What is the solution for y^{''}-9y=cos(3x) ?

The solution for y^{''}-9y=cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-3x}-1/18 cos(3x)

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y^{''}-9y=cos(3x)

y^{^{''}} - 9 y = cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} - \frac{1}{18} cos (3 x)

Solution steps

y^{''} (x) - 9 y = cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} - \frac{1}{18} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 3 x} - \frac{1}{18} cos (3 x)

y^{^{''}} + 9 y = 8 x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 8 y^{^{'}} + 16 y = 20 x + 7

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

y^{^{''}} + 4 y^{^{'}} - 2 y = 2 t^{2} - 3 t + 6

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = x e^{x}

What is the solution for y^{''}-9y=cos(3x) ?
The solution for y^{''}-9y=cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-3x}-1/18 cos(3x)