# Ex 9.6, 14 - Chapter 9 Class 12 Differential Equations

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

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

Chapter 9 Class 12 Differential Equations

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.