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

The solution for y^{''}-y=cos(2x) is y=c_{1}e^x+c_{2}e^{-x}-1/5 cos(2x)

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

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

y = c_{1} e^{x} + c_{2} e^{- x} - \frac{1}{5} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} - \frac{1}{5} cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- x} - \frac{1}{5} cos (2 x)

y^{^{''}} + 2 y^{^{'}} + 2 y = (e^{(- x)}) \cdot cos (x)

y^{^{''}} - 3 y^{^{'}} - 4 y = x e^{x} + 2 cos (x) - sin (x)

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

y^{^{''}} + 4 y = \frac{cos ( 2 x )}{2}

y^{^{''}} - y^{^{'}} - 2 y = \frac{( e ^{t} + e ^{- t} )}{2}

What is the solution for y^{''}-y=cos(2x) ?
The solution for y^{''}-y=cos(2x) is y=c_{1}e^x+c_{2}e^{-x}-1/5 cos(2x)