


Ex 9.6
Ex 9.6, 2
Ex 9.6, 3 Important
Ex 9.6, 4
Ex 9.6, 5 Important
Ex 9.6, 6
Ex 9.6, 7 Important
Ex 9.6, 8 Important
Ex 9.6, 9
Ex 9.6, 10 Deleted for CBSE Board 2022 Exams
Ex 9.6, 11 Deleted for CBSE Board 2022 Exams
Ex 9.6, 12 Important Deleted for CBSE Board 2022 Exams
Ex 9.6, 13
Ex 9.6, 14 Important You are here
Ex 9.6, 15
Ex 9.6, 16 Important
Ex 9.6, 17 Important
Ex 9.6, 18 (MCQ)
Ex 9.6, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams
Last updated at Aug. 20, 2021 by Teachoo
Ex 9.6, 14 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : 1+ 2 +2 = 1 1+ 2 ; =0 when =1 (1 + x2) + 2xy = 1 1 + 2 Divide both sides by (1+ 2) + 2 1 + 2 = 1 1 + 2 .(1 + 2) + 2 1 + 2 y = 1 1 + 2 Comparing with + Py = Q P = 2 1 + 2 & Q = 1 1 + 2 2 Find Integrating factor IF = IF = 2 1 + 2 Let 1+ 2 = t Diff . w.r.t. x 2x = t dx = 2 IF = e 2 2 IF = e IF = e IF = t IF = 1 + x2 Step 4 : Solution of the deferential equation y I.F = . Putting values y (1 + x2) = 1 1 + 2 2 (1 + x2).dx y (1 + x2) = 1 1 + 2 dx y (1 + x2) = tan 1 + Given that y = 0 when x = 1 Putting y = 0 and x = 1 in (1) y (1 + x2) = tan 1 x + c 0(1 + 12) = tan 1 (1)+ c 0 = 4 + C C = 4 Putting value of C in (2) y (1 + x2) = tan-1 x + c y (1 + x2) = tan-1 x