What is the solution for y^{''''}+y=cos(x) ?

The solution for y^{''''}+y=cos(x) is

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y^{''''}+y=cos(x)

y^{^{''''}} + y = cos (x)

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

Solution steps

y^{(4)} (x) + y = cos (x)

Solve linear ODE:

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

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

x^{^{''}} + 2 x^{^{'}} - 8 x = 12 e^{2 t} + 8 t - 10

y^{^{''}} - 4 y^{^{'}} + 5 y = 4 e^{3 t}, y (0) = 2, y^{^{'}} (0) = 7

y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + 6 y = e^{x} + x^{2}

x^{^{''}} + 3 x^{^{'}} + 2 x = sin (3 t)

x^{^{''}} + 8 x^{^{'}} + 16 = 0

What is the solution for y^{''''}+y=cos(x) ?
The solution for y^{''''}+y=cos(x) is