# Ex 9.6, 14 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Aug. 20, 2021 by Teachoo

Ex 9.6

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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

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Ex 9.6, 14 Important You are here

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Ex 9.6, 18 (MCQ)

Ex 9.6, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

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