What is the solution for y^{'''}+9y^'=-18t ?

The solution for y^{'''}+9y^'=-18t is y=c_{1}+c_{2}cos(3t)+c_{3}sin(3t)-t^2

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y^{'''}+9y^'=-18t

y^{^{'''}} + 9 y^{^{'}} = - 18 t

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

Solution steps

y^{'''} (t) + 9 y^{'} (t) = - 18 t

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 5 sec (4 x)

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

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

3 x^{^{''}} + 22 x^{^{'}} + 48 x = 5 cos (4 t)

y^{^{''}} - 4 y^{^{'}} = x \cdot e^{4 x}

What is the solution for y^{'''}+9y^'=-18t ?
The solution for y^{'''}+9y^'=-18t is y=c_{1}+c_{2}cos(3t)+c_{3}sin(3t)-t^2