Slide7.JPG

Slide8.JPG
Slide9.JPG
Slide10.JPG

Go Ad-free

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)โˆ’๐‘ฆใ€— ]๐‘‘๐‘ฅ+๐‘ฅ๐‘‘๐‘ฆ=0;๐‘ฆ=๐œ‹/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)โˆ’๐‘ฆใ€— ]๐‘‘๐‘ฅ+๐‘ฅ๐‘‘๐‘ฆ=0;๐‘ฆ=๐œ‹/4 When ๐‘ฅ=1 Differential equation can be written as [๐‘ฅ sin^2โกใ€–(๐‘ฆ/๐‘ฅ)โˆ’๐‘ฆใ€— ]๐‘‘๐‘ฅ+๐‘ฅ๐‘‘๐‘ฆ=0 x dy = โˆ’ [๐‘ฅ sin^2โกใ€–(๐‘ฆ/๐‘ฅ)โˆ’๐‘ฆใ€— ]dx ๐’…๐’š/๐’…๐’™ = โˆ’[๐’™/๐’™ ใ€–๐ฌ๐ข๐งใ€—^๐Ÿโกใ€–๐’š/๐’™ โˆ’๐’š/๐’™ใ€— ] Let F(x, y) = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ =โˆ’[๐‘ฅ/๐‘ฅ sin^2โกใ€–๐‘ฆ/๐‘ฅ โˆ’๐‘ฆ/๐‘ฅใ€— ] Finding F(๐€x, ๐€y) F(๐œ†x, ๐œ†y) = โˆ’[sin^2โกใ€–๐œ†๐‘ฆ/๐œ†๐‘ฅโˆ’๐œ†๐‘ฆ/๐œ†๐‘ฅใ€— ] = โˆ’[sin^2โกใ€–๐‘ฆ/๐‘ฅโˆ’๐‘ฆ/๐‘ฅใ€— ] = ๐œ†ยฐ F (x, y) โˆด F(x, y) is ๐‘Ž homogenous function of degree zero Putting y = vx Diff w.r.t. x ๐’…๐’š/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 = โˆ’[๐‘ฅ/๐‘ฅ sin^2โกใ€–๐‘ฆ/๐‘ฅ โˆ’๐‘ฆ/๐‘ฅใ€— ] v + (๐’™ ๐’…๐’—)/๐’…๐’™ = โˆ’[ใ€–๐ฌ๐ข๐งใ€—^๐Ÿ ๐’—๐’™/๐’™โˆ’๐’—๐’™/๐’™] v + (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = โˆ’[sin^2 ๐‘ฃโˆ’๐‘ฃ] (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = v โˆ’ sin2 v โˆ’ v (๐‘ฅ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฅ = โˆ’ sin2 v ๐’…๐’—/ใ€–๐’”๐’Š๐’ใ€—^๐Ÿโก๐’— = โˆ’๐’…๐’™/๐’™ Integrating both sides โˆซ1โ–’๐‘‘๐‘ฃ/(sin^2 ๐‘ฃ) = โˆ’โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ โˆซ1โ–’ใ€–๐‘๐‘œ๐‘ ๐‘’๐‘^2 ๐‘ฃ ๐‘‘๐‘ฃใ€— = โˆ’โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ โˆ’ cot v = โˆ’ log |๐‘ฅ|+๐‘ log |๐’™| โˆ’ cot v = C Putting value of v = ๐‘ฆ/๐‘ฅ log |๐‘ฅ| โˆ’ cot (๐‘ฆ/๐‘ฅ) = C Putting x = 1 & y = ๐…/๐Ÿ’ log 1 โˆ’ cot (๐œ‹/4) = C 0 โˆ’ 1 = C C = โˆ’1 Putting value of C in (2) log |๐‘ฅ| โˆ’ cot ๐‘ฆ/๐‘ฅ = โˆ’1 cot ๐‘ฆ/๐‘ฅ โˆ’ log |๐‘ฅ| = 1 cot ๐‘ฆ/๐‘ฅ โˆ’ log |๐‘ฅ| = log e cot ๐‘ฆ/๐‘ฅ = log |๐‘ฅ| + log e cot ๐’š/๐’™ = log |๐’†๐’™|

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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo