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)

x^{^{''}} - 5 x^{^{'}} + 6 x = t^{2} e^{3 t}

\frac{d ^{2} y}{d x ^{2}} - 9 y = sin (3 x)

y^{^{''}} + 16 y = cos (4 t), y (0) = 6, y^{^{'}} (0) = 8

y^{^{''}} + 3 y^{^{'}} + 2 y = 1, + 4 x e^{- 3 x} y (0) = 0

x^{^{''}} + x = t sin (2 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)