What is the solution for 3y^{''}+27y=2cos(3x) ?

The solution for 3y^{''}+27y=2cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+(xsin(3x))/9

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3y^{''}+27y=2cos(3x)

3 y^{^{''}} + 27 y = 2 cos (3 x)

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

Solution steps

3 y^{''} (x) + 27 y = 2 cos (3 x)

Solve linear ODE:

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

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

(3 D^{2} + 3) y = 3 x

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 cos (2 t), y (0) = 1, y^{^{'}} (0) = 1

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

y^{^{''}} - 16 y = sin (2 x) - 3 cos (2 x)

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} lo g_{e} (1 + x^{2})

What is the solution for 3y^{''}+27y=2cos(3x) ?
The solution for 3y^{''}+27y=2cos(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+(xsin(3x))/9