Given differential eqn of,
$y'' + 16y = \cos(4x) - 8e^{3x}$
Initial conditions $y=0$ and $dy/dx = 0$ when $x = 0$
So therefor,
$r^2 + 16 = 0$
where $r$ will be $4i$
so, $Y = C\cos(4x) + D\sin(4x)$
The RHS is the part I have problem with...
This is what I've done so far
$Y_p = Ae^{3x} + B\cos(4x)$
$Y_p' = 3Ae^{3x} - 4B\sin(4x)$
$Y_p'' = 9Ae^{3x} - 16B\cos(4x)$
Sub into eqn...
$9Ae^{3x} - 16B\cos(4x) + 16(Ae^{3x} + B\cos(4x)) = \cos(4x) - 8e^{3x}$
$9Ae^{3x} - 16B\cos(4x) + 16Ae^{3x} + 16B\cos(4x) = \cos(4x) - 8e^{3x}$
$25Ae^{3x} = \cos(4x) - 8e^{3x}$
Taking coeff of $e^{3x}$,
$25A = -8$
$A = -8/25$
From here on Im stuck, not to sure how to find $B$.