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

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


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

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.