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Ex 9.5
Ex 9.5, 2
Ex 9.5, 3 Important
Ex 9.5, 4
Ex 9.5, 5 Important
Ex 9.5, 6
Ex 9.5, 7 Important
Ex 9.5, 8 Important You are here
Ex 9.5, 9
Ex 9.5, 10
Ex 9.5, 11
Ex 9.5, 12 Important
Ex 9.5, 13
Ex 9.5, 14 Important
Ex 9.5, 15
Ex 9.5, 16 Important
Ex 9.5, 17 Important
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