What is the solution for 3y^{''}+27y=2cos(3t) ?

The solution for 3y^{''}+27y=2cos(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(tsin(3t))/9

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3y^{''}+27y=2cos(3t)

3 y^{^{''}} + 27 y = 2 cos (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t sin ( 3 t )}{9}

Solution steps

3 y^{''} (t) + 27 y = 2 cos (3 t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t sin ( 3 t )}{9}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t sin ( 3 t )}{9}

y^{^{''}} + 16 y = 32 t^{2} - 48 t + 100, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 9 y = cosh (3 x)

y^{^{''''}} - 16 y = - 12 t^{3}

y^{^{''''}} - 16 y = 75 sin (t)

What is the solution for 3y^{''}+27y=2cos(3t) ?
The solution for 3y^{''}+27y=2cos(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(tsin(3t))/9