Answer
$y(x)=C_1e^{x}\cos x+C_2xe^{x}\cos x+C_3x^2e^{x}\cos x+C_4e^{x}\sin x+C_5xe^{x}\sin x+C_6x^2e^x\sin x$
Work Step by Step
Given $(D^2-2D+2)^3y=0$
Solve the auxiliary equation for the differential equation.
$(r^2-2r+2)^3=0$
Roots are: $r_1=1-i$ as a multiplicity of 3 and $r_2=1+i$ as a multiplicity of 3.
This implies that there are two independent solutions to the differential equation
$y_1(x)=e^{x}\cos x$
$y_2=xe^{x}\cos x$ and
$y_3(x)=x^2e^{x}\cos x\\
y_4(x)=e^{x}\sin x\\
y_5(x)=xe^{x}\sin x \\
y_6(x)=x^2e^x\sin x$
Therefore, the general equation is equal to $y(x)=C_1e^{x}\cos x+C_2xe^{x}\cos x+C_3x^2e^{x}\cos x+C_4e^{x}\sin x+C_5xe^{x}\sin x+C_6x^2e^x\sin x$