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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 e^{- 2 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{x}

y^{^{''}} = - 3 y^{^{'}} + 2 y + 2 x + 3, y (0) = 2, y (1) = 1

y^{^{''}} + 4 y = 2 cos (x) + 3 sin (x)

y^{^{''}} + y = 3 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

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