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Ex 9.5

Ex 9.5, 1
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Ex 9.5, 8 Important You are here

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Ex 9.5, 18 (MCQ)

Ex 9.5, 19 (MCQ) Important

Last updated at Aug. 11, 2023 by Teachoo

Ex 9.5, 8 For each of the differential equation given in Exercises 1 to 12, find the general solution : (1+๐ฅ^2 )๐๐ฆ+2๐ฅ๐ฆ ๐๐ฅ=cotโกใ๐ฅ ๐๐ฅ(๐ฅโ 0)ใ Given equation (1 + x2)dy + 2xy dx = cot x dx Dividing both sides by dx (1 + x2)๐๐ฆ/๐๐ฅ + 2xy ๐๐ฅ/๐๐ฅ = cot x ๐๐ฅ/๐๐ฅ (1 + x2)๐๐ฆ/๐๐ฅ + 2xy = cot x Dividing both sides by (1 + x2) ๐ ๐/๐ ๐ + ๐๐/((๐ + ๐๐)) y = ๐๐๐โก๐/((๐ + ๐๐)) Comparing (1) with ๐๐ฆ/๐๐ฅ + Py = Q where P = ๐๐/((๐ + ๐^๐)) & Q = ๐๐๐โก๐/((๐ + ๐^๐)) Finding Integrating factor, I.F I.F. = ๐^โซ1โใ๐ ๐๐ฅใ I.F. = ๐^โซ1โใ๐๐/((๐ + ๐^๐ ) ) ๐ ๐ ใ Let t = 1 + x2 dt = 2x dx I.F. = e^โซ1โใ๐๐ก/๐ก ใ = elog |t| = t Putting back t = (1 + x2) = (1 + x2) Solution of the equation is y ร I.F = โซ1โใ๐ร๐ผ.๐น.๐๐ฅ+๐ถใ Putting values, y.(1 + x2) = โซ1โ๐๐๐โก๐/((๐ + ๐๐)) ร (๐+๐๐).dx + c y.(1 + x2) = โซ1โใcotโก๐ฅ ๐๐ฅใ+๐ถ y (1 + x2) = log |sinโก๐ฅ | + C Dividing by (1 + x2) y = (1 + x2)โ1 log |๐ฌ๐ข๐งโก๐ |+๐ช(๐+"x2" )^(โ๐) is the general solution of the given equation Note: This answer does not match with the answer of the book. If we have made any mistake, please comment