What is the solution for y^{'''}-5y^{''}+9y=tcos(2t) ?

The solution for y^{'''}-5y^{''}+9y=tcos(2t) is

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y^{'''}-5y^{''}+9y=tcos(2t)

y^{^{'''}} - 5 y^{^{''}} + 9 y = t cos (2 t)

y = c_{1} e^{- 1.20440 \dots t} + c_{2} e^{1.63555 \dots t} + c_{3} e^{4.56884 \dots t} - \frac{8 t sin ( 2 t )}{905} + \frac{29 t cos ( 2 t )}{905} - \frac{21108}{819025} sin (2 t) + \frac{44}{819025} cos (2 t)

Solution steps

y^{'''} (t) - 5 y^{''} (t) + 9 y = t cos (2 t)

Solve linear ODE:

y = c_{1} e^{- 1.20440 \dots t} + c_{2} e^{1.63555 \dots t} + c_{3} e^{4.56884 \dots t} - \frac{8 t sin ( 2 t )}{905} + \frac{29 t cos ( 2 t )}{905} - \frac{21108}{819025} sin (2 t) + \frac{44}{819025} cos (2 t)

y = c_{1} e^{- 1.20440 \dots t} + c_{2} e^{1.63555 \dots t} + c_{3} e^{4.56884 \dots t} - \frac{8 t sin ( 2 t )}{905} + \frac{29 t cos ( 2 t )}{905} - \frac{21108}{819025} sin (2 t) + \frac{44}{819025} cos (2 t)

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

y^{^{''}} + 6 y^{^{'}} + 5 y = sin (x)

y^{^{''}} + 4 y = 4 cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} = sin (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{t}{e ^{t}} + \frac{e ^{t}}{t}

What is the solution for y^{'''}-5y^{''}+9y=tcos(2t) ?
The solution for y^{'''}-5y^{''}+9y=tcos(2t) is