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Solving Linear differential equations - Equation given
Ex 9.5, 19 (MCQ) Important
Misc 14 (MCQ) Important
Ex 9.5, 2
Ex 9.5, 10
Ex 9.5, 3 Important
Ex 9.5, 4
Misc 15 (MCQ)
Ex 9.5, 13
Ex 9.5, 8 Important
Misc 10 Important
Misc 11
Ex 9.5, 14 Important
Ex 9.5, 6
Ex 9.5, 5 Important
Ex 9.5, 9
Ex 9.5, 7 Important
Ex 9.5, 15
Example 14
Ex 9.5, 1 Important
Ex 9.5, 12 Important
Ex 9.5, 11
Example 16
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Example 22 Important
Solving Linear differential equations - Equation given
Last updated at Aug. 14, 2023 by Teachoo
Example 17 Find the particular solution of the differential equation ๐๐ฆ/๐๐ฅ+๐ฆ cotโกใ๐ฅ=2๐ฅ+๐ฅ^2 cotโก๐ฅ(๐ฅโ 0) ใ given that ๐ฆ=0 ๐คโ๐๐ ๐ฅ=๐/2 ๐๐ฆ/๐๐ฅ+๐ฆ cotโกใ๐ฅ=2๐ฅ+๐ฅ^2 cotโก๐ฅ ใ Differential equation is of the form ๐ ๐/๐ ๐+๐ท๐=๐ธ where P = cot x & Q = 2x + x2 cot x IF = ๐^โซ1โใ๐ ๐๐ฅใ IF = ๐^โซ1โใ๐๐จ๐ญโก๐ ๐ ๐ใ IF = ใ๐^logโกsinโก๐ฅ ใ^" " IF = sin x Solution is y (IF) =โซ1โใ(๐ร๐ผ๐น) ๐๐ฅ+๐ใ y sin x = โซ1โใ๐ฌ๐ข๐งโก๐ร(๐๐+๐^(๐ ) ๐๐จ๐ญโก๐ ) ๐ ๐ใ + C y sin x = โซ1โใ(2๐ฅ sinโก๐ฅ+๐ฅ^(2 ) sinโกใ๐ฅ cotโก๐ฅ ใ ) ๐๐ฅใ + C y sinโก๐ฅ = โซ1โใ2๐ฅ sinโก๐ฅ ๐๐ฅ+ใ โซ1โใ๐ฅ^2 sinโก๐ฅ cotโก๐ฅ ๐๐ฅ+ใ ๐ถ y sinโก๐ฅ = 2โซ1โใ๐ฌ๐ข๐งโก๐ (๐) ๐ ๐ใ+โซ1โใ๐ฅ^2 sinโก๐ฅ cotโก๐ฅ ๐๐ฅ+ใ ๐ถ Integrating by parts with โซ1โใ๐(๐ฅ) ๐(๐ฅ) ๐๐ฅ=๐(๐ฅ) โซ1โใ๐(๐ฅ) ๐๐ฅ โโซ1โใ[๐^โฒ (๐ฅ) โซ1โใ๐(๐ฅ) ๐๐ฅ] ๐๐ฅใใใใ Take f (x) = sin x & g (x) = ๐ฅ y sinโก๐ฅ = 2 [๐ฌ๐ข๐งโก๐ โซ1โใ๐ ๐ ๐โใ โซ1โใ[๐๐๐โกใ๐ โซ1โใ๐ ๐ ๐ใ ใ ] ๐ ๐ใ] + โซ1โใ๐ฅ^2 sinโก๐ฅ ใ cotโก๐ฅ dx + C y sinโก๐ฅ = 2 [sinโก๐ฅ [๐ฅ^2/2]โโซ1โใ๐๐๐โกใ๐ ใ [๐ฅ^2/2]๐ ๐ใ] + โซ1โใ๐ฅ^2 sinโก๐ฅ ใ cotโก๐ฅ dx y sin x = x2sin x โ โซ1โ๐^๐ cos x dx + โซ1โใ๐^๐ ๐๐๐โก๐ ใ ๐๐๐โก๐ dx + C y sin x = x2sin x โ โซ1โ๐ฅ^2 cos x dx + โซ1โใ๐ฅ^2 sinโก๐ฅ ใรcosโก๐ฅ/sinโก๐ฅ dx + C y sin x = x2sin x โ โซ1โ๐ฅ^2 cos x dx + โซ1โใ๐ฅ^2 cosโก๐ฅ ใ dx + C y sin x = x2 sin x + C Given that y = 0 when x = ๐/2 Putting ๐=๐ /๐ and y = 0 in (1) (0) sin ๐/2=(๐/2)^2 sinโกใ(๐/2)+Cใ 0 =๐^2/4 (1)+C ใโ๐ ใ^๐/๐=๐ Putting value in C in (1) y sin x = x2 sin x + C y sin x = ๐^๐ ๐๐๐โกใ๐ โใ ๐ ^๐/๐ Dividing both sides by sin x (๐ฆ sinโก๐ฅ)/sinโก๐ฅ =(๐ฅ^2 sinโก๐ฅ)/sinโก๐ฅ โ๐^2/(4 sinโก๐ฅ ) ๐=๐^๐โ๐ ^๐/ใ๐ ๐ฌ๐ข๐งใโก๐ where sinโกใ๐ฅโ 0ใ y sinโก๐ฅ = 2 [๐ฌ๐ข๐งโก๐ โซ1โใ๐ ๐ ๐โใ โซ1โใ[๐๐๐โกใ๐ โซ1โใ๐ ๐ ๐ใ ใ ] ๐ ๐ใ] + โซ1โใ๐ฅ^2 sinโก๐ฅ ใ cotโก๐ฅ dx + C