# Ex 9.5, 9 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 9 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. 𝑦 𝑑𝑥+𝑥𝑙𝑜𝑔 𝑦𝑥𝑑𝑦−2𝑥 𝑑𝑦=0 Step 1 : Find 𝑑𝑦𝑑𝑥 𝑦 𝑑𝑥+𝑥𝑙𝑜𝑔 𝑦𝑥𝑑𝑦−2𝑥 𝑑𝑦=0 dy 𝑥 log 𝑦𝑥−2𝑥 = − y dx 𝑑𝑦𝑑𝑥 = −𝑦𝑥 log 𝑦𝑥 − 2𝑥 𝑑𝑦𝑑𝑥 = −𝑦−𝑥 2 − log 𝑦𝑥 𝑑𝑦𝑑𝑥 = 𝑦𝑥2 − log 𝑦𝑥 Step 2. Putting F(x , y) = 𝑑𝑦𝑑𝑥 and finding F(𝜆x, 𝜆y) F(x, y) = 𝑦𝑥2 − log 𝑦𝑥 𝐹(𝜆𝑥,𝜆𝑦) = 𝜆𝑦𝜆𝑥2 − log 𝜆𝑦𝜆𝑥 = 𝑦𝑥2 − log 𝑦𝑥 = 𝜆° 𝐹(𝑥, 𝑦) Thus, F(x, y) is a homogenous equation function of order zero Therefore 𝑑𝑦𝑑𝑥 is a homogenous differential equation Step 3 : Solving 𝑑𝑦𝑑𝑥 by putting y = vx Putting y = vx Diff w.r.t.x 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v 𝑑𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = x 𝑑𝑣𝑑𝑥 + v Putting value of 𝑑𝑦𝑑𝑥 and y = vx in (1) 𝑑𝑦𝑑𝑥 = 𝑦𝑥2 − log 𝑦𝑥 v + 𝑥 𝑑𝑣𝑑𝑥 = 𝑣𝑥𝑥2 − 𝑙𝑜𝑔 𝑣𝑥𝑥 v + 𝑥 𝑑𝑣𝑑𝑥 = 𝑣2 − log𝑣 𝑥 𝑑𝑣𝑑𝑥 = 𝑣2 − log𝑣 − v 𝑥 𝑑𝑣𝑑𝑥 = 𝑣 − 2𝑣 + 𝑣 log𝑣2 − log𝑣 𝑥 𝑑𝑣𝑑𝑥 = 𝑣 log𝑣 − 𝑣2 − log𝑣 2 − log𝑣𝑣 log𝑣 dv = 𝑑𝑥𝑥 Integrating both sides 2 − log𝑣𝑣 log𝑣 − 𝑣 dv = 𝑑𝑥𝑥 2 − log𝑣−𝑣 (1 − log𝑣) dv = log x + log c 1 + 1 − log𝑣−𝑣 (1 − log𝑣) 𝑑𝑣= log x + log c 1(−𝑣)(1 − log𝑣)𝑑𝑣 − 1𝑣𝑑𝑣 = log x + log c 1𝑣( log𝑣 − 1)𝑑𝑣 − 1𝑣𝑑𝑣 = log x + log c 𝑑𝑣𝑣( log𝑣 − 1) – log v = log x + log c Put t = log v − 1 dt = 1𝑣 dv So, our equation becomes 𝑑𝑡𝑡 − log v = log x + log c log t − log v = log x + log c Putting value of t log (log v − 1) −log v + = log x + log c log (log v − 1) = log x + log c + log v log (log v − 1) = log C xv Putting value of v = 𝑦𝑥 log log 𝑦𝑥−1= log𝑥 𝑐 𝑦𝑥 log log 𝑦𝑥−1= log 𝑐𝑦 log 𝑦𝑥 − 1 = cy cy = log 𝒚𝒙 − 1

Chapter 9 Class 12 Differential Equations

Serial order wise

About the Author

Davneet Singh

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.