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

y_c = A e^2x + B e^-x 

Particular solution is of the form 

y_p = C cos2x + D sin2x

 ( y_p ) = - 2 C sin2x + 2 D cos2x  ..................................(2)

d²/dx² ( y_p ) = - 4 C cos2x - 4 D sin2x = -4 y_p ..........................(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 = y_c + y_p = A e^2x + B e^-x  - (1/16) cos2x - (3/16 ) sin2x