Ex 9.3, 5 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Variable separation - Equation given
Ex 9.3, 23 (MCQ)
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Chapter 9 Class 12 Differential Equations
Concept wise
- Order and Degree
- Gen and Particular Solution
- Formation of Differntial equation when general solution given
-
Variable separation - Equation given
- Variable separation - Statement given
- Solving homogeneous differential equation
- Solving Linear differential equations - Equation given
- Solving Linear differential equations - Statement given
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
