What is the solution for y^{''}-y^'-6y=8cos(3x) ?

The solution for y^{''}-y^'-6y=8cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-4/39 sin(3x)-20/39 cos(3x)

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y^{''}-y^'-6y=8cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 8 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{4}{39} sin (3 x) - \frac{20}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 8 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{4}{39} sin (3 x) - \frac{20}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{4}{39} sin (3 x) - \frac{20}{39} cos (3 x)

y^{^{''}} + 60 y^{^{'}} + 900 y = - 360 x e^{- 4 x}

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

y^{^{''}} - y = 12 x^{2} + e^{x}

y^{^{''}} - 2 y^{^{'}} + y = sin (x) + 3 cos (2 x)

y^{^{''}} + 4 y = t^{2} + 10 e^{t}

What is the solution for y^{''}-y^'-6y=8cos(3x) ?
The solution for y^{''}-y^'-6y=8cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-4/39 sin(3x)-20/39 cos(3x)