# Ex 9.6, 15 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.6, 15 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : / 3 cot =sin 2 ; =2 when = /2 / 3 cot =sin 2 / + ( 3 cot x) y = (sin 2x) Comparing with / + Py = Q P = 3 cot x & Q = sin 2x Finding Integrating factor (IF) IF = e^ 1 = e^ 1 3 cot = e^( 3 1 ) = e^(3 log |sin | ) = e^log |sin |^( 3) = e^log 1/|sin^3 | = e^log | ^3 |^3 = ^3 I.F = ^3 Solution of differential equation y IF = 1 . Putting values. y cosec3 x = 1 sin 2 . ^3 y cosec3x = 1 (2 sin cos )/sin^3 dx y cosec3x = 1 (2 cos )/sin^2 dx Let I = 2 1 cos /sin^2 Put t = sin x Diff w.r.t x / = cos x dx = /cos I = 2 1 cos /( ^2 ) /( ) I = 2 1 /( ^2 ) = 2 /( 1 ) ^( 1)+ = 2/ + Put value t = sin x = ( 2)/sin + = 2 + Put value of I in (2) y cosec3 x = 2 cosec x + c y = ( 2 )/( ^2 ) + /( ^3 ) y = 2 sin2 x + C sin3 x Putting x = /2 , y = 2 in (3) 2 = 2 Sin2 /2 + C Sin3 /2 2 = 2 (1)2 + C(1)3 2 = 2 + C C = 2 + 2 C = 4 Put value of C in (3) y = 2 sin2 x + C sin3 x y = 2 sin2 x + 4 sin3 x y = 4 sin3 x 2 sin2 x

Chapter 9 Class 12 Differential Equations

Serial order wise

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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.