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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 17 Find the particular solution of the differential equation ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฆ cotโกใ€–๐‘ฅ=2๐‘ฅ+๐‘ฅ^2 cotโก๐‘ฅ(๐‘ฅโ‰ 0) ใ€— given that ๐‘ฆ=0 ๐‘คโ„Ž๐‘’๐‘› ๐‘ฅ=๐œ‹/2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฆ cotโกใ€–๐‘ฅ=2๐‘ฅ+๐‘ฅ^2 cotโก๐‘ฅ ใ€— Differential equation is of the form ๐’…๐’š/๐’…๐’™+๐‘ท๐’š=๐‘ธ where P = cot x & Q = 2x + x2 cot x IF = ๐‘’^โˆซ1โ–’ใ€–๐‘ ๐‘‘๐‘ฅใ€— IF = ๐’†^โˆซ1โ–’ใ€–๐œ๐จ๐ญโก๐’™ ๐’…๐’™ใ€— IF = ใ€–๐‘’^logโกsinโก๐‘ฅ ใ€—^" " IF = sin x Solution is y (IF) =โˆซ1โ–’ใ€–(๐‘„ร—๐ผ๐น) ๐‘‘๐‘ฅ+๐‘ใ€— y sin x = โˆซ1โ–’ใ€–๐ฌ๐ข๐งโก๐’™ร—(๐Ÿ๐’™+๐’™^(๐Ÿ ) ๐œ๐จ๐ญโก๐’™ ) ๐’…๐’™ใ€— + C y sin x = โˆซ1โ–’ใ€–(2๐‘ฅ sinโก๐‘ฅ+๐‘ฅ^(2 ) sinโกใ€–๐‘ฅ cotโก๐‘ฅ ใ€— ) ๐‘‘๐‘ฅใ€— + C y sinโก๐‘ฅ = โˆซ1โ–’ใ€–2๐‘ฅ sinโก๐‘ฅ ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’ใ€–๐‘ฅ^2 sinโก๐‘ฅ cotโก๐‘ฅ ๐‘‘๐‘ฅ+ใ€— ๐ถ y sinโก๐‘ฅ = 2โˆซ1โ–’ใ€–๐ฌ๐ข๐งโก๐’™ (๐’™) ๐’…๐’™ใ€—+โˆซ1โ–’ใ€–๐‘ฅ^2 sinโก๐‘ฅ cotโก๐‘ฅ ๐‘‘๐‘ฅ+ใ€— ๐ถ Integrating by parts with โˆซ1โ–’ใ€–๐‘“(๐‘ฅ) ๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ=๐‘“(๐‘ฅ) โˆซ1โ–’ใ€–๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ โˆ’โˆซ1โ–’ใ€–[๐‘“^โ€ฒ (๐‘ฅ) โˆซ1โ–’ใ€–๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ] ๐‘‘๐‘ฅใ€—ใ€—ใ€—ใ€— Take f (x) = sin x & g (x) = ๐‘ฅ y sinโก๐‘ฅ = 2 [๐ฌ๐ข๐งโก๐’™ โˆซ1โ–’ใ€–๐’™ ๐’…๐’™โˆ’ใ€— โˆซ1โ–’ใ€–[๐’„๐’๐’”โกใ€–๐’™ โˆซ1โ–’ใ€–๐’™ ๐’…๐’™ใ€— ใ€— ] ๐’…๐’™ใ€—] + โˆซ1โ–’ใ€–๐‘ฅ^2 sinโก๐‘ฅ ใ€— cotโก๐‘ฅ dx + C y sinโก๐‘ฅ = 2 [sinโก๐‘ฅ [๐‘ฅ^2/2]โˆ’โˆซ1โ–’ใ€–๐’„๐’๐’”โกใ€–๐’™ ใ€— [๐‘ฅ^2/2]๐’…๐’™ใ€—] + โˆซ1โ–’ใ€–๐‘ฅ^2 sinโก๐‘ฅ ใ€— cotโก๐‘ฅ dx y sin x = x2sin x โˆ’ โˆซ1โ–’๐’™^๐Ÿ cos x dx + โˆซ1โ–’ใ€–๐’™^๐Ÿ ๐’”๐’Š๐’โก๐’™ ใ€— ๐’„๐’๐’•โก๐’™ dx + C y sin x = x2sin x โˆ’ โˆซ1โ–’๐‘ฅ^2 cos x dx + โˆซ1โ–’ใ€–๐‘ฅ^2 sinโก๐‘ฅ ใ€—ร—cosโก๐‘ฅ/sinโก๐‘ฅ dx + C y sin x = x2sin x โˆ’ โˆซ1โ–’๐‘ฅ^2 cos x dx + โˆซ1โ–’ใ€–๐‘ฅ^2 cosโก๐‘ฅ ใ€— dx + C y sin x = x2 sin x + C Given that y = 0 when x = ๐œ‹/2 Putting ๐’™=๐…/๐Ÿ and y = 0 in (1) (0) sin ๐œ‹/2=(๐œ‹/2)^2 sinโกใ€–(๐œ‹/2)+Cใ€— 0 =๐œ‹^2/4 (1)+C ใ€–โˆ’๐…ใ€—^๐Ÿ/๐Ÿ’=๐‚ Putting value in C in (1) y sin x = x2 sin x + C y sin x = ๐’™^๐Ÿ ๐’”๐’Š๐’โกใ€–๐’™ โˆ’ใ€— ๐…^๐Ÿ/๐Ÿ’ Dividing both sides by sin x (๐‘ฆ sinโก๐‘ฅ)/sinโก๐‘ฅ =(๐‘ฅ^2 sinโก๐‘ฅ)/sinโก๐‘ฅ โˆ’๐œ‹^2/(4 sinโก๐‘ฅ ) ๐’š=๐’™^๐Ÿโˆ’๐…^๐Ÿ/ใ€–๐Ÿ’ ๐ฌ๐ข๐งใ€—โก๐’™ where sinโกใ€–๐‘ฅโ‰ 0ใ€— y sinโก๐‘ฅ = 2 [๐ฌ๐ข๐งโก๐’™ โˆซ1โ–’ใ€–๐’™ ๐’…๐’™โˆ’ใ€— โˆซ1โ–’ใ€–[๐’„๐’๐’”โกใ€–๐’™ โˆซ1โ–’ใ€–๐’™ ๐’…๐’™ใ€— ใ€— ] ๐’…๐’™ใ€—] + โˆซ1โ–’ใ€–๐‘ฅ^2 sinโก๐‘ฅ ใ€— cotโก๐‘ฅ dx + C

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.