Step 1:

$y=(A+Bx)e^{ 3x}$-----(i)

$\large\frac{dy}{dx}$$=3e^{3x} (A+Bx)+Be^{3x}$-----(ii)

=>$\large\frac{dy}{dx}$$=3y+Be^{3x} $-----(iii)

=>$Be^{3x}=\large\frac{dy}{dx}$$-3y$

Step 2:

Now from (iii)

$\large\frac{d^2y}{dx^2}$$=3 \large\frac{dy}{dx}$$+3 Be^{3x}$

$\therefore \large\frac{d^2y}{dx^2}$$= 3 \large\frac{dy}{dx} $$+ 3 \bigg[\large\frac{dy}{dx}$$-3y\bigg]$

$\therefore \large\frac{d^2y}{dx^2} $$-6 \large\frac{dy}{dx} $$+9y=0$ is the required D.E