Answer
See below
Work Step by Step
We are given $L_1=D+x\\
L_2=D+(2x-1)$
1. Determine Ly: $$L_1y = y' + xy\\
L_2y=y'+(2x-1)y$$
2. Substitute y = $2e^{3x}$:
$$L_1(L_2y)=L_1[y'+(2x-1)y]\\
=y''+2y+(2x-1)y'+[y'+(2x-1)y]x\\
=y''+(3x-1)y'+(2x^2-x+2)y$$
$$L_2(L_1y)=L_2(y'+xy)\\
=y''+y+xy'+(2x-1)(y'+xy)\\
=y''+(3x-1)y'-(2x^2-x+1)y$$
Then:
$$L_1L_2=D^2+D(3x-1)+(2x^2-x+2)\\
L_2L_1=D^2+D(3x-1)+(2x^2-x+1)$$
Hence, $L_1L_2 \ne L_2L_1$
