Question

Solve `(D^3+D^2+D+1)y=sin^2x`

Answer 1

`(D^3+D^2+D+1)y=sin^2x`

For complementary solution ,
𝒇(𝑫)=𝟎
`therefore (D^3+D^2+D+1)=0`
Roots are : D = -1 , +i , -i
The complementary solution of given diff eqn is ,
`y_c=c_1cosx+c_2sinx+c_3e^(-x)`
For complementary solution ,

`y_p=1/(f(D))x=1/(D^3+D^2+D+1)sin^2x=1/(2(D^3+D^2+D+1))(1-cos2x)`

`=1/(2(D^3+D^2+D+1))e^(0x)-1/(2(D^3+D^2+D+1))cos2x`

`=1/2-1/2xx1/(-D-4+d+1)cos2x`

`=1/2+1/6 1/(D+1)cos2x`

`=1/2+1/6 1/(D+1) (D-1)/(D-1)cos2x`

`=1/2+1/6 (D-1)/(D^2-1)cos2x`

`=1/2+1/6(-2sin2x-cos2x)/(-4-1)cos2x`

`y_p=1/2+1/30(2sin2x-cos2x)`

The general solution of given diff. eqn is given by,

`y_g=y_c+y_p=c_1cosx+c_2sinx+c_3e^(-x)+1/2+1/30(2sin2x+cos2x)`