Given $y= e^{3x}[C \cos 2x+D \sin 2x]$-----(1)

Step 1:

$\large\frac{dy}{dx}$$=3e^{3x}[C \cos 2x+D \sin 2x]+e^{3x}[-2C \sin 2+ 2D \cos 2x]$

$\therefore \large\frac{dy}{dx}$$=3y+2e^{3x}[-C \sin 2x+ D \cos 2x]$-----(ii)

Step 2:

$\large\frac{d^2y}{dx^2}$$=3 \large\frac{dy}{dx} \normalsize +6e^{3x}[-C \sin 2x+D \cos 2x]+4e^{3x}[-2C \cos 2x- 2D \sin 2x]$

Step 3:

$\large\frac{d^2y}{dx^2}$$=3\large \frac{dy}{dx}$$+6e^{3x}[-C \sin 2x+D \cos 2x]-4y$-----(iii)

Step 4:

From (ii) $2e^{3x}(-c \sin 2x +D \cos 2x)= \large\frac{dy}{dx}$$-3y$

Step 5:

Substitute in (iii)

$\large\frac{d^2y}{dx^2}$$=3\large \frac{dy}{dx}$$+3\bigg[\large\frac{dy}{dx}$$-3y\bigg]-4y$

$\large\frac{d^2y}{dx^2}$$-6\large \frac{dy}{dx}$$+13y=0$ is the required D.E