# Ex 9.4, 14 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 9.4, 14 For 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 π π/π π = x π π/π π + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯ = π¦/π₯βπππ ππ(π¦/π₯) v + x π π/π π = ππ/π β cosec (ππ/π) v + x ππ£/ππ₯ = v β cosec v (π₯ ππ£)/ππ₯ = v β cosec v β v (π₯ ππ£)/ππ₯ = β cosec v (βπ π)/(πππππ π) = π π/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|