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)

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)