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

Ex 9.5

Ex 9.5, 1
Important

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

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 You are here

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 May 29, 2023 by Teachoo

Ex 9.5, 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