The differential equation of the family of curves \(\mathrm{y}=e^{x}(\mathrm{~A} \cos x+\mathrm{B} \sin x)\), where
\(\mathrm{A}\) and \(\mathrm{B}\) are arbitrary constants is

Given \(y=e^{x}(A \cos x+B \sin x)\)...(1)
\(\frac{d y}{d x} =e^{x}(-A \sin x+B \cos x)+e^{x}(A \cos x+B \sin x) \)
\( \frac{d^{2} y}{d x^{2}} =e^{x}(-A \cos x-B \sin x)+e^{x}(-A \sin x~+\) \(B \cos x)+e^{x}(-A \sin x+B \cos x) \)
\( -e^{x}(A \cos x+B \sin x) \)
\( =2 e^{x}(-A \sin x+B \cos x)\)
Thus we get
\(
\begin{aligned} y &=e^{x}(A \cos x+B \sin x) \\ \frac{d y}{d x} &=e^{x}(-A \sin x+A \cos x+B \cos x+B \sin x) \end{aligned}
\)
\( \frac{d^{2} y}{d x^{2}}=e^{x}(-2 A \sin x+2 B \cos x) \)
\( \therefore \frac{d^{2} y}{d x^{2}}-\frac{2 d y}{d x}+2 y \)
\( =e^{x}[-2 A \sin x+2 B \cos x+2 A \sin x-2 A \cos x~-\) \(2 B \cos x-2 B \sin x+2 A \cos x+2 B \sin x] \)
\( =0\)