What is the solution for y^{''}-y^'-6y=2cos(3x) ?

The solution for y^{''}-y^'-6y=2cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-1/39 sin(3x)-5/39 cos(3x)

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y^{''}-y^'-6y=2cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 2 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{39} sin (3 x) - \frac{5}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 2 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{39} sin (3 x) - \frac{5}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{39} sin (3 x) - \frac{5}{39} cos (3 x)

2 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 6 y = 14 e^{x}

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 40 y = - \frac{20}{3} t - \frac{4}{3}

y^{^{''}} + 2 y^{^{'}} + 2 y = 10 sin (2 t)

y^{^{''}} + 2 y^{^{'}} = 4 x

y^{^{''}} + 3 y = x^{3} - 1

What is the solution for y^{''}-y^'-6y=2cos(3x) ?
The solution for y^{''}-y^'-6y=2cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-1/39 sin(3x)-5/39 cos(3x)