What is the solution for y^{''}-3y=cos(3x) ?

The solution for y^{''}-3y=cos(3x) is y=c_{1}e^{sqrt(3)x}+c_{2}e^{-sqrt(3)x}-1/12 cos(3x)

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y^{''}-3y=cos(3x)

y^{^{''}} - 3 y = cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} - \frac{1}{12} cos (3 x)

Solution steps

y^{''} (x) - 3 y = cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} - \frac{1}{12} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} - \frac{1}{12} cos (3 x)

y^{^{''}} - 4 y = 4 t - 8 e^{- 2 t}

y^{^{''}} + y^{^{'}} = cos (3 t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x ^{4}}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4, y^{^{'}} (0) = - 1

y^{^{''}} + 9 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-3y=cos(3x) ?
The solution for y^{''}-3y=cos(3x) is y=c_{1}e^{sqrt(3)x}+c_{2}e^{-sqrt(3)x}-1/12 cos(3x)