Question

Solve `(D^2+2)y=e^xcosx+x^2e^(3x)`

Answer 1

`(D^2+2)y=e^xcosx+x^2e^(3x)`
For complementary solution,

𝒇(𝑫)=𝟎

`therefore(D^2+2)=0`
Roots are : D = √𝟐𝒊 ,−√𝟐𝒊

Roots of given diff. eqn are complex.
The complementary solution of given diff. eqn is given by,

`therefore y_c=c_1cossqrt(2x)+c_2sinsqrt(2x)`
For particular integral ,

`y_p=1/(f(D))x=1/(D^2+1)e^xcosx+1/(D^2+1)x^2e^(3x)`

`=e^x1/((D+1)^2+1)cosx+1/(D^2+1)x^2e^(3x)`

`=e^x1/(D^2+2D+3)cosx+e^(3x)1/((D+3)^2+2)x^2`

`=e^x1/2(D-1)/(D^2-1)cosx+e^(3x)1/(D^2+6D+11)x^2`

`=e^x1/4(sinxcosx)+e^(3x)/11[1+(6D+D^2)/11]^(-1)x^2`

`=e^x1/4(sinxcosx)+e^(3x)/11[1+(6D+D^2)/11+(36D^2)/121+..]x^2`

`therefore y_p=e^x1/4(sinx+cosx)+e^(3x)/11[x^2-(12x)/11+50/121]`

`y_g=y_c+y_p=c_1cossqrt(2x)+c_2sinsqrt(2x)+e^x1/4(sinx+cosx)+e^(3x)/11[x^2-(12x)/11+50/121]`