What is the solution for y^{''}-y^'-6y=4cos(3x) ?

The solution for y^{''}-y^'-6y=4cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-2/39 sin(3x)-10/39 cos(3x)

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y^{''}-y^'-6y=4cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 4 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{2}{39} sin (3 x) - \frac{10}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 4 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{2}{39} sin (3 x) - \frac{10}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{2}{39} sin (3 x) - \frac{10}{39} cos (3 x)

y^{^{'''}} + 3 y^{^{''}} - 4 y^{^{'}} - 12 y = 4 cos (x)

y^{^{'''}} - a y^{^{'}} = 1 + cosh (a x)

y^{^{''}} + 5 y^{^{'}} + 8 y = 24 + 23 x + 12 x^{2}

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

y^{^{''}} - 2 y^{^{'}} + y = e^{x} \frac{1}{x ^{2} + 1}

What is the solution for y^{''}-y^'-6y=4cos(3x) ?
The solution for y^{''}-y^'-6y=4cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-2/39 sin(3x)-10/39 cos(3x)