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a, b, c, d are arbitrary constants. Then the corresponding differential equation to $y=a c^x+b e^{-x}+c \cos x+d \sin x$ is
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$\mathrm{y}^{(4)}=\mathrm{y}$
$\because y=a e^x+b e^{-x}+c \cos x+d \sin x$
$\begin{aligned} & \Rightarrow y^{\prime}=a e^x-b e^{-x}-c \sin x+d \cos x \\ & \Rightarrow y^{\prime \prime}=a e^x+b e^{-x}-c \cos x-d \sin x \\ & \Rightarrow y^{\prime \prime \prime}=a e^x-b e^{-x}+c \sin x-d \cos x \\ & \Rightarrow y^4=a e^x+b e^{-x}+c \cos x+d \sin x \\ & \Rightarrow y^4=y .\end{aligned}$
$\begin{aligned} & \Rightarrow y^{\prime}=a e^x-b e^{-x}-c \sin x+d \cos x \\ & \Rightarrow y^{\prime \prime}=a e^x+b e^{-x}-c \cos x-d \sin x \\ & \Rightarrow y^{\prime \prime \prime}=a e^x-b e^{-x}+c \sin x-d \cos x \\ & \Rightarrow y^4=a e^x+b e^{-x}+c \cos x+d \sin x \\ & \Rightarrow y^4=y .\end{aligned}$
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