What is the solution for y^{''}-y^'-6y=32cos(3x) ?

The solution for y^{''}-y^'-6y=32cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-16/39 sin(3x)-80/39 cos(3x)

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y^{''}-y^'-6y=32cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 32 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{16}{39} sin (3 x) - \frac{80}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 32 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{16}{39} sin (3 x) - \frac{80}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{16}{39} sin (3 x) - \frac{80}{39} cos (3 x)

y^{^{'''}} - 3 y^{^{'}} + 2 y = e^{- 2 x}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 5 y^{^{'}} - 6 y = 10, y (0) = 2, y^{^{'}} (0) = 1

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

What is the solution for y^{''}-y^'-6y=32cos(3x) ?
The solution for y^{''}-y^'-6y=32cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-16/39 sin(3x)-80/39 cos(3x)