Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.4, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : [๐ฅ sin^2โกใ(๐ฆ/๐ฅ)โ๐ฆใ ]๐๐ฅ+๐ฅ๐๐ฆ=0;๐ฆ=๐/4 When ๐ฅ=1 Ex 9.4, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : [๐ฅ sin^2โกใ(๐ฆ/๐ฅ)โ๐ฆใ ]๐๐ฅ+๐ฅ๐๐ฆ=0;๐ฆ=๐/4 When ๐ฅ=1 Differential equation can be written as [๐ฅ sin^2โกใ(๐ฆ/๐ฅ)โ๐ฆใ ]๐๐ฅ+๐ฅ๐๐ฆ=0 x dy = โ [๐ฅ sin^2โกใ(๐ฆ/๐ฅ)โ๐ฆใ ]dx ๐๐/๐๐ = โ[๐/๐ ใ๐ฌ๐ข๐งใ^๐โกใ๐/๐ โ๐/๐ใ ] Let F(x, y) = ๐๐ฆ/๐๐ฅ =โ[๐ฅ/๐ฅ sin^2โกใ๐ฆ/๐ฅ โ๐ฆ/๐ฅใ ] Finding F(๐x, ๐y) F(๐x, ๐y) = โ[sin^2โกใ๐๐ฆ/๐๐ฅโ๐๐ฆ/๐๐ฅใ ] = โ[sin^2โกใ๐ฆ/๐ฅโ๐ฆ/๐ฅใ ] = ๐ยฐ F (x, y) โด F(x, y) is ๐ homogenous function of degree zero Putting y = vx Diff w.r.t. x ๐๐/๐๐ = x ๐๐/๐๐ + v Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐๐ฆ/๐๐ฅ = โ[๐ฅ/๐ฅ sin^2โกใ๐ฆ/๐ฅ โ๐ฆ/๐ฅใ ] v + (๐ ๐๐)/๐๐ = โ[ใ๐ฌ๐ข๐งใ^๐ ๐๐/๐โ๐๐/๐] v + (๐ฅ ๐๐ฃ)/๐๐ฅ = โ[sin^2 ๐ฃโ๐ฃ] (๐ฅ ๐๐ฃ)/๐๐ฅ = v โ sin2 v โ v (๐ฅ ๐๐ฃ)/๐๐ฅ = โ sin2 v ๐๐/ใ๐๐๐ใ^๐โก๐ = โ๐๐/๐ Integrating both sides โซ1โ๐๐ฃ/(sin^2 ๐ฃ) = โโซ1โ๐๐ฅ/๐ฅ โซ1โใ๐๐๐ ๐๐^2 ๐ฃ ๐๐ฃใ = โโซ1โ๐๐ฅ/๐ฅ โ cot v = โ log |๐ฅ|+๐ log |๐| โ cot v = C Putting value of v = ๐ฆ/๐ฅ log |๐ฅ| โ cot (๐ฆ/๐ฅ) = C Putting x = 1 & y = ๐/๐ log 1 โ cot (๐/4) = C 0 โ 1 = C C = โ1 Putting value of C in (2) log |๐ฅ| โ cot ๐ฆ/๐ฅ = โ1 cot ๐ฆ/๐ฅ โ log |๐ฅ| = 1 cot ๐ฆ/๐ฅ โ log |๐ฅ| = log e cot ๐ฆ/๐ฅ = log |๐ฅ| + log e cot ๐/๐ = log |๐๐|

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.