What is the solution for -y^{'''}-y^'=2cos(t) ?

The solution for -y^{'''}-y^'=2cos(t) is y=c_{1}+c_{2}cos(t)+c_{3}sin(t)+tcos(t)

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

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

y = c_{1} + c_{2} cos (t) + c_{3} sin (t) + t cos (t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} cos (t) + c_{3} sin (t) + t cos (t)

y = c_{1} + c_{2} cos (t) + c_{3} sin (t) + t cos (t)

What is the solution for -y^{'''}-y^'=2cos(t) ?
The solution for -y^{'''}-y^'=2cos(t) is y=c_{1}+c_{2}cos(t)+c_{3}sin(t)+tcos(t)