What is the solution for 6y^{''}+6y^'+73y=8cos(4x) ?

The solution for 6y^{''}+6y^'+73y=8cos(4x) is

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6y^{''}+6y^'+73y=8cos(4x)

6 y^{^{''}} + 6 y^{^{'}} + 73 y = 8 cos (4 x)

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{143 x}{2 3}) + c_{2} sin (\frac{143 x}{2 3})) + \frac{192}{1105} sin (4 x) - \frac{184}{1105} cos (4 x)

Solution steps

6 y^{''} (x) + 6 y^{'} (x) + 73 y = 8 cos (4 x)

Solve linear ODE:

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{143 x}{2 3}) + c_{2} sin (\frac{143 x}{2 3})) + \frac{192}{1105} sin (4 x) - \frac{184}{1105} cos (4 x)

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{143 x}{2 3}) + c_{2} sin (\frac{143 x}{2 3})) + \frac{192}{1105} sin (4 x) - \frac{184}{1105} cos (4 x)

\frac{d ^{2} y}{d t ^{2}} + 5 y = 5 sin (t)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 6 y = - e^{x} + 12 x

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 cos (x) - 4 sin (x)

\frac{1}{2} y^{^{''}} + 2 y = 4 tan (2 t) - \frac{1}{3} e^{t}

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

What is the solution for 6y^{''}+6y^'+73y=8cos(4x) ?
The solution for 6y^{''}+6y^'+73y=8cos(4x) is