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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = cos (2 t) + sin (t)

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

y^{^{''}} - 6 y + 9 = e^{3 x}

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

y^{^{''}} + 81 y = 2 sin (9 t)

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