$y=e^{x}(a \cos x+b \sin x)$
$y=e^{x}(a \cos x+b \sin x)$ ...(1)
Differentiating both sides with respect to x, we get:
$y^{\prime}=e^{x}(a \cos x+b \sin x)+e^{x}(-a \sin x+b \cos x)$
$\Rightarrow y^{\prime}=e^{x}[(a+b) \cos x-(a-b) \sin x]$ ...(2)
Again, differentiating with respect to x, we get:
$y^{\prime \prime}=e^{x}[(a+b) \cos x-(a-b) \sin x]+e^{x}[-(a+b) \sin x-(a-b) \cos x]$
$y^{\prime \prime}=e^{x}[2 b \cos x-2 a \sin x]$
$y^{\prime \prime}=2 e^{x}(b \cos x-a \sin x)$
$\Rightarrow \frac{y^{\prime \prime}}{2}=e^{x}(b \cos x-a \sin x)$ ...(3)
Adding equations (1) and (3), we get:
$y+\frac{y^{\prime \prime}}{2}=e^{x}[(a+b) \cos x-(a-b) \sin x]$
$\Rightarrow y+\frac{y^{\prime \prime}}{2}=y^{\prime}$
$\Rightarrow 2 y+y^{\prime \prime}=2 y^{\prime}$
$\Rightarrow y^{\prime \prime}-2 y^{\prime}+2 y=0$
This is the required differential equation of the given curve.