# Ex 9.5, 14 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 14For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition :ππ¦/ππ₯βπ¦/π₯+πππ ππ(π¦/π₯)=0;π¦=0 When π₯=1 Differential equation is ππ¦/ππ₯ = π¦/π₯βπππ ππ(π¦/π₯) Let F(x, y) = ππ¦/ππ₯ = π¦/π₯βπππ ππ(π¦/π₯) Finding F(πx, πy) F(πx, πy) = ("π" π¦)/("π" π₯)βπππ ππ(("π" π¦)/("π" π₯)) = π¦/π₯ β cosec (π¦/π₯) = πΒ° F(x, y) β΄ F(x, y) is π homogenous function of degree zero F(πx, πy) = πΒ° F(x , y) Putting y = vx Diff w.r.t. x ππ¦/ππ₯ = x ππ£/ππ₯ + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯ = π¦/π₯βπππ ππ(π¦/π₯) v + x ππ£/ππ₯ = π£π₯/π₯ β cosec (π£π₯/π₯) v + x ππ£/ππ₯ = v β cosec v (π₯ ππ£)/ππ₯ = v β cosec v β v (π₯ ππ£)/ππ₯ = β cosec v (βππ£)/(πππ ππ π£) = ππ₯/π₯ Integrating both sides β«1βγ(βππ£)/(πππ ππ π£) " = " β«1βππ₯/π₯γ β«1βγβsinβ‘π£ ππ£γ=logβ‘γ|π₯|+πγ Put value of v = π¦/π₯ cos π¦/π₯ = log |π₯| + C Putting x = 1 & y = 0 cos 0/1 = log 1 + C 1 = 0 + C C = 1 Putting value in (2) cos π¦/2 = log |π₯| + 1 cos π¦/π₯ = log |π₯| + log e cos π/π = log |ππ|

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.