What is the solution for y^{''}+9y=x-2cos(3x) ?

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

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y^{''}+9y=x-2cos(3x)

y^{^{''}} + 9 y = x - 2 cos (3 x)

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

Solution steps

y^{''} (x) + 9 y = x - 2 cos (3 x)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 t

y^{^{''}} + 4 y^{^{'}} = sin (2 x)

(D^{2} + 13 D + 13) x = 12 t^{2} + 2 t - cos (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y = x e^{\frac{x}{2}}

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