What is the solution for y^{''}-4y=4cos(2x) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} + 4 y = (\frac{x}{4} - \frac{1}{8}) e^{4 x}

y^{^{'''}} + 4 y^{^{'}} = sec (2 t)

D (D - 1)^{3} (y) = (x^{2} + 2 x) e^{x}

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

y^{^{''}} + y^{^{'}} + 64 y = 5 cos (2 t)

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