Find the differential equation for the family of curves given by $y = e^{x}(a \cos x + b \sin x)$ by eliminating the arbitrary constants $a$ and $b$.

(D) Given the equation: $y = e^{x}(a \cos x + b \sin x)$ ............$(1)$
Differentiating both sides with respect to $x$ using the product rule:
$y' = e^{x}(a \cos x + b \sin x) + e^{x}(-a \sin x + b \cos x)$
$y' = y + e^{x}(-a \sin x + b \cos x)$ ............$(2)$
Differentiating again with respect to $x$:
$y'' = y' + [e^{x}(-a \sin x + b \cos x) + e^{x}(-a \cos x - b \sin x)]$
$y'' = y' + (y' - y) + e^{x}(-a \cos x - b \sin x)$
$y'' = 2y' - y - e^{x}(a \cos x + b \sin x)$
Since $e^{x}(a \cos x + b \sin x) = y$,we substitute this back:
$y'' = 2y' - y - y$
$y'' = 2y' - 2y$
$y'' - 2y' + 2y = 0$
This is the required differential equation.