If (D$^2$ + 16)y = cos 4x
Here the auxiliary equation is m$^2$ + 16 = 0
$\Rightarrow $ m = ± 4
$\therefore $ Complementary function
= (A cos 4x + B sin 4x)
& Particular Integral (P.I.)
$= \frac{1}{D^2 + 16}. \, cos4x$
But $\frac{1}{D^2 \, + \, a^2} cos ax = \frac{x}{2a}sin ax$
$\therefore \, \, P.I.=\frac{x}{2 \times 4}. \, sin 4x = \frac{x}{8}sin4x$
$\therefore $ Solution y = Complementary function
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, $+ Particular Integral
$\Rightarrow $ y = A cos 4x + B sin 4x + $\frac{x}{8}$sin 4 x