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(A) Given the equation: $y = a e^x + b e^{-x} + c \cos x + d \sin x$ Taking the first derivative: $y' = a e^x - b e^{-x} - c \sin x + d \cos x$ Taking

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$y^{(4)} = y$

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(A) Given the equation: $y = a e^x + b e^{-x} + c \cos x + d \sin x$ Taking the first derivative: $y' = a e^x - b e^{-x} - c \sin x + d \cos x$ Taking the second derivative: $y'' = a e^x + b e^{-x} - c \cos x - d \sin x$ Taking the third derivative: $y''' = a e^x - b e^{-x} + c \sin x - d \cos x$ Taking the fourth derivative: $y^{(4)} = a e^x + b e^{-x} + c \cos x + d \sin x$ Comparing the fourth derivative with the original equation,we get $y^{(4)} = y$,which can be written as $y^{(4)} - y = 0$.

If $a, b, c, d$ are arbitrary constants,then the corresponding differential equation for $y = a e^x + b e^{-x} + c \cos x + d \sin x$ is

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