Answer to Question #271617 in Differential Equations for lucifer

Question

((D-3D'-2)^3)z=6(e^2x)sin(3x+y)

Answer 1

Complementary Function:

solution for $((D-3D'-2)^3)z=0$ :

$C.F.=f(-3x-y)+xg(-3x-y)+x^2h(-3x-y)$

Particular Integral:

$P.I.=\frac{1}{(D-3D'-2)^3}6e^{2x}sin(3x+y)=6e^{2x}\frac{1}{(D+2-3D'-2)^3}sin(3x+y)=$

$=6e^{2x}x^2\frac{1}{D-3D'}sin(3x+y)=6e^{2x}x^3\frac{1}{3}sin(3x+y)=2e^{2x}x^3sin(3x+y)$

$z=C.F.+P.I.$

$z=f(-3x-y)+xg(-3x-y)+x^2h(-3x-y)+2e^{2x}x^3sin(3x+y)$