Check sibling questions

Ex 9.4, 5 - Find general solution: (ex + e-x) dy - (ex-e-x)dx

Ex 9.4, 5 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 5 - Chapter 9 Class 12 Differential Equations - Part 3

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!


Transcript

Ex 9.4, 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 y = log (𝒆^𝒙−𝒆^(−𝒙)) + C As 𝑒^𝑥−𝑒^(−𝑥) > 0 So, its always positive Removing the modulus

Ask a doubt (live)
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.