Check sibling questions

Ex 9.5, 14 - Find particular solution: dy/dx - y/x + cosec = 0 - Ex 9.5

Ex 9.5, 14 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 14 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.5, 14 - Chapter 9 Class 12 Differential Equations - Part 4

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


Transcript

Ex 9.4, 14 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : 𝑑𝑦/𝑑𝑥−𝑦/𝑥+𝑐𝑜𝑠𝑒𝑐(𝑦/𝑥)=0;𝑦=0 When 𝑥=1 Differential equation is 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥−𝑐𝑜𝑠𝑒𝑐(𝑦/𝑥) Let F(x, y) = 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥−𝑐𝑜𝑠𝑒𝑐(𝑦/𝑥) Finding F(𝜆x, 𝜆y) F(𝜆x, 𝜆y) = ("𝜆" 𝑦)/("𝜆" 𝑥)−𝑐𝑜𝑠𝑒𝑐(("𝜆" 𝑦)/("𝜆" 𝑥)) = 𝑦/𝑥 − cosec (𝑦/𝑥) = 𝜆° F(x, y) ∴ F(x, y) is 𝑎 homogenous function of degree zero F(𝜆x, 𝜆y) = 𝜆° F(x , y) Putting y = vx Diff w.r.t. x 𝑑𝑦/𝑑𝑥 = x 𝑑𝑣/𝑑𝑥 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥−𝑐𝑜𝑠𝑒𝑐(𝑦/𝑥) v + x 𝑑𝑣/𝑑𝑥 = 𝑣𝑥/𝑥 − cosec (𝑣𝑥/𝑥) v + x 𝑑𝑣/𝑑𝑥 = v − cosec v (𝑥 𝑑𝑣)/𝑑𝑥 = v − cosec v − v (𝑥 𝑑𝑣)/𝑑𝑥 = − cosec v (−𝑑𝑣)/(𝑐𝑜𝑠𝑒𝑐 𝑣) = 𝑑𝑥/𝑥 Integrating both sides ∫1▒〖(−𝑑𝑣)/(𝑐𝑜𝑠𝑒𝑐 𝑣) " = " ∫1▒𝑑𝑥/𝑥〗 ∫1▒〖−sin⁡𝑣 𝑑𝑣〗=log⁡〖|𝑥|+𝑐〗 Put value of v = 𝑦/𝑥 cos 𝑦/𝑥 = log |𝑥| + C Putting x = 1 & y = 0 cos 0/1 = log 1 + C 1 = 0 + C C = 1 Putting value in (2) cos 𝑦/2 = log |𝑥| + 1 cos 𝑦/𝑥 = log |𝑥| + log e cos 𝒚/𝒙 = log |𝒆𝒙|

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.