JEE Class main Answered
( D2 - D - 2 ) y = sin2x .....................(1)
where D = d/dx and D2 = d2/dx2
characterstic equation for above homogeneous part of differential equation is
r2 - r -2 = 0
Above characterstic equation has roots r = 2 and r = -1
Then complimentary solution is
yc = A e2x + B e-x
Particular solution is of the form
yp = C cos2x + D sin2x
( yp ) = - 2 C sin2x + 2 D cos2x ..................................(2)
d2/dx2 ( yp ) = - 4 C cos2x - 4 D sin2x = -4 yp ..........................(3)
If we substitute eqn.(2) and (3) in eqn.(1) , then we get
2 C sin2x + 2 D cos2c -6C cos2x - 6 D sin2x = sin2x
By equating coeffieints , we get the following equations from above expression.
2C-6D = 1
-6C+2D = 0
By solving above equations , we get
C = -1/16 and D = -3/16
General solution y is
y = yc + yp = A e2x + B e-x - (1/16) cos2x - (3/16 ) sin2x