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

The solution for y^{''}-4y=cos(2x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-1/8 cos(2x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{1}{8} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{1}{8} cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{1}{8} cos (2 x)

\frac{1}{4} y^{^{''}} + 4 y = 3 tan (4 t) - \frac{1}{2} e^{2 t}

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

y^{^{''}} + 4 y = (x^{2} - 3) \cdot sin (2 x)

4 y^{^{''}} + 4 y^{^{'}} + y = 4 e^{- \frac{x}{4}}

y^{^{''}} + 8 y^{^{'}} + 17 y = 2 e^{- 3 x}

What is the solution for y^{''}-4y=cos(2x) ?
The solution for y^{''}-4y=cos(2x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-1/8 cos(2x)