What is the solution for (D^4-1)y=e^{-x} ?

The solution for (D^4-1)y=e^{-x} is y=c_{1}e^x+c_{2}e^{-x}+c_{3}cos(x)+c_{4}sin(x)-(e^{-x}x)/4

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(D^4-1)y=e^{-x}

(D^{4} - 1) y = e^{- x}

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) - \frac{e ^{- x} x}{4}

Solution steps

(D^{4} - 1) y = e^{- x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) - \frac{e ^{- x} x}{4}

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) - \frac{e ^{- x} x}{4}

y^{^{''}} + y = x sin (2 x)

y^{^{''}} + 4 y = 3 cos (2.1 x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x}, y (\frac{π}{2}) = 0, y (0) = 0

y^{^{''''}} + 2 y^{^{''}} + y = (x - 4)^{2}

y^{^{''}} + 16 y = 6 sin (4 t), y (0) = 2, y^{^{'}} (0) = 0

What is the solution for (D^4-1)y=e^{-x} ?
The solution for (D^4-1)y=e^{-x} is y=c_{1}e^x+c_{2}e^{-x}+c_{3}cos(x)+c_{4}sin(x)-(e^{-x}x)/4