Ex 9.6, 15 - Chapter 9 Class 12 Differential Equations

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