−𝒙)) + C

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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 and Computer Science at Teachoo