What is the solution for y^{''}-4y=cos(3x) ?

The solution for y^{''}-4y=cos(3x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-1/13 cos(3x)

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y^{''}-4y=cos(3x)

y^{^{''}} - 4 y = cos (3 x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{1}{13} cos (3 x)

Solution steps

y^{''} (x) - 4 y = cos (3 x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{1}{13} cos (3 x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{1}{13} cos (3 x)

y^{^{''}} + 3 y^{^{'}} + 21 y = 9 cos (4 t)

4 y^{^{''}} + 7 y^{^{'}} - y = 15

y^{^{''}} + 7 y^{^{'}} + 12 y = 5 e^{- 6 x}

y^{^{''}} - 4 y^{^{'}} - 45 y = (7 + 2 x)

y^{^{''}} + 81 y = 2 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}-4y=cos(3x) ?
The solution for y^{''}-4y=cos(3x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-1/13 cos(3x)