# Ex 9.5, 13 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 𝑥 sin2 𝑦𝑥−𝑦𝑑𝑥+𝑥𝑑𝑦=0;𝑦= 𝜋4 When 𝑥=1 Differential equation can be written as 𝑥 sin2 𝑦𝑥−𝑦𝑑𝑥+𝑥𝑑𝑦=0 x dy = − 𝑥 sin2 𝑦𝑥−𝑦dx 𝑑𝑦𝑑𝑥 = − 𝑥𝑥 sin2 𝑦𝑥 − 𝑦𝑥 Let F(x, y) = 𝑑𝑦𝑑𝑥 =− 𝑥𝑥 sin2 𝑦𝑥 − 𝑦𝑥 Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = − sin2 𝜆𝑦𝜆𝑥− 𝜆𝑦𝜆𝑥 = − sin2 𝑦𝑥− 𝑦𝑥 = 𝜆° F (x, y) ∴ F(x, y) is 𝑎 homogenous function of degree zero Putting y = vx Diff w.r.t. x 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v Putting value of 𝑑𝑦𝑑𝑥 and y = vx in (1) 𝑑𝑦𝑑𝑥 = − 𝑥𝑥 sin2 𝑦𝑥 − 𝑦𝑥 v + 𝑥 𝑑𝑣𝑑𝑥 = − sin2 𝑣𝑥𝑥− 𝑣𝑥𝑥 v + 𝑥 𝑑𝑣𝑑𝑥 = − sin2𝑣−𝑣 𝑥 𝑑𝑣𝑑𝑥 = v − sin2 v − v 𝑥 𝑑𝑣𝑑𝑥 = − sin2 v 𝑑𝑣 sin2𝑣 = − 𝑑𝑥𝑥 Integrating both sides 𝑑𝑣 sin2𝑣 = − 𝑑𝑥𝑥 𝑐𝑜𝑠𝑒 𝑐2 𝑣 𝑑𝑣 = − 𝑑𝑥𝑥 − cot v = − log 𝑥+𝑐 log 𝑥 − cot v = C Putting value of v = 𝑦𝑥 log 𝑥 − cot 𝑦𝑥 = C Putting x = 1 & y = 𝜋4 log 1 − cot 𝜋4 = C 0 − 1 = C C = −1 Putting value of C in (2) log 𝑥 − cot 𝑦𝑥 = −1 cot 𝑦𝑥 − log 𝑥 = 1 cot 𝑦𝑥 − log 𝑥 = log e cot 𝑦𝑥 = log 𝑥 + log e cot 𝒚𝒙 = log 𝒆𝒙

Chapter 9 Class 12 Differential Equations

Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.