text: # Solve ( D 3 + D 2 + D + 1 ) Y = Sin 2 X - Applied Mathematics 2
  
  Sum
  
  Solve (D^3+D^2+D+1)y=sin^2x
  
  #### Solution
  
  (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)
  
  Concept: Linear Differential Equation with Constant Coefficient‐ Complementary Function
  Is there an error in this question or solution?