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Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 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 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.